701. A laboratory compares an ordinary weak diamagnetic metal and a specimen in an ideal superconducting Meissner state.
Specimen P is an ordinary weak diamagnet with susceptibility \(\chi_P=-2.0\times10^{-5}\). Specimen Q is in the ideal perfect-diamagnetic Meissner state, represented in the stated NCERT model by \(\chi_Q=-1\) and \(\mu_{r,Q}=0\). Both are examined under conditions appropriate to these simplified descriptions. Field penetration is considered in the bulk, away from specimen surfaces, and the applied magnetic intensity is prescribed in the ideal constitutive comparison.
Which option correctly compares the corresponding relative permeabilities and the bulk magnetic induction in the two specimens?
ⓐ. P has \(\mu_r=1.00002\) and enhanced induction; Q has \(\mu_r=1\) and merely a slightly weakened induction.
ⓑ. P and Q both have \(\mu_r=0\), because all diamagnetic materials completely expel applied magnetic fields.
ⓒ. P has \(\mu_r=0.99998\) and slightly reduced induction; Q has \(\mu_r=-1\) and an internal field pointing opposite to the applied field.
ⓓ. P has \(\mu_r=0.99998\) and slightly reduced induction; Q has \(\mu_r=0\) and zero bulk induction in the ideal Meissner state.
Correct Answer: P has \(\mu_r=0.99998\) and slightly reduced induction; Q has \(\mu_r=0\) and zero bulk induction in the ideal Meissner state.
Explanation: \(\textbf{Use the susceptibility–permeability relation:}\)
For a material described by the stated isotropic susceptibility model,
\[
\mu_r=1+\chi
\]
An ordinary weak diamagnet has a small negative susceptibility.
Consequently, its relative permeability is slightly less than unity.
\(\textbf{Calculate the ordinary diamagnet's permeability:}\)
For P,
\[
\mu_{r,P}=1-2.0\times10^{-5}
\]
\[
\mu_{r,P}=0.99998
\]
The corresponding induction is
\[
B_P=\mu_0\mu_{r,P}H
\]
Thus, the induction is only slightly smaller than the value \(\mu_0H\) for the same prescribed H.
Ordinary diamagnetism does not imply complete exclusion of magnetic induction from the specimen.
\(\textbf{Apply the ideal Meissner-state description:}\)
For Q, the stated perfect-diamagnetic model gives
\[
\chi_Q=-1
\]
Therefore,
\[
\mu_{r,Q}=1+\chi_Q
\]
\[
\mu_{r,Q}=0
\]
In the ideal bulk Meissner state,
\[
B_Q=0
\]
The magnetic field is expelled from the bulk superconducting region.
\(\textbf{Distinguish bulk exclusion from exterior magnetic fields:}\)
The ideal zero-field conclusion concerns the superconducting bulk under the stated conditions.
It does not mean the magnetic field outside the specimen vanishes.
Field lines can be redirected around the superconducting body.
The ordinary diamagnet and the ideal Meissner-state superconductor therefore share an opposing magnetic response but differ greatly in its extent.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_{r,P}=0.99998,\qquad
\mu_{r,Q}=0,\qquad
B_P=0.99998\,\mu_0H,\qquad
B_Q=0\ \text{in the ideal bulk Meissner state}}
\]
An ordinary diamagnet slightly opposes an applied magnetic field.
The perfect-diamagnetic Meissner state excludes magnetic induction from the superconducting bulk in the ideal description.
The distinction is not merely a difference in the signs of susceptibility; both are negative.
It is the limiting magnitude of the response that distinguishes the ordinary and perfect cases.
702. An ordinary paramagnetic specimen consists of atoms possessing magnetic moments that can orient partially in an applied magnetic field.
The specimen is maintained at a fixed absolute temperature while the external magnetic intensity is gradually increased from zero to a very large value.
A graph has applied magnetic intensity H on the horizontal axis and positive bulk magnetisation M on the vertical axis. It begins at the origin with increasing M, bends progressively so that the increase in M per additional unit H becomes smaller, and approaches an approximately horizontal limiting region at sufficiently high H. The specimen is assumed to remain paramagnetic over the full experiment, without a ferromagnetic transition.
Consider the following statements.
Statement I: The limiting behaviour is consistent with increasing alignment of pre-existing microscopic magnetic moments until little additional alignment remains possible.
Statement II: The field-dependent flattening necessarily proves the development of ferromagnetic domains.
Statement III: A simple linear relation \(M=\chi H\) with one fixed positive susceptibility cannot describe the entire curve, including the limiting region.
Statement IV: The existence of a limiting magnetisation means that the atomic magnetic moments themselves have vanished.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. I and III only
ⓒ. II and IV only
ⓓ. I, III and IV only
Correct Answer: I and III only
Explanation: Paramagnetic atoms or ions possess microscopic magnetic moments capable of partial alignment in an applied field.
Without a substantial applied field, thermal motion tends to randomise their orientations.
As the applied magnetic intensity increases, the magnetic influence increasingly favours alignment.
The bulk magnetisation consequently increases.
However, the number of microscopic magnetic moments and their magnitudes are finite.
Once nearly all moments are aligned as far as the model permits, increasing the field cannot produce the same additional alignment as before.
The approach towards saturation is therefore consistent with the mechanism in Statement I.
Statement I is correct.
The graph does not establish ferromagnetism.
Ordinary paramagnetism can exhibit alignment saturation without developing the cooperative domain structure characteristic of a ferromagnet.
Statement II is incorrect.
A fixed linear susceptibility predicts magnetisation proportional to applied intensity throughout its range of validity.
Such a straight-line relation cannot continue indefinitely into a finite saturation plateau.
Statement III is correct.
Saturation does not mean the microscopic magnetic moments disappear.
It means that their orientations have become increasingly constrained by the applied field.
Statement IV is incorrect.
The graph identifies a limitation of the weak-field linear-response approximation.
It does not justify relabelling the material or discarding the microscopic moments that produce its magnetisation.
703. A ferromagnetic specimen is heated gradually in a weak, fixed applied magnetic field.
At lower temperatures the specimen shows strong magnetisation associated with ordered magnetic domains. As its temperature increases, the magnetisation associated with this cooperative ordering progressively weakens. Near a characteristic temperature \(T_C\), ferromagnetic ordering disappears. Above \(T_C\), the specimen shows an ordinary weak paramagnetic response to the still-applied field. The specimen's chemical composition remains unchanged throughout the experiment.
Which option correctly explains the physical transition and its relation to microscopic magnetic moments?
ⓐ. Above \(T_C\), all atoms lose their electrons, so the specimen becomes nonmagnetic in the absence of any possible field-induced response.
ⓑ. \(T_C\) marks the temperature at which negative diamagnetic susceptibility first becomes possible; above it the specimen must repel a magnet strongly.
ⓒ. The ferromagnetic domain ordering is disrupted by sufficient heating; above \(T_C\), the specimen is paramagnetic in the stated description even though microscopic magnetic moments can still exist.
ⓓ. The strong magnetisation below \(T_C\) arises only from the applied field creating new atomic magnetic moments; the disappearance of domains above \(T_C\) means that the specimen must become a perfect diamagnet.
Correct Answer: The ferromagnetic domain ordering is disrupted by sufficient heating; above \(T_C\), the specimen is paramagnetic in the stated description even though microscopic magnetic moments can still exist.
Explanation: Ferromagnetic behaviour depends on cooperative alignment of microscopic magnetic moments.
At lower temperatures, this collective interaction can sustain magnetically ordered domains.
An applied magnetic field favours alignment and growth of domains oriented appropriately relative to that field.
The resulting bulk magnetisation can be much greater than in an ordinary paramagnet.
Increasing temperature strengthens the randomising effects of thermal motion.
At sufficiently high temperature, the cooperative ferromagnetic order is no longer maintained.
The characteristic transition temperature is called the Curie temperature, denoted \(T_C\).
Above this transition, the specimen is described as paramagnetic in the standard NCERT treatment.
Its magnetic moments need not vanish merely because long-range cooperative ordering has disappeared.
Instead, the remaining moments respond more weakly to the applied field and are subject to thermal disorder.
The material's magnetic classification can therefore change with temperature without a change in chemical identity.
The transition is not caused by every atom losing electrons.
Nor does the loss of ferromagnetic order imply the onset of perfect diamagnetism.
The observations distinguish loss of collective magnetic order from destruction of microscopic magnetic moments.
704. Two long solenoids have identical winding densities and carry equal steady currents. Solenoid P contains an ordinary weak diamagnetic material, while solenoid Q contains an ordinary weak paramagnetic material.
At the specified operating temperature, their magnetic susceptibilities are
\[
\chi_P=-0.002,\qquad \chi_Q=+0.003
\]
Both materials are isotropic and linearly responding, and the applied magnetic intensity inside each solenoid is
\[
H=1000\,\text{A m}^{-1}
\]
A positively charged particle enters the uniform central region of each solenoid with the same momentum perpendicular to the local magnetic field. Its circular-orbit radii are \(r_P\) and \(r_Q\), respectively.
Assume the particles remain within the approximately uniform field regions.
Which option correctly gives the two magnetisations, the ratio \(r_P/r_Q\), and the direction of the material response?
ⓐ. \(M_P=-2.0\,\text{A m}^{-1}\), \(M_Q=+3.0\,\text{A m}^{-1}\); \(r_P/r_Q=998/1003\); P opposes and Q reinforces the applied field.
ⓑ. \(M_P=-2.0\,\text{A m}^{-1}\), \(M_Q=+3.0\,\text{A m}^{-1}\); \(r_P/r_Q=1003/998\); P opposes and Q reinforces the applied field.
ⓒ. \(M_P=+2.0\,\text{A m}^{-1}\), \(M_Q=-3.0\,\text{A m}^{-1}\); \(r_P/r_Q=1003/998\); P reinforces and Q opposes the applied field.
ⓓ. \(M_P=-2.0\,\text{A m}^{-1}\), \(M_Q=+3.0\,\text{A m}^{-1}\); \(r_P/r_Q=1003/998\); both materials have magnetisation along the applied field.
Correct Answer: \(M_P=-2.0\,\text{A m}^{-1}\), \(M_Q=+3.0\,\text{A m}^{-1}\); \(r_P/r_Q=1003/998\); P opposes and Q reinforces the applied field.
Explanation: \(\textbf{Determine the magnetic response of each specimen:}\)
For an ordinary linear magnetic material,
\[
M=\chi H
\]
The sign of susceptibility determines whether magnetisation is parallel or antiparallel to the applied magnetic intensity.
For P,
\[
M_P=(-0.002)(1000)
\]
\[
M_P=-2.0\,\text{A m}^{-1}
\]
Its magnetisation opposes the applied field.
For Q,
\[
M_Q=(+0.003)(1000)
\]
\[
M_Q=+3.0\,\text{A m}^{-1}
\]
Its magnetisation reinforces the applied field.
\(\textbf{Determine the total magnetic induction:}\)
The field in matter is
\[
B=\mu_0(H+M)
\]
Therefore,
\[
B_P=\mu_0(1000-2)
\]
\[
B_P=998\mu_0
\]
For Q,
\[
B_Q=\mu_0(1000+3)
\]
\[
B_Q=1003\mu_0
\]
The applied magnetic intensity is identical, but the inductions are slightly different because the material contributions differ.
\(\textbf{Connect magnetic induction with charged-particle motion:}\)
For perpendicular entry into a uniform magnetic field,
\[
|q|vB=\frac{mv^2}{r}
\]
Consequently,
\[
r=\frac{p}{|q|B}
\]
where \(p=mv\) is the particle momentum in the nonrelativistic treatment.
The particle momentum and charge magnitude are the same in both trials.
Therefore,
\[
\frac{r_P}{r_Q}=\frac{B_Q}{B_P}
\]
Substituting,
\[
\frac{r_P}{r_Q}=\frac{1003\mu_0}{998\mu_0}
\]
\[
\frac{r_P}{r_Q}=\frac{1003}{998}
\]
\[
\frac{r_P}{r_Q}\approx1.00501
\]
\(\textbf{Final answer:}\)
\[
\boxed{M_P=-2.0\,\text{A m}^{-1},\qquad
M_Q=+3.0\,\text{A m}^{-1},\qquad
\frac{r_P}{r_Q}=\frac{1003}{998}}
\]
The diamagnetic material slightly reduces the magnetic induction relative to the same applied intensity in vacuum.
The paramagnetic material slightly increases it.
A particle with fixed momentum consequently follows a slightly larger circular path in P.
The experiment connects material susceptibility with a measurable consequence of the total magnetic induction.
It is important to use \(B\), rather than \(H\) alone, in the magnetic Lorentz-force relation.
705. A simplified ferromagnetic specimen has \(100\) equal-volume domains. Every domain has magnetisation magnitude
\[
M_s=5.0\times10^5\,\text{A m}^{-1}
\]
The domains point only along the four coordinate directions shown:
| Domain direction | Number of domains |
|---|
| Positive \(x\) | 40 |
| Negative \(x\) | 20 |
| Positive \(y\) | 25 |
| Negative \(y\) | 15 |
The total specimen volume is \(2.0\times10^{-6}\,\text{m}^3\).
The specimen is held rigidly in an external uniform magnetic field
\[
\vec B=(0.30\hat i-0.40\hat j)\,\text{T}
\]
Assume the domain populations and directions do not change during the instantaneous torque measurement.
What are the specimen's resultant magnetic moment, magnetic torque and orientation-dependent magnetic potential energy?
ⓐ. \(\vec m=(0.20\hat i-0.10\hat j)\,\text{A m}^2\); \(\vec\tau=-0.050\hat k\,\text{N m}\); \(U=-0.100\,\text{J}\).
ⓑ. \(\vec m=(0.20\hat i+0.10\hat j)\,\text{A m}^2\); \(\vec\tau=+0.11\hat k\,\text{N m}\); \(U=+0.020\,\text{J}\).
ⓒ. \(\vec m=(0.20\hat i+0.10\hat j)\,\text{A m}^2\); \(\vec\tau=-0.050\hat k\,\text{N m}\); \(U=-0.020\,\text{J}\).
ⓓ. \(\vec m=(0.20\hat i+0.10\hat j)\,\text{A m}^2\); \(\vec\tau=-0.11\hat k\,\text{N m}\); \(U=-0.020\,\text{J}\).
Correct Answer: \(\vec m=(0.20\hat i+0.10\hat j)\,\text{A m}^2\); \(\vec\tau=-0.11\hat k\,\text{N m}\); \(U=-0.020\,\text{J}\).
Explanation: \(\textbf{Combine the domain magnetisations vectorially:}\)
Magnetisation is magnetic moment per unit volume.
Because all domains occupy equal volumes, the bulk magnetisation is the arithmetic vector average of their individual magnetisations.
The positive- and negative-\(x\) domains contribute oppositely:
\[
M_x=\frac{40-20}{100}M_s
\]
\[
M_x=0.20(5.0\times10^5)
\]
\[
M_x=1.0\times10^5\,\text{A m}^{-1}
\]
Similarly, the \(y\)-component is
\[
M_y=\frac{25-15}{100}M_s
\]
\[
M_y=0.10(5.0\times10^5)
\]
\[
M_y=5.0\times10^4\,\text{A m}^{-1}
\]
Therefore,
\[
\vec M=(1.0\times10^5\hat i+5.0\times10^4\hat j)\,\text{A m}^{-1}
\]
\(\textbf{Recover the specimen's total magnetic moment:}\)
The moment is
\[
\vec m=V\vec M
\]
Using the specified volume,
\[
\vec m=(2.0\times10^{-6})(1.0\times10^5\hat i+5.0\times10^4\hat j)
\]
\[
\vec m=(0.20\hat i+0.10\hat j)\,\text{A m}^2
\]
\(\textbf{Calculate the magnetic torque:}\)
In the applied uniform field,
\[
\vec\tau=\vec m\times\vec B
\]
Substituting,
\[
\vec\tau=(0.20\hat i+0.10\hat j)\times(0.30\hat i-0.40\hat j)
\]
The nonzero cross products are
\[
(0.20\hat i)\times(-0.40\hat j)=-0.080\hat k
\]
and
\[
(0.10\hat j)\times(0.30\hat i)=-0.030\hat k
\]
Therefore,
\[
\vec\tau=(-0.080-0.030)\hat k
\]
\[
\vec\tau=-0.11\hat k\,\text{N m}
\]
\(\textbf{Calculate the orientation-dependent energy:}\)
The potential energy is
\[
U=-\vec m\cdot\vec B
\]
\[
U=-[(0.20)(0.30)+(0.10)(-0.40)]
\]
\[
U=-(0.060-0.040)
\]
\[
U=-0.020\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\vec m=(0.20\hat i+0.10\hat j)\,\text{A m}^2,\qquad
\vec\tau=-0.11\hat k\,\text{N m},\qquad U=-0.020\,\text{J}}
\]
Different ferromagnetic domains can possess substantial magnetisation while partly cancelling one another macroscopically.
Here the cancellation must be carried out separately along two perpendicular directions.
The resulting magnetic moment is neither parallel nor antiparallel to the applied field.
Its component perpendicular to the field determines torque, while its component parallel to the field determines orientation energy.
The domain populations are treated as fixed only for the specified instantaneous measurement.
706. A student wants to determine whether increasing temperature weakens the magnetisation of an ordinary paramagnetic specimen.
At \(300\,\text{K}\), the specimen is placed in a long solenoid carrying \(1.0\,\text{A}\), and its magnetisation is measured. In a second trial, the specimen is heated to \(450\,\text{K}\), but the solenoid current is also increased to \(2.0\,\text{A}\). The measured magnetisation is now greater than before. The solenoid geometry and specimen are unchanged, and the sample remains an ordinary paramagnet throughout.
Assertion (A): The two measurements do not establish that heating by itself increased the paramagnetic magnetisation.
Reason (R): Both the magnetising field and temperature were changed, and the stronger applied field can promote moment alignment even while greater thermal agitation opposes that alignment.
Select the correct answer.
ⓐ. Both A and R are true, and R correctly explains A.
ⓑ. Both A and R are true, but R does not correctly explain A.
ⓒ. A is true, but R is false.
ⓓ. A is false, but R is true.
Correct Answer: Both A and R are true, and R correctly explains A.
Explanation: An ordinary paramagnet contains microscopic magnetic moments that can become partially aligned by an applied magnetic field.
This alignment contributes to its macroscopic magnetisation.
Thermal agitation tends to randomise the directions of those moments.
At a fixed applied field within the ordinary response regime, increasing temperature generally weakens the paramagnetic alignment.
However, the experiment did not hold the applied field fixed.
The current through the same long solenoid was doubled between trials.
Its magnetising intensity therefore increased because, for the same winding density, the applied intensity is proportional to solenoid current.
The specimen was simultaneously exposed to a greater temperature and a stronger magnetising field.
These changes influence the moment alignment in competing directions.
A greater final magnetisation is consequently compatible with the stronger applied field having more than compensated for the heating effect.
It does not demonstrate that thermal agitation itself strengthened the magnetic alignment.
The Assertion is true.
The Reason is also true and directly explains why the measurements cannot isolate temperature as the causal variable.
A controlled test of the temperature effect would compare magnetisation at different temperatures while maintaining the same applied magnetising intensity and otherwise comparable conditions.
The observed change in magnetisation cannot be attributed to temperature alone when the applied magnetising field also changes.
707. An ordinary paramagnetic core is placed inside a sufficiently long solenoid having a fixed winding density \(n\). Throughout the operating range,
\[
\chi=\frac CT,\qquad H=nI,\qquad B=\mu_0(1+\chi)H
\]
The core remains in its linear Curie-law regime.
At temperature \(T_1\), its susceptibility is \(0.020\), and the solenoid carries current \(I_1\).
The temperature is then doubled. The solenoid current is adjusted to preserve exactly the SAME total magnetic induction \(B\) inside the material as in the initial state.
What are the required ratio \(I_2/I_1\) and the resulting magnetisation ratio \(M_2/M_1\)?
ⓐ. \(\displaystyle\frac{I_2}{I_1}=1,\quad\frac{M_2}{M_1}=\frac12\)
ⓑ. \(\displaystyle\frac{I_2}{I_1}=\frac{101}{102},\quad\frac{M_2}{M_1}=\frac{101}{204}\)
ⓒ. \(\displaystyle\frac{I_2}{I_1}=\frac{102}{101},\quad\frac{M_2}{M_1}=\frac{51}{101}\)
ⓓ. \(\displaystyle\frac{I_2}{I_1}=2,\quad\frac{M_2}{M_1}=1\)
Correct Answer: \(\displaystyle\frac{I_2}{I_1}=\frac{102}{101},\quad\frac{M_2}{M_1}=\frac{51}{101}\)
Explanation: \(\textbf{Apply Curie's law after heating:}\)
For the stated paramagnet,
\[
\chi=\frac CT
\]
The final temperature is
\[
T_2=2T_1
\]
Therefore,
\[
\chi_2=\chi_1\frac{T_1}{T_2}
\]
\[
\chi_2=0.020\left(\frac12\right)
\]
\[
\chi_2=0.010
\]
The relative permeabilities at the two operating points are
\[
\mu_{r1}=1+\chi_1=1.020
\]
and
\[
\mu_{r2}=1+\chi_2=1.010
\]
\(\textbf{Use the condition of constant total induction:}\)
The solenoid's applied magnetic intensity is
\[
H=nI
\]
Thus, the total induction is
\[
B=\mu_0(1+\chi)nI
\]
The winding density and the required induction are unchanged.
Consequently,
\[
(1+\chi_1)I_1=(1+\chi_2)I_2
\]
\(\textbf{Determine the required current adjustment:}\)
Rearranging,
\[
\frac{I_2}{I_1}=\frac{1+\chi_1}{1+\chi_2}
\]
Substituting,
\[
\frac{I_2}{I_1}=\frac{1.020}{1.010}
\]
\[
\frac{I_2}{I_1}=\frac{102}{101}
\]
The current must increase slightly to compensate for the reduction in susceptibility.
\(\textbf{Determine what happens to magnetisation:}\)
The magnetisation relation is
\[
M=\chi H=\chi nI
\]
Therefore,
\[
\frac{M_2}{M_1}=\frac{\chi_2nI_2}{\chi_1nI_1}
\]
\[
\frac{M_2}{M_1}=\frac{0.010}{0.020}\frac{102}{101}
\]
\[
\frac{M_2}{M_1}=\frac12\frac{102}{101}
\]
\[
\frac{M_2}{M_1}=\frac{51}{101}
\]
\(\textbf{Check the distinction between the two controlled quantities:}\)
The total induction is deliberately maintained at its original magnitude.
That condition does not mean the material magnetisation remains unchanged.
The weaker material susceptibility requires a small increase in applied intensity, while the induced magnetisation still decreases substantially.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{I_2}{I_1}=\frac{102}{101},\qquad
\frac{M_2}{M_1}=\frac{51}{101}}
\]
Keeping the total magnetic induction fixed is different from keeping the magnetising current fixed.
The current must be adjusted because heating changes the material's contribution to the field.
The required increase is small for a weak paramagnet.
Even after the adjustment, the magnetisation falls to slightly more than half its initial value.
The result follows from applying both the constitutive relation and Curie's law under the same fixed-induction constraint.
708. A long solenoid has \(200\) turns, length \(0.50\,\text{m}\), cross-sectional area \(5.0\times10^{-3}\,\text{m}^2\), and maintained current \(2.0\,\text{A}\).
It is completely filled with an ordinary linear paramagnetic core of susceptibility
\[
\chi=+0.020
\]
Assume the applied magnetic intensity inside the long solenoid is \(H=nI\), where \(n=N/L\), and the uniformly magnetised core has volume \(AL\).
The magnetic moment of the free-current winding and the induced magnetic moment of the core are assumed to add in the usual long-solenoid dipole model.
The combined apparatus is placed with its axis perpendicular to an independent uniform external field of magnitude \(0.25\,\text{T}\).
The solenoid current is subsequently reversed while maintaining its magnitude, and the core remains in the same reversible linear-response regime.
Which option correctly gives the induced core moment, the original external-field torque magnitude and the torque immediately after current reversal?
ⓐ. Core moment \(0.040\,\text{A m}^2\); original torque \(0.510\,\text{N m}\); torque reverses with unchanged magnitude.
ⓑ. Core moment \(0.040\,\text{A m}^2\); original torque \(0.500\,\text{N m}\); torque remains unchanged after reversal.
ⓒ. Core moment \(2.00\,\text{A m}^2\); original torque \(1.00\,\text{N m}\); torque reverses with unchanged magnitude.
ⓓ. Core moment \(0.040\,\text{A m}^2\); original torque \(0.510\,\text{N m}\); torque remains in its original direction because the core is paramagnetic.
Correct Answer: Core moment \(0.040\,\text{A m}^2\); original torque \(0.510\,\text{N m}\); torque reverses with unchanged magnitude.
Explanation: \(\textbf{Determine the magnetising intensity inside the solenoid:}\)
The winding density is
\[
n=\frac NL
\]
\[
n=\frac{200}{0.50}
\]
\[
n=400\,\text{m}^{-1}
\]
The applied magnetic intensity is
\[
H=nI
\]
\[
H=(400)(2.0)
\]
\[
H=800\,\text{A m}^{-1}
\]
\(\textbf{Calculate the core magnetisation:}\)
The core is an ordinary linear paramagnet:
\[
M=\chi H
\]
Substituting,
\[
M=(0.020)(800)
\]
\[
M=16\,\text{A m}^{-1}
\]
Its magnetisation points along the initial solenoid field.
\(\textbf{Recover the induced magnetic moment of the core:}\)
Magnetisation is magnetic moment per unit volume:
\[
m_{\mathrm{core}}=MV
\]
The core volume is
\[
V=AL
\]
\[
V=(5.0\times10^{-3})(0.50)
\]
\[
V=2.5\times10^{-3}\,\text{m}^3
\]
Thus,
\[
m_{\mathrm{core}}=(16)(2.5\times10^{-3})
\]
\[
m_{\mathrm{core}}=0.040\,\text{A m}^2
\]
\(\textbf{Include the current-winding moment:}\)
The solenoid's free-current moment is
\[
m_{\mathrm{winding}}=NIA
\]
\[
m_{\mathrm{winding}}=(200)(2.0)(5.0\times10^{-3})
\]
\[
m_{\mathrm{winding}}=2.00\,\text{A m}^2
\]
Both moments are initially parallel.
Therefore,
\[
m_{\mathrm{total}}=2.00+0.040
\]
\[
m_{\mathrm{total}}=2.040\,\text{A m}^2
\]
\(\textbf{Determine the torque from the independent external field:}\)
The total moment is perpendicular to the external field.
Hence,
\[
\tau=m_{\mathrm{total}}B_e
\]
\[
\tau=(2.040)(0.25)
\]
\[
\tau=0.510\,\text{N m}
\]
\(\textbf{Reverse the solenoid current:}\)
The current-winding moment reverses.
The paramagnetic core's induced magnetisation also reverses because it remains aligned with the reversed magnetising intensity.
Therefore, both contributions to the total magnetic moment reverse.
The torque in the unchanged external field reverses direction while retaining the magnitude \(0.510\,\text{N m}\).
\(\textbf{Final answer:}\)
\[
\boxed{m_{\mathrm{core}}=0.040\,\text{A m}^2,\qquad
|\tau|=0.510\,\text{N m},\qquad
\vec\tau_{\mathrm{after}}=-\vec\tau_{\mathrm{before}}}
\]
The solenoid winding possesses a magnetic moment because electric current circulates through its turns.
The paramagnetic core adds a smaller field-induced moment.
Both respond to reversal of the magnetising current in the specified reversible regime.
The external torque acts on the total dipole moment, not merely on the conducting winding.
This differs from a hard permanent-magnet core whose remanent magnetisation might not reverse with the exciting current.
709. An infinitely long straight wire lies along the \(z\)-axis. At a point on the positive \(x\)-axis, \(0.050\,\text{m}\) from the wire, its magnetic field is measured as
\[
\vec B=+40\,\mu\text{T}\hat j
\]
Two very small ordinary weak magnetic specimens are then placed separately at the same radial distance from the wire.
Specimen P is observed to move weakly towards the wire when released. Specimen Q moves weakly away from the wire.
The current is then reversed without changing its magnitude.
Assume that the specimens remain in their ordinary reversible response regimes and that the spatial distribution of magnetic-field magnitude is unchanged.
Take \(\mu_0/(2\pi)=2.0\times10^{-7}\,\text{N A}^{-2}\).
Which option correctly identifies the wire current, the two material responses and what happens after current reversal?
ⓐ. The original wire current is \(10\,\text{A}\) along negative \(z\); P is diamagnetic, Q is paramagnetic; both reverse their radial motion.
ⓑ. The original wire current is \(10\,\text{A}\) along positive \(z\); P is paramagnetic, Q is diamagnetic; both reverse their radial motion.
ⓒ. The wire carries \(10\,\text{A}\) along positive \(z\); P is paramagnetic and Q diamagnetic; their magnetisations reverse, but their radial motion tendencies remain unchanged.
ⓓ. The original wire current is \(20\,\text{A}\) along positive \(z\); both samples are paramagnetic because the wire creates a nonuniform magnetic field.
Correct Answer: The wire carries \(10\,\text{A}\) along positive \(z\); P is paramagnetic and Q diamagnetic; their magnetisations reverse, but their radial motion tendencies remain unchanged.
Explanation: \(\textbf{Recover the current magnitude from the measured field:}\)
For an infinitely long straight conductor,
\[
B=\frac{\mu_0I}{2\pi r}
\]
Rearranging,
\[
I=\frac{Br}{\mu_0/(2\pi)}
\]
Substituting,
\[
I=\frac{(40\times10^{-6})(0.050)}{2.0\times10^{-7}}
\]
\[
I=10\,\text{A}
\]
\(\textbf{Determine its direction:}\)
At a point along positive \(x\), current along positive \(z\) produces magnetic field along positive \(y\).
The observed \(+\hat j\) field therefore establishes that the original wire current flows along positive \(z\).
\(\textbf{Interpret the field-strength gradient:}\)
The magnetic field magnitude is
\[
B(r)=\frac{\mu_0I}{2\pi r}
\]
It is greater nearer the wire and smaller farther away.
Therefore, moving radially towards the wire means moving towards stronger field.
Moving away means moving towards weaker field.
\(\textbf{Identify the specimens from their motion:}\)
An ordinary paramagnetic specimen is weakly attracted towards a region of greater magnetic-field magnitude.
An ordinary diamagnetic specimen is weakly repelled towards a region of smaller magnetic-field magnitude.
Consequently, P is paramagnetic and Q is diamagnetic.
\(\textbf{Analyse reversal of the wire current:}\)
Current reversal reverses the magnetic-field direction at every fixed point.
The field magnitude remains unchanged because the current magnitude and geometry are fixed.
The local magnetic field at the test position becomes
\[
\vec B'=-40\,\mu\text{T}\hat j
\]
P's induced magnetisation, which is parallel to the field, also reverses.
Q's induced magnetisation, which is antiparallel to the field, reverses as well.
However, the location of stronger and weaker field magnitudes has not changed.
P therefore still tends towards the wire, while Q still tends away from it.
\(\textbf{Final answer:}\)
\[
I=10\,\text{A along }+z,\qquad
P:\ \text{paramagnetic},\qquad
Q:\ \text{diamagnetic},\qquad
\text{radial tendencies unchanged after reversal}
\]
The right-hand rule determines the direction of the current from the measured field vector.
The inverse-distance relation determines the radial field-strength variation.
The specimens' movements then distinguish positive and negative magnetic susceptibility.
Reversing the field vector does not reverse the locations of strong and weak magnetic field.
The experiment therefore separates a change of magnetisation direction from a change in the direction of translational motion.
710. A previously magnetised ferromagnetic bar retains a magnetic moment even after the external magnetising field has been removed.
At that stage its uniform remanent magnetisation is measured as
\[
\vec M_r=2.0\times10^5\hat k\,\text{A m}^{-1}
\]
and its volume is
\[
V=2.0\times10^{-6}\,\text{m}^3
\]
A probe measures the bar's distant magnetic field at an axial position \(0.20\,\text{m}\) from its centre.
The bar is sufficiently small compared with this distance for the short-dipole approximation to apply.
Take \(\mu_0/(4\pi)=10^{-7}\,\text{T m A}^{-1}\).
A student claims that the measured moment must be zero because the externally applied magnetising intensity is now zero and the relation \(M=\chi H\) gives zero at \(H=0\).
Which option gives the bar's actual moment and distant axial field, together with the correct assessment of the student's claim?
ⓐ. \(\vec m=\vec0\) and \(\vec B=\vec0\); the student's claim is correct for every ferromagnetic material.
ⓑ. \(\vec m=0.40\hat k\,\text{A m}^2\), \(\vec B=+10\,\mu\text{T}\hat k\); the claim is false because remanent magnetisation depends on magnetic history.
ⓒ. \(\vec m=0.40\hat k\,\text{A m}^2\), \(\vec B=-10\,\mu\text{T}\hat k\); the claim is incorrect because the axial field always points opposite to the bar's moment.
ⓓ. \(\vec m=0.20\hat k\,\text{A m}^2\), \(\vec B=+5.0\,\mu\text{T}\hat k\); the claim is correct only when the field probe is removed.
Correct Answer: \(\vec m=0.40\hat k\,\text{A m}^2\), \(\vec B=+10\,\mu\text{T}\hat k\); the claim is false because remanent magnetisation depends on magnetic history.
Explanation: \(\textbf{Determine the magnetic moment from the measured remanence:}\)
Magnetisation is magnetic moment per unit volume:
\[
\vec M=\frac{\vec m}{V}
\]
Therefore,
\[
\vec m=V\vec M
\]
Substituting,
\[
\vec m=(2.0\times10^{-6})(2.0\times10^5\hat k)
\]
\[
\vec m=0.40\hat k\,\text{A m}^2
\]
\(\textbf{Calculate the distant axial field:}\)
For a short magnetic dipole,
\[
B_a=\frac{\mu_0}{4\pi}\frac{2m}{r^3}
\]
The probe distance is
\[
r=0.20\,\text{m}
\]
Hence,
\[
B_a=10^{-7}\frac{2(0.40)}{(0.20)^3}
\]
\[
B_a=10^{-7}\frac{0.80}{0.008}
\]
\[
B_a=1.0\times10^{-5}\,\text{T}
\]
\[
B_a=10\,\mu\text{T}
\]
On the dipole axis, the field points along the magnetic moment:
\[
\vec B_a=+10\,\mu\text{T}\hat k
\]
\(\textbf{Examine the student's use of susceptibility:}\)
The simple relation
\[
M=\chi H
\]
describes a linear magnetic response with an appropriate fixed susceptibility.
It is useful for ordinary weak diamagnetic and paramagnetic materials under the stated conditions.
A ferromagnet can retain magnetisation after the external magnetising field is removed.
Its magnetic state can depend on its previous magnetisation history.
Therefore, the measured nonzero remanence must not be discarded by applying a history-independent relation through the origin.
\(\textbf{Final answer:}\)
\[
\boxed{\vec m=0.40\hat k\,\text{A m}^2,\qquad
\vec B_a=+10\,\mu\text{T}\hat k}
\]
The bar remains a source of magnetic field even after the external magnetising apparatus has been switched off.
Its remanent magnetisation produces a nonzero magnetic moment.
That moment gives a measurable distant dipole field.
The contradiction arises from applying a weak-material linear-response relation to a ferromagnetic state where the magnetisation depends on magnetic history.
The measured remanence, rather than an unjustified constant-susceptibility assumption, controls the result.
711. A sample of nickel is examined at two temperatures on opposite sides of its ferromagnetic transition. The same weak magnetising apparatus is used in both trials.
The experimental observations are:
| Observation | Lower temperature | Higher temperature |
|---|
| Magnetisation while field is applied | Strong | Weak and along the field |
| Magnetisation after field removal | Appreciable remanence | Negligible |
| Magnetic induction in a separate uniform-field test region | \(8.0\,\text{mT}\) | \(2.0\,\text{mT}\) |
A charged particle of fixed mass, charge magnitude and perpendicular-entry momentum is sent separately through each uniform-field test region.
Which option gives the ratio of the particle's higher-temperature orbit radius to its lower-temperature radius and the most defensible explanation of the material's thermal change?
ⓐ. Radius ratio \(1:4\); heating strengthens cooperative domain ordering and therefore reduces the orbit radius.
ⓑ. Radius ratio \(4:1\); the higher-temperature behaviour is consistent with loss of ferromagnetic order and a paramagnetic response above the Curie temperature.
ⓒ. Radius ratio \(4:1\); the observations prove that the microscopic magnetic moments cease to exist above the Curie temperature.
ⓓ. Radius ratio \(1:1\); the charged-particle orbit radius depends only on the apparatus current, not on the measured induction.
Correct Answer: Radius ratio \(4:1\); the higher-temperature behaviour is consistent with loss of ferromagnetic order and a paramagnetic response above the Curie temperature.
Explanation: \(\textbf{Separate the measured induction from the material mechanism:}\)
The charged particle is tested in two regions with specified magnetic inductions.
For motion perpendicular to a uniform field,
\[
r=\frac{p}{|q|B}
\]
The particle momentum and charge magnitude are unchanged.
Therefore, the orbit radius is inversely proportional to the measured induction.
\(\textbf{Compare the two orbit radii:}\)
Let \(r_L\) and \(r_H\) denote the lower- and higher-temperature radii.
Then,
\[
\frac{r_H}{r_L}=\frac{B_L}{B_H}
\]
Using the measured field values,
\[
\frac{r_H}{r_L}=\frac{8.0\,\text{mT}}{2.0\,\text{mT}}
\]
\[
\frac{r_H}{r_L}=4
\]
Thus, the orbit radius is four times greater in the higher-temperature test region.
\(\textbf{Interpret the temperature-dependent material response:}\)
The lower-temperature specimen shows strong magnetisation and appreciable retention after the magnetising field is removed.
Those observations are consistent with ferromagnetic ordering.
At the higher temperature, the response is much weaker and the magnetisation after field removal is negligible.
Such behaviour is consistent with the material having crossed its Curie temperature into the paramagnetic regime.
The thermal transition disrupts cooperative magnetic ordering.
It does not require all microscopic atomic or ionic magnetic moments to disappear.
\(\textbf{Check the role of the test particle:}\)
The particle's orbit measures the actual magnetic induction in the specified region.
Its radius cannot be predicted solely from the current in the magnetising apparatus when the material contribution changes.
The ratio follows from the supplied induction measurements rather than from an assumption of constant permeability.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{r_H}{r_L}=4}
\]
The charged-particle measurement provides an independent physical consequence of the change in magnetic induction.
The magnetic-material observations identify the likely change from ferromagnetic to paramagnetic response.
Both forms of evidence are needed for the complete answer.
The classification concerns collective magnetic ordering, whereas the orbit-radius calculation concerns the measured total magnetic induction.
712. A moving-coil galvanometer is designed with two distinct magnetic components: a magnet assembly that provides a persistent magnetic field, and a shaped core inside the moving coil that helps produce an approximately radial field geometry.
The magnet assembly must retain its magnetisation when the galvanometer is stored without any exciting current. The shaped central core should support a strong, well-directed magnetic field and respond readily when its magnetic state is changed during manufacturing or adjustment. The coil is suspended by a torsion spring, and its normal operating current is small.
Consider the following statements.
Statement I: A suitable hard magnetic material may be used for the persistent-field magnet assembly, while soft iron is suitable for the shaped central core.
Statement II: The radial-field geometry helps maintain an approximately constant magnetic torque per unit coil current throughout the intended range of angular deflection.
Statement III: Replacing the soft iron core with a strongly remanent hard magnetic core must necessarily improve scale linearity, regardless of the resulting field geometry.
Statement IV: If the radial field magnitude is increased while \(N\), \(A\) and the torsional constant remain unchanged, the galvanometer's current sensitivity increases.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. I, III and IV only
ⓒ. II, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: The permanent-field component and the shaped central core perform different functions.
A permanent magnet should retain a substantial magnetic moment when the external magnetising influence has been removed.
That makes an appropriate hard magnetic material useful for the persistent-field assembly.
Soft iron is well suited for shaping and concentrating magnetic field within the instrument.
Its relatively soft magnetic response is compatible with the role of a core rather than a permanent-magnet source.
Statement I is correct.
In a radial-field galvanometer, the magnetic field remains approximately parallel to the moving coil's plane as the coil rotates.
The magnetic torque for a fixed current is therefore approximately independent of deflection.
This gives the usual linear relation between current and angular displacement when the spring torque is proportional to displacement.
Statement II is correct.
Magnetic hardness alone does not determine the spatial geometry of a magnetic field.
Replacing the core with a hard material could change the field distribution and would not guarantee a radial field.
The stronger remanence of a hard material is therefore not proof of improved scale linearity.
Statement III is incorrect.
For a radial-field galvanometer, current sensitivity is
\[
S_I=\frac{\phi}{I}=\frac{NAB}{\kappa}
\]
where \(\kappa\) is the torsional constant.
Increasing \(B\) while the other quantities remain fixed increases the deflection produced by the same winding current.
Statement IV is correct.
The design requires both suitable magnetic materials and suitable field geometry.
A strong magnetic field that varies unfavourably with coil orientation need not produce the same calibration quality as a properly shaped radial field.
713. A hypothetical composite consists of equal volumes of two noninteracting, ordinary weak magnetic materials. Both constituents experience the same prescribed uniform magnetic intensity \(H=1000\,\text{A m}^{-1}\).
Constituent P is diamagnetic, with temperature-independent susceptibility over the stated range:
\[
\chi_P=-8.0\times10^{-5}
\]
Constituent Q is paramagnetic and obeys Curie's law. Its susceptibility at \(200\,\text{K}\) is
\[
\chi_Q(200\,\text{K})=+1.2\times10^{-4}
\]
Assume both components remain chemically stable and in their respective linear-response regimes throughout.
Because their volumes are equal and they experience the same H, the composite's average magnetisation is
\[
M_{\mathrm{avg}}=\frac{M_P+M_Q}{2}
\]
At what temperature does the net magnetisation become zero, and what are its values at \(200\,\text{K}\) and \(400\,\text{K}\)?
ⓐ. Zero at \(300\,\text{K}\); \(M_{\mathrm{avg}}=-0.020\,\text{A m}^{-1}\) at \(200\,\text{K}\); \(M_{\mathrm{avg}}=+0.010\,\text{A m}^{-1}\) at \(400\,\text{K}\).
ⓑ. Zero at \(400\,\text{K}\); \(M_{\mathrm{avg}}=+0.020\,\text{A m}^{-1}\) at \(200\,\text{K}\); \(M_{\mathrm{avg}}=0\) at \(400\,\text{K}\).
ⓒ. Zero at \(300\,\text{K}\); \(M_{\mathrm{avg}}=+0.020\,\text{A m}^{-1}\) at \(200\,\text{K}\); \(M_{\mathrm{avg}}=-0.010\,\text{A m}^{-1}\) at \(400\,\text{K}\).
ⓓ. Zero at \(300\,\text{K}\); both constituents individually cease to exhibit magnetic response at that temperature.
Correct Answer: Zero at \(300\,\text{K}\); \(M_{\mathrm{avg}}=+0.020\,\text{A m}^{-1}\) at \(200\,\text{K}\); \(M_{\mathrm{avg}}=-0.010\,\text{A m}^{-1}\) at \(400\,\text{K}\).
Explanation: \(\textbf{Write the two constituent magnetisations:}\)
For a linearly responding material,
\[
M=\chi H
\]
The diamagnetic constituent has a fixed negative susceptibility.
Therefore,
\[
M_P=(-8.0\times10^{-5})(1000)
\]
\[
M_P=-0.080\,\text{A m}^{-1}
\]
\(\textbf{Determine the paramagnetic temperature relation:}\)
For Q, Curie's law gives
\[
\chi_Q(T)=\frac CT
\]
At \(200\,\text{K}\),
\[
C=\chi_QT
\]
\[
C=(1.2\times10^{-4})(200)
\]
\[
C=2.4\times10^{-2}\,\text{K}
\]
Thus,
\[
\chi_Q(T)=\frac{2.4\times10^{-2}}{T}
\]
\(\textbf{Find the condition for zero average magnetisation:}\)
The net magnetisation vanishes when
\[
M_P+M_Q=0
\]
Since H is the same for both materials,
\[
\chi_P+\chi_Q(T_0)=0
\]
Therefore,
\[
\chi_Q(T_0)=8.0\times10^{-5}
\]
Using Curie's law,
\[
\frac{2.4\times10^{-2}}{T_0}=8.0\times10^{-5}
\]
\[
T_0=300\,\text{K}
\]
\(\textbf{Calculate the composite response at }200\,\text{K}\textbf{:}\)
At this temperature,
\[
M_Q=(1.2\times10^{-4})(1000)
\]
\[
M_Q=+0.120\,\text{A m}^{-1}
\]
The equal-volume average is
\[
M_{\mathrm{avg},200}=\frac{-0.080+0.120}{2}
\]
\[
M_{\mathrm{avg},200}=+0.020\,\text{A m}^{-1}
\]
\(\textbf{Calculate the composite response at }400\,\text{K}\textbf{:}\)
The paramagnetic susceptibility becomes
\[
\chi_Q(400)=1.2\times10^{-4}\frac{200}{400}
\]
\[
\chi_Q(400)=6.0\times10^{-5}
\]
Thus,
\[
M_Q(400)=(6.0\times10^{-5})(1000)
\]
\[
M_Q(400)=+0.060\,\text{A m}^{-1}
\]
The composite average is
\[
M_{\mathrm{avg},400}=\frac{-0.080+0.060}{2}
\]
\[
M_{\mathrm{avg},400}=-0.010\,\text{A m}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{T_0=300\,\text{K},\qquad
M_{\mathrm{avg},200}=+0.020\,\text{A m}^{-1},\qquad
M_{\mathrm{avg},400}=-0.010\,\text{A m}^{-1}}
\]
The temperature-dependent paramagnetic response initially dominates the equal-volume composite.
As temperature rises, thermal randomisation weakens that positive contribution.
At \(300\,\text{K}\), it exactly balances the negative diamagnetic contribution.
At \(400\,\text{K}\), the diamagnetic contribution dominates.
Neither constituent has to change its own magnetic classification for the sign of the composite's net magnetisation to change.
The equal-volume and common-field assumptions are essential to the stated averaging relation.
714. Two specimens P and Q are made from the same ordinary weak paramagnetic material. Both are at the same temperature and experience the same uniform applied magnetic intensity.
Their susceptibility is
\[
\chi=+0.002
\]
and the applied magnetic intensity is
\[
H=1000\,\text{A m}^{-1}
\]
Specimen P has volume \(1.0\times10^{-5}\,\text{m}^3\), while specimen Q has volume \(4.0\times10^{-5}\,\text{m}^3\).
Assume both are sufficiently small for the distant point-dipole approximation to apply at the observation positions, and that their induced magnetic moments point along positive \(z\).
A magnetic probe is placed the same sufficiently large distance from the centre of each specimen on its positive axial line.
Which option correctly compares their magnetisations, magnetic moments, magnetic inductions inside the material under the stated equal-H model, and distant axial fields?
ⓐ. Q has four times P's magnetisation and four times its axial field, but both have the same magnetic moment.
ⓑ. Both have magnetisation \(2.0\,\text{A m}^{-1}\) and equal total induction under the prescribed H; Q has four times P's magnetic moment and four times its distant axial field.
ⓒ. Both have magnetisation \(2.0\,\text{A m}^{-1}\); Q has four times P's total induction and the same distant axial field.
ⓓ. Both have magnetisation \(0.002\,\text{A m}^{-1}\); Q has four times P's magnetic moment, but the distant axial fields are equal because susceptibility is unchanged.
Correct Answer: Both have magnetisation \(2.0\,\text{A m}^{-1}\) and equal total induction under the prescribed H; Q has four times P's magnetic moment and four times its distant axial field.
Explanation: \(\textbf{Determine the intrinsic magnetic response:}\)
For either specimen,
\[
M=\chi H
\]
The material, temperature and applied intensity are identical.
Therefore,
\[
M_P=M_Q=(0.002)(1000)
\]
\[
M_P=M_Q=2.0\,\text{A m}^{-1}
\]
Magnetisation is the magnetic moment per unit volume, not the total moment.
\(\textbf{Calculate the two total magnetic moments:}\)
The induced moment is
\[
m=MV
\]
For P,
\[
m_P=(2.0)(1.0\times10^{-5})
\]
\[
m_P=2.0\times10^{-5}\,\text{A m}^2
\]
For Q,
\[
m_Q=(2.0)(4.0\times10^{-5})
\]
\[
m_Q=8.0\times10^{-5}\,\text{A m}^2
\]
Thus,
\[
\frac{m_Q}{m_P}=4
\]
\(\textbf{Compare the induction inside the specimens:}\)
For the stated equal-H constitutive comparison,
\[
B=\mu_0(H+M)
\]
Both H and M have the same values in P and Q.
Consequently,
\[
B_P=B_Q=\mu_0(1000+2)
\]
\[
B_P=B_Q=1002\mu_0
\]
Different specimen volumes do not by themselves alter this local constitutive relation.
\(\textbf{Compare the distant axial dipole fields:}\)
The axial field of a small magnetic dipole is
\[
B_a=\frac{\mu_0}{4\pi}\frac{2m}{r^3}
\]
The two probes are at the same distance r from their respective specimen centres.
Therefore,
\[
\frac{B_{a,Q}}{B_{a,P}}=\frac{m_Q}{m_P}
\]
\[
\frac{B_{a,Q}}{B_{a,P}}=4
\]
The axial field directions are the same because both induced moments point along positive \(z\).
\(\textbf{Final answer:}\)
\[
\boxed{M_P=M_Q=2.0\,\text{A m}^{-1},\qquad
m_P=2.0\times10^{-5}\,\text{A m}^2,\qquad
m_Q=8.0\times10^{-5}\,\text{A m}^2,\qquad
B_P=B_Q=1002\mu_0,\qquad
\frac{B_{a,Q}}{B_{a,P}}=4}
\]
Magnetisation characterises the local material response per unit volume.
The total magnetic moment also depends on how much material is present.
The prescribed equal-H model consequently gives equal magnetisation and equal local induction for the two specimens.
The larger specimen nevertheless has a greater total induced dipole moment.
At equally distant axial observation points, that greater moment produces a stronger far-field dipole contribution.
A susceptibility measurement and a magnetic-moment measurement do not describe the same physical quantity.
715. A ferromagnetic specimen is subjected to the following thermal and magnetic sequence:
Stage P: The specimen is below its Curie temperature and has strongly ordered magnetic domains, but its bulk magnetisation is nearly zero because domains of different orientations largely compensate. Stage Q: A magnetic field is applied while the temperature remains below the Curie temperature, producing substantial net magnetisation. Stage R: The field is removed, and some remanent magnetisation remains. Stage S: The specimen is heated sufficiently above its Curie temperature, disrupting the cooperative ferromagnetic ordering. Stage T: The specimen is cooled back below its Curie temperature in the absence of any applied magnetic field.
Consider the following statements.
Statement I: In Stage P, nearly zero bulk magnetisation does not imply that the individual domains have zero magnetisation.
Statement II: In Stage S, the disappearance of ferromagnetic ordering does not require every microscopic magnetic moment to vanish.
Statement III: During Stage T, ferromagnetic domains may form again without necessarily recovering the same macroscopic magnetisation direction and magnitude that existed in Stage R.
Statement IV: The remanent magnetisation in Stage R proves that the specimen must remain ferromagnetic even when heated to arbitrarily high temperatures.
Which statements are correct?
ⓐ. I, II and IV only
ⓑ. I, III and IV only
ⓒ. II, III and IV only
ⓓ. I, II and III only
Correct Answer: I, II and III only
Explanation: The sequence distinguishes microscopic magnetic ordering from the net magnetisation of the complete specimen.
Below the Curie temperature, a ferromagnet can consist of domains that are individually strongly magnetised.
Different domain directions can compensate, yielding very little bulk magnetisation.
Statement I is correct.
Applying a magnetic field favours domains aligned with that field and can produce a substantial macroscopic response.
After removing the field, some of the favourable alignment may persist as remanent magnetisation.
Heating the specimen sufficiently above its Curie temperature disrupts the cooperative ordering responsible for ferromagnetism.
The material then exhibits a paramagnetic response in the standard NCERT description.
Microscopic magnetic moments can still exist in the paramagnetic state.
Statement II is correct.
When the specimen is cooled again below the transition temperature, ferromagnetic ordering can reappear.
However, cooling in the absence of an applied field does not require the reformed domains to reproduce the exact former distribution.
The final bulk magnetisation may be small if newly formed domain directions largely compensate.
Statement III is correct.
The previous existence of remanence is not evidence that ferromagnetic order survives unlimited heating.
The Curie temperature places a thermal limit on the cooperative ordering.
Statement IV is incorrect.
The thermal cycle demonstrates that a material can change its magnetic state while its chemical identity remains unchanged.
It also shows that magnetic history matters when predicting remanence after field removal and subsequent heating or cooling.
716. Two ferromagnetic specimens P and Q are initially magnetised strongly in the positive direction.
Their measured behaviour is:
| Test | Specimen P | Specimen Q |
|---|
| Immediately after the magnetising field is removed | Substantial positive remanence | Substantial positive remanence |
| After applying a small oppositely directed field | Remanence is greatly reduced | Most of the positive magnetisation persists |
| After applying a considerably stronger oppositely directed field | Magnetisation is reversed | Magnetisation can then be reversed |
The observations are qualitative; no complete magnetisation-versus-field loops have been measured.
Which conclusion is best justified?
ⓐ. P is magnetically harder than Q because its magnetisation responds to the smaller reverse field.
ⓑ. P is relatively soft, Q relatively hard; both exhibit remanence, which by itself does not determine resistance to reversal.
ⓒ. P is paramagnetic and Q diamagnetic because neither specimen remains permanently unchanged under reverse fields.
ⓓ. P and Q must have the same magnetic hardness because both initially retain substantial magnetisation after removing the magnetising field.
Correct Answer: P is relatively soft, Q relatively hard; both exhibit remanence, which by itself does not determine resistance to reversal.
Explanation: Both specimens initially retain substantial magnetisation after the original magnetising field is removed.
This observation demonstrates remanent magnetic behaviour.
It does not by itself establish how difficult their magnetic states are to reverse.
The response to an oppositely directed field supplies additional evidence.
P loses much of its retained magnetisation under a relatively small reverse field.
This is consistent with comparatively soft magnetic behaviour.
Q retains most of its positive magnetisation under the same weak reverse field.
A substantially stronger reverse field is needed to change its retained magnetic state significantly.
This is consistent with comparatively hard magnetic behaviour.
A hard magnetic material is useful when magnetisation should persist despite moderate opposing magnetic influences.
A soft magnetic material is preferable in applications requiring relatively easy changes of magnetisation.
The words *soft* and *hard* describe magnetic behaviour rather than ordinary mechanical hardness.
Both specimens remain ferromagnetic under the specified tests.
The observations do not justify assigning exact numerical material constants without further measurements.
They do, however, support a comparison of resistance to magnetic reversal.
The critical reasoning is to distinguish retention after removing a field from resistance to an applied opposing field.
717. A radial-field moving-coil galvanometer has winding resistance \(G=60\,\Omega\), full-scale current \(I_g=2.0\,\text{mA}\), and a fixed permitted maximum angular deflection.
It is converted into a \(6.0\,\text{V}\) voltmeter using a series multiplier.
The magnetic field in the galvanometer is then increased to \(1.5\) times its original value, while the number of turns, coil area, torsional constant, winding resistance and maximum permitted deflection remain unchanged.
A new series multiplier is selected so that the modified instrument still has a \(6.0\,\text{V}\) full-scale range.
Which option correctly gives the original and new multipliers and compares the current and voltage sensitivities of the two completed voltmeters?
ⓐ. Original multiplier \(2940\,\Omega\), new multiplier \(4440\,\Omega\); both current and voltage sensitivities increase by \(50\%\).
ⓑ. Original multiplier \(2940\,\Omega\), new multiplier \(1940\,\Omega\); current sensitivity increases by \(50\%\), and the completed voltmeter's voltage sensitivity decreases.
ⓒ. Original multiplier \(3000\,\Omega\), new multiplier \(4500\,\Omega\); current sensitivity remains unchanged because the full-scale voltage is fixed.
ⓓ. Original multiplier \(2940\,\Omega\), new multiplier \(4440\,\Omega\); intrinsic current sensitivity increases by \(50\%\), but voltage sensitivity of the completed voltmeter remains unchanged.
Correct Answer: Original multiplier \(2940\,\Omega\), new multiplier \(4440\,\Omega\); intrinsic current sensitivity increases by \(50\%\), but voltage sensitivity of the completed voltmeter remains unchanged.
Explanation: \(\textbf{Determine the original voltmeter resistance:}\)
For a galvanometer converted into a voltmeter,
\[
V_{\mathrm{FS}}=I_g(G+R)
\]
where \(R\) is the series multiplier.
Thus,
\[
G+R_1=\frac{V_{\mathrm{FS}}}{I_g}
\]
\[
G+R_1=\frac{6.0}{2.0\times10^{-3}}
\]
\[
G+R_1=3000\,\Omega
\]
Subtracting the winding resistance,
\[
R_1=3000-60
\]
\[
R_1=2940\,\Omega
\]
\(\textbf{Determine how the stronger field changes the galvanometer:}\)
In a radial magnetic field,
\[
NIAB=\kappa\phi
\]
The current sensitivity is
\[
S_I=\frac{\phi}{I}=\frac{NAB}{\kappa}
\]
With \(N\), \(A\) and \(\kappa\) unchanged, increasing \(B\) by a factor of \(1.5\) increases the current sensitivity by the same factor.
The current needed for the original full-scale angle therefore decreases:
\[
I'_g=\frac{I_g}{1.5}
\]
\[
I'_g=\frac{2.0}{1.5}\,\text{mA}
\]
\[
I'_g=\frac43\,\text{mA}
\]
\(\textbf{Redesign the multiplier for the unchanged voltage range:}\)
The new voltmeter must still reach full-scale deflection at \(6.0\,\text{V}\).
Therefore,
\[
G+R_2=\frac{6.0}{(4/3)\times10^{-3}}
\]
\[
G+R_2=4500\,\Omega
\]
Hence,
\[
R_2=4500-60
\]
\[
R_2=4440\,\Omega
\]
\(\textbf{Compare the two sensitivities:}\)
The intrinsic current sensitivity increases by
\[
\frac{S'_I}{S_I}=1.5
\]
For either completed voltmeter,
\[
S_V=\frac{\phi}{V}
\]
Both have the same permitted full-scale deflection and the same full-scale voltage.
Consequently,
\[
\frac{S'_V}{S_V}=1
\]
The larger multiplier offsets the increase in galvanometer current sensitivity when an external voltage is applied.
\(\textbf{Final answer:}\)
\[
\boxed{R_1=2940\,\Omega,\qquad R_2=4440\,\Omega,\qquad
\frac{S'_I}{S_I}=1.5,\qquad \frac{S'_V}{S_V}=1}
\]
The magnetic modification increases the coil's angular response to a given current.
It also reduces the current needed for full-scale deflection.
To preserve the original voltage range, the total voltmeter resistance must therefore increase.
The final voltage sensitivity remains unchanged because the full-scale voltage and angular range have both been preserved.
This illustrates why increasing current sensitivity does not necessarily increase the voltage sensitivity of a converted instrument.
718. A galvanometer has winding resistance \(100\,\Omega\) and full-scale current \(10\,\text{mA}\).
It is converted into a two-range voltmeter using a selector that connects either a \(900\,\Omega\) or a \(2900\,\Omega\) series multiplier.
An unknown source has constant emf \(\mathcal E\) and internal resistance \(r\).
The voltmeter is connected directly across the source in two separate steady-state trials:
| Selected multiplier | Observed terminal voltage |
|---|
| \(900\,\Omega\) | \(4.0\,\text{V}\) |
| \(2900\,\Omega\) | \(6.0\,\text{V}\) |
Assume the galvanometer calibration is accurate and the source characteristics remain unchanged between trials.
What are the source emf, its internal resistance and the ratio of the winding currents in the two trials?
ⓐ. \(\mathcal E=6.0\,\text{V}\), \(r=500\,\Omega\); current ratio \(2:1\).
ⓑ. \(\mathcal E=8.0\,\text{V}\), \(r=1000\,\Omega\); current ratio \(1:2\).
ⓒ. \(\mathcal E=8.0\,\text{V}\), \(r=1000\,\Omega\); current ratio \(2:1\).
ⓓ. \(\mathcal E=10.0\,\text{V}\), \(r=1500\,\Omega\); current ratio \(2:1\).
Correct Answer: \(\mathcal E=8.0\,\text{V}\), \(r=1000\,\Omega\); current ratio \(2:1\).
Explanation: \(\textbf{Identify the voltmeter input resistance in each trial:}\)
The galvanometer and the selected multiplier are connected in series.
For the first trial,
\[
R_{V1}=G+R_1
\]
\[
R_{V1}=100+900
\]
\[
R_{V1}=1000\,\Omega
\]
For the second trial,
\[
R_{V2}=100+2900
\]
\[
R_{V2}=3000\,\Omega
\]
\(\textbf{Recover the currents actually drawn from the source:}\)
The voltmeter is the only external load.
Therefore, each winding carries the total current drawn from the source.
Using the observed terminal voltages,
\[
I_1=\frac{V_1}{R_{V1}}
\]
\[
I_1=\frac{4.0}{1000}
\]
\[
I_1=4.0\,\text{mA}
\]
Similarly,
\[
I_2=\frac{6.0}{3000}
\]
\[
I_2=2.0\,\text{mA}
\]
Hence,
\[
I_1:I_2=2:1
\]
\(\textbf{Connect terminal voltage with the source emf:}\)
For a source delivering current I,
\[
V=\mathcal E-Ir
\]
Equivalently,
\[
\mathcal E=V+Ir
\]
The first trial gives
\[
\mathcal E=4.0+0.004r
\]
The second gives
\[
\mathcal E=6.0+0.002r
\]
\(\textbf{Eliminate the unknown emf:}\)
Equating the two expressions,
\[
4.0+0.004r=6.0+0.002r
\]
\[
0.002r=2.0
\]
\[
r=1000\,\Omega
\]
\(\textbf{Recover the emf and check both measurements:}\)
Using the first trial,
\[
\mathcal E=4.0+(0.004)(1000)
\]
\[
\mathcal E=8.0\,\text{V}
\]
For the second trial,
\[
V_2=8.0-(0.002)(1000)
\]
\[
V_2=6.0\,\text{V}
\]
The result agrees with the independent measurement.
\(\textbf{Final answer:}\)
\[
\boxed{\mathcal E=8.0\,\text{V},\qquad r=1000\,\Omega,\qquad I_1:I_2=2:1}
\]
A real voltmeter changes the current drawn from a source because it has finite input resistance.
Changing its series multiplier therefore changes the source terminal voltage.
Two different voltmeter readings do not necessarily indicate that the source emf itself has changed.
Here the two loading conditions provide enough information to determine both source emf and internal resistance.
The full-scale currents are not reached in either trial, so both measurements remain within the galvanometer's permitted range.
719. A moving-coil galvanometer has resistance \(G=50\,\Omega\) and full-scale winding current \(I_g=2.0\,\text{mA}\).
It must be converted into a \(1.0\,\text{A}\) ammeter.
The laboratory has suitable low-value resistors, but each individual resistor is rated to dissipate no more than \(0.060\,\text{W}\).
The designer considers using either one shunt resistor or two identical resistors connected in parallel with one another, with their combined branch connected across the galvanometer.
Assume all resistors are ohmic and the required full-scale ammeter reading is maintained.
Which arrangement is suitable?
ⓐ. One \(0.1002\,\Omega\) shunt rated \(0.060\,\text{W}\), because the galvanometer itself dissipates almost all the electrical power.
ⓑ. Two \(0.2004\,\Omega\) resistors in parallel across the galvanometer, each dissipating approximately \(0.0499\,\text{W}\) at full scale.
ⓒ. Two \(0.1002\,\Omega\) resistors in parallel across the galvanometer, each dissipating approximately \(0.0499\,\text{W}\) at full scale.
ⓓ. Two \(0.2004\,\Omega\) resistors connected in series with the galvanometer, because this arrangement limits its winding current.
Correct Answer: Two \(0.2004\,\Omega\) resistors in parallel across the galvanometer, each dissipating approximately \(0.0499\,\text{W}\) at full scale.
Explanation: \(\textbf{Determine the galvanometer's full-scale terminal voltage:}\)
At maximum permitted deflection,
\[
V_g=I_gG
\]
Substituting,
\[
V_g=(2.0\times10^{-3})(50)
\]
\[
V_g=0.100\,\text{V}
\]
A shunt connected across the winding must have this same potential difference at full scale.
\(\textbf{Determine the required bypass current:}\)
The complete ammeter must carry
\[
I_A=1.0\,\text{A}
\]
Only \(0.002\,\text{A}\) may pass through the galvanometer.
Thus,
\[
I_s=I_A-I_g
\]
\[
I_s=1.0-0.002
\]
\[
I_s=0.998\,\text{A}
\]
\(\textbf{Determine the equivalent shunt resistance:}\)
By Ohm's law,
\[
S_{\mathrm{eq}}=\frac{V_g}{I_s}
\]
\[
S_{\mathrm{eq}}=\frac{0.100}{0.998}
\]
\[
S_{\mathrm{eq}}\approx0.10020\,\Omega
\]
\(\textbf{Check whether a single resistor is suitable:}\)
At full scale, its dissipated power would be
\[
P_s=V_gI_s
\]
\[
P_s=(0.100)(0.998)
\]
\[
P_s=0.0998\,\text{W}
\]
This exceeds the available individual resistor rating of \(0.060\,\text{W}\).
Therefore, a single available resistor with the correct resistance is not thermally suitable.
\(\textbf{Design the two-resistor parallel shunt:}\)
For two identical resistors of resistance \(R\) in parallel,
\[
S_{\mathrm{eq}}=\frac R2
\]
Therefore,
\[
R=2S_{\mathrm{eq}}
\]
\[
R\approx0.20040\,\Omega
\]
Each parallel resistor carries half the bypass current:
\[
I_{\mathrm{each}}=\frac{0.998}{2}
\]
\[
I_{\mathrm{each}}=0.499\,\text{A}
\]
\(\textbf{Check the heating of each resistor:}\)
Each resistor has the same \(0.100\,\text{V}\) terminal voltage.
Therefore,
\[
P_{\mathrm{each}}=VI_{\mathrm{each}}
\]
\[
P_{\mathrm{each}}=(0.100)(0.499)
\]
\[
P_{\mathrm{each}}=0.0499\,\text{W}
\]
This is below the specified \(0.060\,\text{W}\) rating.
\(\textbf{Final answer:}\)
\[
\boxed{R_{\mathrm{each}}\approx0.2004\,\Omega,\qquad
P_{\mathrm{each}}=0.0499\,\text{W}\lt 0.060\,\text{W}}
\]
The shunt protects the sensitive galvanometer by carrying almost all the measured current.
However, the shunt itself must be able to dissipate the electrical power associated with that current.
The correct equivalent resistance alone is not sufficient for a practical design.
Dividing the shunt into two equal parallel resistors preserves the necessary equivalent resistance while sharing the heating.
Connecting the resistors in series with the galvanometer would not provide the required ammeter conversion.
720. An ideal \(12.0\,\text{V}\) source supplies two resistors of \(2.0\,\text{k}\Omega\) each in series.
A galvanometer having resistance \(100\,\Omega\) and full-scale current \(1.0\,\text{mA}\) is converted into a voltmeter by adding a \(4900\,\Omega\) series multiplier.
The two 2.0-kilohm resistors form a potential divider across the ideal source. The voltmeter is connected directly across the LOWER resistor only. The other resistor remains above this parallel combination. The voltmeter has a linear pointer scale, and the measured voltage is within its full-scale range.
What voltage does the connected meter indicate, what would the potential difference across the lower resistor be without the meter, and what percentage error is introduced by the connection, relative to the unmeasured value?
ⓐ. Meter indicates \(6.0\,\text{V}\); unmeasured value \(6.0\,\text{V}\); zero loading error.
ⓑ. Meter indicates \(5.0\,\text{V}\); unmeasured value \(6.0\,\text{V}\); approximately \(20.0\%\) underestimation.
ⓒ. Meter indicates \(4.8\,\text{V}\); unmeasured value \(6.0\,\text{V}\); \(20.0\%\) underestimation.
ⓓ. Meter indicates \(5.0\,\text{V}\); unmeasured value \(6.0\,\text{V}\); approximately \(16.7\%\) underestimation.
Correct Answer: Meter indicates \(5.0\,\text{V}\); unmeasured value \(6.0\,\text{V}\); approximately \(16.7\%\) underestimation.
Explanation: \(\textbf{Determine the converted voltmeter resistance:}\)
The multiplier is connected in series with the galvanometer.
Its total input resistance is
\[
R_V=G+R
\]
\[
R_V=100+4900
\]
\[
R_V=5000\,\Omega
\]
The full-scale voltage is
\[
V_{\mathrm{FS}}=I_gR_V
\]
\[
V_{\mathrm{FS}}=(1.0\times10^{-3})(5000)
\]
\[
V_{\mathrm{FS}}=5.0\,\text{V}
\]
\(\textbf{Determine the undisturbed divider voltage:}\)
Without the voltmeter, the two \(2000\,\Omega\) resistors are connected in series.
They divide the ideal \(12.0\,\text{V}\) source equally:
\[
V_{\mathrm{lower,ideal}}=12.0\frac{2000}{2000+2000}
\]
\[
V_{\mathrm{lower,ideal}}=6.0\,\text{V}
\]
\(\textbf{Calculate the resistance of the loaded lower branch:}\)
Connecting the voltmeter adds a parallel path across the lower resistor.
The parallel resistance is
\[
R_p=\frac{(2000)(5000)}{2000+5000}
\]
\[
R_p=\frac{10000}{7}\,\Omega
\]
\(\textbf{Determine the actual divider voltage:}\)
The upper \(2000\,\Omega\) resistor is now in series with \(R_p\).
Therefore,
\[
V_{\mathrm{measured}}=12.0\frac{R_p}{2000+R_p}
\]
Substituting,
\[
V_{\mathrm{measured}}=
12.0\frac{10000/7}{2000+10000/7}
\]
\[
V_{\mathrm{measured}}=
12.0\frac{10000}{24000}
\]
\[
V_{\mathrm{measured}}=5.0\,\text{V}
\]
The voltmeter reaches its full-scale deflection but does not exceed it.
\(\textbf{Calculate the loading error:}\)
The absolute error relative to the undisturbed divider voltage is
\[
\Delta V=6.0-5.0
\]
\[
\Delta V=1.0\,\text{V}
\]
The percentage underestimation is
\[
\epsilon=\frac{\Delta V}{V_{\mathrm{lower,ideal}}}\times100
\]
\[
\epsilon=\frac{1.0}{6.0}\times100
\]
\[
\epsilon\approx16.7\%
\]
\(\textbf{Final answer:}\)
\[
\boxed{V_{\mathrm{measured}}=5.0\,\text{V},\qquad
V_{\mathrm{lower,ideal}}=6.0\,\text{V},\qquad
\epsilon\approx16.7\%\ \text{underestimation}}
\]
The voltmeter is not an ideal infinite-resistance instrument.
Its finite input resistance changes the effective lower-arm resistance of the divider.
The source therefore establishes a different voltage distribution after the meter is connected.
A correct instrument reading can still differ from the potential difference that existed before the measurement.
The error is caused by circuit loading, not by an incorrect galvanometer calibration.
