301. Bar magnets are often stored in pairs with soft iron keepers across their ends. The keepers are used mainly to
ⓐ. increase the geometric length of each magnet
ⓑ. make magnetic field lines intersect inside the keeper
ⓒ. convert \(N\)-poles into \(S\)-poles permanently
ⓓ. provide a closed magnetic path
Correct Answer: provide a closed magnetic path
Explanation: A soft iron keeper provides an easy magnetic path between opposite poles of a magnet or a pair of magnets. This helps complete the magnetic circuit and reduces the tendency of the magnet to lose magnetisation over time. The keeper becomes temporarily magnetised while in contact with the magnet. It is not used to change the identity of poles. Proper storage is especially useful for maintaining the strength of permanent magnets.
302. Consider the following statements about care of magnets.
Statement I: Heating can weaken a magnet.
Statement II: Hammering or dropping can reduce magnetisation.
Statement III: A soft iron keeper helps preserve magnetisation during storage.
ⓐ. II and III only
ⓑ. I, II and III
ⓒ. I and III only
ⓓ. I and II only
Correct Answer: I, II and III
Explanation: Statement I is true because heating increases thermal disorder and can disturb magnetic domain alignment. Statement II is true because mechanical shocks may also disturb the domain arrangement. Statement III is true because a soft iron keeper provides a favourable magnetic path and helps preserve magnetisation. These are practical consequences of the domain model of ferromagnetism. Magnet care is mainly about maintaining ordered domain contribution and reducing unnecessary demagnetising influences.
303. A short bar magnet of moment \(m\) is placed with \(\vec{m}\parallel\vec{B}_H\). At a neutral point on its equatorial line, the distance from the centre is \(r\). The relation used to find \(r\) is
ⓐ. \(\oint \vec{B}\cdot d\vec{A}=B_H\)
ⓑ. \(mB_H\sin r=0\)
ⓒ. \(\frac{\mu_0}{4\pi}\frac{2m}{r^3}=B_H\)
ⓓ. \(\frac{\mu_0}{4\pi}\frac{m}{r^3}=B_H\)
Correct Answer: \(\frac{\mu_0}{4\pi}\frac{m}{r^3}=B_H\)
Explanation: When \(\vec{m}\parallel\vec{B}_H\), neutral points occur on the equatorial line of the short magnet. On the equatorial line, the dipole field magnitude is \(\frac{\mu_0}{4\pi}\frac{m}{r^3}\) and its direction is opposite to \(\vec{m}\). Since \(\vec{m}\) is parallel to \(\vec{B}_H\), the equatorial field can oppose Earth's horizontal field. At a neutral point, the two magnitudes must be equal. The axial factor \(2\) is not used because the point is on the equatorial line.
304. A short magnet has \(m=0.80\,A\,m^2\). It is placed with \(\vec{m}\parallel\vec{B}_H\), where \(B_H=4.0\times10^{-5}\,T\). Taking \(\frac{\mu_0}{4\pi}=10^{-7}\,T\,m\,A^{-1}\), the distance of an equatorial neutral point from the centre is
ⓐ. \(0.100\,m\)
ⓑ. \(0.400\,m\)
ⓒ. \(0.200\,m\)
ⓓ. \(0.126\,m\)
Correct Answer: \(0.126\,m\)
Explanation: \( \textbf{Given:} \) \(m=0.80\,A\,m^2\), \(B_H=4.0\times10^{-5}\,T\), and \(\frac{\mu_0}{4\pi}=10^{-7}\,T\,m\,A^{-1}\).
Since \(\vec{m}\parallel\vec{B}_H\), the neutral point is on the equatorial line.
At a neutral point:
\[
B_{equatorial}=B_H
\]
\[
\frac{\mu_0}{4\pi}\frac{m}{r^3}=B_H
\]
Solve for \(r^3\):
\[
r^3=\frac{\frac{\mu_0}{4\pi}m}{B_H}
\]
\( \textbf{Substitution:} \)
\[
r^3=\frac{(10^{-7})(0.80)}{4.0\times10^{-5}}
\]
\[
r^3=\frac{0.80}{4.0}\times10^{-2}
\]
\[
r^3=0.20\times10^{-2}=2.0\times10^{-3}
\]
\[
r=\sqrt[3]{2.0\times10^{-3}}\,m
\]
\[
r\approx0.126\,m
\]
\( \textbf{Final answer:} \) The neutral point distance is approximately \(0.126\,m\).
The equatorial formula is used because the magnet's moment is parallel to Earth's horizontal field.
305. A material has \(\chi_m=-1.2\times10^{-5}\), and another has \(\chi_m=3.0\times10^{-4}\). In a non-uniform magnetic field, their expected responses are respectively
ⓐ. weak attraction and weak repulsion
ⓑ. identical motion toward stronger field
ⓒ. weak repulsion and weak attraction
ⓓ. strong permanent magnetism and no response
Correct Answer: weak repulsion and weak attraction
Explanation: The first material has negative susceptibility, so it is diamagnetic. Diamagnetic materials develop magnetisation opposite to the applied field and are weakly repelled from stronger field regions. The second material has small positive susceptibility, so it is paramagnetic. Paramagnetic materials develop weak magnetisation along the applied field and are weakly attracted toward stronger field regions. The sign of \(\chi_m\) decides the direction of response relative to the applied field.
306. A current loop of area \(A\) carries current \(I\) and is placed in a uniform magnetic field \(B\). The loop is turned from \(\theta=0^\circ\) to \(\theta=90^\circ\), where \(\theta\) is the angle between \(\vec{m}\) and \(\vec{B}\). The increase in potential energy is
ⓐ. \(IAB\)
ⓑ. \(0\)
ⓒ. \(2IAB\)
ⓓ. \(-IAB\)
Correct Answer: \(IAB\)
Explanation: \( \textbf{Magnetic moment of loop:} \)
\[
m=IA
\]
\( \textbf{Potential energy of a magnetic dipole:} \)
\[
U=-mB\cos\theta
\]
\( \textbf{Initial angle:} \) \(\theta_i=0^\circ\).
\[
U_i=-mB\cos0^\circ=-mB
\]
\( \textbf{Final angle:} \) \(\theta_f=90^\circ\).
\[
U_f=-mB\cos90^\circ=0
\]
\( \textbf{Increase in energy:} \)
\[
\Delta U=U_f-U_i
\]
\[
\Delta U=0-(-mB)=mB
\]
Using \(m=IA\):
\[
\Delta U=IAB
\]
\( \textbf{Final answer:} \) The increase in potential energy is \(IAB\).
The result is positive because the loop is moved away from its minimum-energy alignment.
307. A ferromagnetic material has high retentivity, high coercivity, and a wide hysteresis loop. A second material has high permeability, low coercivity, and a narrow hysteresis loop. The first and second materials are better suited respectively for
ⓐ. permanent magnet and transformer core
ⓑ. transformer core and permanent magnet
ⓒ. temporary relay core and permanent magnet
ⓓ. diamagnetic shielding and compass correction
Correct Answer: permanent magnet and transformer core
Explanation: High retentivity helps a material keep magnetisation, and high coercivity helps it resist demagnetisation. These are desirable features of a permanent magnet. A wide loop indicates larger hysteresis loss, which is not suitable for repeated magnetic cycling. The second material has high permeability and low coercivity, so it is easily magnetised and demagnetised. Its narrow hysteresis loop makes it suitable for transformer cores because energy loss per cycle is small.
308. A closed surface is drawn around a bar magnet placed inside a high-permeability shield. The magnetic field distribution changes, but the net magnetic flux through the closed surface remains
ⓐ. \(\frac{q}{\varepsilon_0}\)
ⓑ. \(B_H\tan I\)
ⓒ. \(0\)
ⓓ. \(mB\sin\theta\)
Correct Answer: \(0\)
Explanation: Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero. This remains true even when magnetic materials change the distribution of magnetic field lines. A high-permeability shield can guide field lines and reduce the field in a protected region, but it does not create isolated magnetic monopoles. Since magnetic field lines remain closed, the algebraic flux through the closed surface cancels. The result is about closed-surface flux, not about the local field being zero everywhere.
309. A magnetised steel needle is heated strongly, then hammered, and finally placed in an alternating magnetic field of gradually decreasing strength. The combined effect is most likely to
ⓐ. increase its retentivity without limit
ⓑ. reduce its magnetisation significantly
ⓒ. make its susceptibility exactly dimensionless for the first time
ⓓ. create a single isolated magnetic pole
Correct Answer: reduce its magnetisation significantly
Explanation: Heating increases thermal agitation and can disturb domain alignment. Hammering can also disturb the ordered arrangement of domains. An alternating magnetic field can repeatedly reverse the preferred direction of domain alignment and reduce the net magnetisation. Together, these processes are common demagnetising methods. They weaken the ordered magnetic state but do not create isolated magnetic poles.
310. A magnetising field changes a ferromagnetic specimen from an unmagnetised state to saturation. After the field is removed, the specimen still has a large magnetisation. This retained value is linked most directly with
ⓐ. retentivity
ⓑ. magnetic declination
ⓒ. angle of dip
ⓓ. Gauss's law for magnetism
Correct Answer: retentivity
Explanation: When a ferromagnetic material is taken to saturation, many domains have become strongly aligned. If the applied magnetic intensity is then reduced to zero, some magnetisation may remain. This residual magnetisation is measured by retentivity. A large retained value means the material can keep magnetisation well. Declination and dip are elements of Earth's magnetism, while Gauss's law for magnetism concerns net magnetic flux through closed surfaces.
311. In a shielded room, a sensitive magnetic sensor reads a much smaller magnetic field than outside. The shield is most likely effective because
ⓐ. the external field has stopped forming closed field lines
ⓑ. the sensor has become non-magnetic permanently
ⓒ. a high-permeability wall carries most field lines through itself
ⓓ. the wall material has zero volume
Correct Answer: a high-permeability wall carries most field lines through itself
Explanation: A magnetic shield reduces the magnetic field in a protected region by giving field lines a favourable path through the shield material. High permeability is the key property that makes this path effective. The field inside the shielded room is therefore much weaker than the field outside. The sensor itself does not need to lose its magnetic properties for the reading to fall. Magnetic shielding changes field distribution while preserving the closed-loop nature of magnetic field lines.
312. Consider the following statements across magnetism and matter.
Statement I: A neutral point is produced by vector cancellation of magnetic fields.
Statement II: A transformer core should have a narrow hysteresis loop.
Statement III: A magnetic shield works best when its material has high permeability.
ⓐ. I and III only
ⓑ. I, II and III
ⓒ. I and II only
ⓓ. II and III only
Correct Answer: I, II and III
Explanation: Statement I is true because a neutral point occurs where two magnetic fields are equal in magnitude and opposite in direction. Statement II is true because a narrow hysteresis loop reduces energy loss per cycle in a transformer core. Statement III is true because high-permeability material guides magnetic field lines through the shield and weakens the field inside the protected region. These statements connect field superposition, material response, and device design. They belong to different parts of the chapter but use the same vector-field and material-response ideas.
313. A material is placed in a uniform external magnetic field. Inside the material, magnetic field lines are found to be slightly less dense than in the surrounding space. The material is most likely
ⓐ. a hard permanent magnet only
ⓑ. strongly ferromagnetic
ⓒ. diamagnetic
ⓓ. a perfect paramagnet with infinite susceptibility
Correct Answer: diamagnetic
Explanation: Diamagnetic materials develop induced magnetisation opposite to the applied magnetic field. This weak opposition slightly reduces the magnetic field inside the material compared with the surrounding field. In a field-line picture, this appears as field lines being slightly expelled or less dense inside the diamagnetic substance. Ferromagnetic materials instead concentrate field lines strongly inside them. The effect is weak for ordinary diamagnetic materials, so the phrase “slightly less dense” is important.
314. Study the field-line behaviour in different materials.
| Row | Material response | Field-line behaviour inside material |
|---|
| P | Diamagnetic | Slightly less dense than outside |
| Q | Paramagnetic | Slightly more dense than outside |
| R | Ferromagnetic | Highly concentrated inside |
| S | Diamagnetic | Highly concentrated because \(\mu_r\gg1\) |
The unsupported row is
ⓐ. Row Q
ⓑ. Row P
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Diamagnetic materials have \(\chi_m\lt0\) and \(\mu_r\) slightly less than \(1\), so field lines are slightly less dense inside them. Paramagnetic materials have small positive \(\chi_m\), so field lines are slightly more concentrated than in free space. Ferromagnetic materials can have very large permeability, so field lines become highly concentrated inside them. Row S wrongly assigns a ferromagnetic field-line pattern to a diamagnetic material. The density of field lines should be read together with the sign and magnitude of susceptibility.
315. A weakly magnetic solid is repelled from a region of stronger magnetic field and shows almost no change in susceptibility when its temperature changes moderately. This behaviour best matches
ⓐ. ferromagnetism below Curie temperature
ⓑ. paramagnetism obeying Curie's law
ⓒ. a saturated hard magnet only
ⓓ. diamagnetism
Correct Answer: diamagnetism
Explanation: Diamagnetic materials are weakly repelled from stronger magnetic field regions. Their susceptibility is small and negative. Diamagnetic susceptibility is usually almost independent of temperature over ordinary ranges. Paramagnetic susceptibility, in contrast, generally decreases with increasing temperature according to Curie's law. Ferromagnetic materials show much stronger attraction and domain-based behaviour, so the weak repulsion and near temperature-independence identify diamagnetism.
316. A graph of magnetisation \(M\) against magnetic intensity \(H\) for a ferromagnetic material is described below.
Starting from the unmagnetised state, \(M\) rises rapidly for small \(H\), then rises more slowly, and finally becomes nearly constant even when \(H\) is increased further.
The nearly constant part of the graph represents
ⓐ. magnetic saturation
ⓑ. zero susceptibility for all fields
ⓒ. neutral point formation
ⓓ. magnetic declination
Correct Answer: magnetic saturation
Explanation: In a ferromagnetic material, domains initially respond strongly to an applied magnetic intensity. At first, favourable domains grow and domain walls move, so magnetisation rises rapidly. As more domains become aligned, fewer additional domains are available to contribute. Eventually, the magnetisation approaches a limiting value called saturation. Increasing \(H\) further then produces only a small additional change in \(M\), so the curve becomes nearly flat.
317. In the early part of a ferromagnetic magnetisation curve, the rapid increase of \(M\) with \(H\) is mainly due to
ⓐ. disappearance of all microscopic magnetic moments
ⓑ. conversion of magnetic field into electric field
ⓒ. growth and alignment of favourable domains
ⓓ. formation of isolated magnetic monopoles
Correct Answer: growth and alignment of favourable domains
Explanation: Ferromagnetic materials contain domains in which many microscopic magnetic moments are already aligned. In an initially unmagnetised specimen, these domains are arranged in different directions, giving little net magnetisation. When an external magnetising field is applied, domains favourably oriented with the field can grow or rotate. This produces a rapid increase in net magnetisation. The process changes the domain arrangement, not the existence of magnetic moments themselves.
318. A ferromagnetic specimen is already close to saturation. If the applied magnetic intensity \(H\) is increased further, the magnetisation \(M\) changes only slightly because
ⓐ. most domains are already nearly aligned
ⓑ. the magnetic field inside must be zero
ⓒ. \(\chi_m\) has become negative for all ferromagnets
ⓓ. the specimen has become diamagnetic
Correct Answer: most domains are already nearly aligned
Explanation: Saturation means that the material has reached a state where most magnetic domains are already aligned as much as possible with the applied field. Once this state is nearly reached, increasing \(H\) cannot produce much extra alignment. Therefore \(M\) increases only slowly or becomes almost constant. This does not mean that the material has become diamagnetic. It means the ferromagnetic response has reached its alignment limit.
319. Match the magnetic material with a suitable example.
| Column I | Column II |
|---|
| P. Diamagnetic | 1. Aluminium |
| Q. Paramagnetic | 2. Iron |
| R. Ferromagnetic | 3. Bismuth |
| S. Ferromagnetic example group | 4. Iron, cobalt, nickel |
ⓐ. P-4, Q-1, R-3, S-2
ⓑ. P-3, Q-1, R-2, S-4
ⓒ. P-1, Q-3, R-2, S-4
ⓓ. P-3, Q-2, R-1, S-4
Correct Answer: P-3, Q-1, R-2, S-4
Explanation: Bismuth is a common example of a diamagnetic material. Aluminium is a common example of a paramagnetic material. Iron is ferromagnetic, and the standard ferromagnetic group includes iron, cobalt, and nickel. The examples match the sign and strength of magnetic response: weak negative for diamagnetic, weak positive for paramagnetic, and strong positive for ferromagnetic. Examples are useful only when they are tied to the response type, not memorised as isolated names.
320. A material shows a small positive slope in its \(M-H\) graph at room temperature. When temperature is increased, the slope decreases. The material is best described as
ⓐ. ferromagnetic with a wide hysteresis loop necessarily
ⓑ. paramagnetic
ⓒ. a material with \(\mu_r\lt1\)
ⓓ. diamagnetic with strong negative susceptibility
Correct Answer: paramagnetic
Explanation: The slope of an \(M-H\) graph in the linear region represents magnetic susceptibility. A small positive slope means a small positive susceptibility, which is typical of paramagnetic materials. When temperature increases, thermal agitation reduces the alignment of atomic magnetic moments, so the susceptibility decreases. This matches Curie's law behaviour, \(\chi_m=\frac{C}{T}\), in the suitable range. Diamagnetic materials would show small negative susceptibility, not a positive slope.
