201. Study the table for a short bar magnet in Earth's horizontal magnetic field.
| Row | Magnet orientation | Line on which cancellation is possible |
|---|
| P | \(\vec{m}\parallel\vec{B}_H\) | Equatorial line |
| Q | \(\vec{m}\parallel\vec{B}_H\) | Axial line because fields oppose there |
| R | \(\vec{m}\) opposite to \(\vec{B}_H\) | Axial line |
| S | \(\vec{m}\) opposite to \(\vec{B}_H\) | No neutral point can ever occur |
The unsupported rows are
ⓐ. R and S
ⓑ. Q and S
ⓒ. P and R
ⓓ. P and Q
Correct Answer: Q and S
Explanation: Row P is supported because if \(\vec{m}\parallel\vec{B}_H\), the equatorial field of the magnet is opposite to \(\vec{m}\) and can cancel Earth's field. Row R is supported because if \(\vec{m}\) is opposite to \(\vec{B}_H\), the axial field is along \(\vec{m}\) and can oppose Earth's field. Row Q is unsupported because axial field reinforces Earth's field when \(\vec{m}\parallel\vec{B}_H\). Row S is unsupported because neutral points can occur in the opposite-orientation case, but on the axial line. The cancellation line is decided by the direction of the dipole field relative to \(\vec{B}_H\).
202. At a point near a bar magnet, Earth's horizontal field is \(B_H=3.6\times10^{-5}\,T\) toward north. The magnet's field at the same point is \(3.6\times10^{-5}\,T\) toward south. The resultant field is
ⓐ. \(3.6\times10^{-5}\,T\) toward south
ⓑ. \(3.6\times10^{-5}\,T\) toward north
ⓒ. \(0\)
ⓓ. \(7.2\times10^{-5}\,T\) toward north
Correct Answer: \(0\)
Explanation: \( \textbf{Given:} \) Earth's field \(B_H=3.6\times10^{-5}\,T\) toward north.
\( \textbf{Magnet's field:} \) \(B_m=3.6\times10^{-5}\,T\) toward south.
The two fields are along the same line but in opposite directions.
Take north as positive:
\[
B_H=+3.6\times10^{-5}\,T
\]
\[
B_m=-3.6\times10^{-5}\,T
\]
\( \textbf{Resultant field:} \)
\[
B_{net}=B_H+B_m
\]
\[
B_{net}=3.6\times10^{-5}-3.6\times10^{-5}
\]
\[
B_{net}=0
\]
\( \textbf{Final answer:} \) The resultant field is \(0\).
The point is neutral because the two equal fields act in opposite directions.
203. A compass is moved slowly near a bar magnet kept in Earth's field. At one point, the needle becomes uncertain and does not settle along a stable direction. The best explanation is that
ⓐ. the resultant magnetic field may be nearly zero there
ⓑ. the compass needle has become non-magnetic permanently
ⓒ. Earth's magnetic field has changed into electric field
ⓓ. magnetic field lines must intersect at that point
Correct Answer: the resultant magnetic field may be nearly zero there
Explanation: A compass needle aligns with the resultant magnetic field at its location. Near a bar magnet, the resultant field is the vector sum of the magnet's field and Earth's field. If these two fields nearly cancel, the resultant field becomes very small. A compass placed there may not settle clearly because there is no strong direction to follow. This is a practical indication of a neutral or nearly neutral region, not evidence that field lines intersect.
204. Consider the following statements about neutral points near a bar magnet.
Statement I: They are explained by superposition of magnetic fields.
Statement II: At a neutral point, the magnet's field and Earth's field are equal in magnitude and opposite in direction.
Statement III: At a neutral point, both individual fields must be absent.
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, II and III
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Statement I is true because the resultant field near a bar magnet in Earth's field is obtained by vector addition. Statement II is also true because neutral point formation requires equal and opposite fields. Statement III is false because the individual fields need not be zero. Earth's field and the magnet's field may both be present, but their vector sum becomes zero. Neutrality is a cancellation condition, not an absence of all magnetic influence.
205. Magnetisation of a material describes
ⓐ. magnetic moment per unit volume of the material
ⓑ. magnetic flux through a closed surface only
ⓒ. resistance per unit length of the material
ⓓ. electric charge per unit area of the material
Correct Answer: magnetic moment per unit volume of the material
Explanation: Magnetisation tells how strongly a material is magnetised in terms of magnetic dipole moment per unit volume. It is represented by the vector \(\vec{M}\). If many microscopic magnetic moments inside a material align in a common direction, the net magnetic moment per unit volume becomes significant. This idea connects the magnetic behaviour of matter with microscopic dipoles. It should not be confused with magnetic flux, which is related to field passing through an area.
206. The relation defining magnetisation is
ⓐ. \(M=\frac{V}{m_{net}}\)
ⓑ. \(M=m_{net}V\)
ⓒ. \(M=\frac{m_{net}}{V}\)
ⓓ. \(M=m_{net}+V\)
Correct Answer: \(M=\frac{m_{net}}{V}\)
Explanation: \( \textbf{Definition:} \) Magnetisation is magnetic moment per unit volume.
If the net magnetic moment of a sample is \(m_{net}\) and its volume is \(V\), then:
\[
M=\frac{m_{net}}{V}
\]
This relation shows that magnetisation is a density-like quantity for magnetic moment.
A larger net magnetic moment gives larger \(M\) if the volume is fixed.
For the same \(m_{net}\), a larger volume gives a smaller magnetisation.
\( \textbf{Final answer:} \) The correct relation is \(M=\frac{m_{net}}{V}\).
The product \(m_{net}V\) would not represent moment per unit volume.
207. A sample has net magnetic moment \(m_{net}=0.040\,A\,m^2\) and volume \(2.0\times10^{-4}\,m^3\). Its magnetisation is
ⓐ. \(500\,A\,m^{-1}\)
ⓑ. \(100\,A\,m^{-1}\)
ⓒ. \(200\,A\,m^{-1}\)
ⓓ. \(80\,A\,m^{-1}\)
Correct Answer: \(200\,A\,m^{-1}\)
Explanation: \( \textbf{Given:} \) \(m_{net}=0.040\,A\,m^2\) and \(V=2.0\times10^{-4}\,m^3\).
\( \textbf{Required:} \) Magnetisation \(M\).
\( \textbf{Definition:} \)
\[
M=\frac{m_{net}}{V}
\]
\( \textbf{Substitution:} \)
\[
M=\frac{0.040\,A\,m^2}{2.0\times10^{-4}\,m^3}
\]
\( \textbf{Numerical simplification:} \)
\[
\frac{0.040}{2.0\times10^{-4}}=200
\]
\( \textbf{Unit simplification:} \)
\[
\frac{A\,m^2}{m^3}=A\,m^{-1}
\]
\[
M=200\,A\,m^{-1}
\]
\( \textbf{Final answer:} \) The magnetisation is \(200\,A\,m^{-1}\).
The unit \(A\,m^{-1}\) follows from magnetic moment divided by volume.
208. The SI unit of magnetisation \(\vec{M}\) is
ⓐ. \(A\,m^{-1}\)
ⓑ. \(T\,m^{-1}\)
ⓒ. \(N\,m^{-1}\)
ⓓ. \(A\,m^{2}\)
Correct Answer: \(A\,m^{-1}\)
Explanation: Magnetisation is defined as magnetic moment per unit volume. Magnetic moment has unit \(A\,m^2\), and volume has unit \(m^3\). Dividing gives \(\frac{A\,m^2}{m^3}=A\,m^{-1}\). Thus the SI unit of magnetisation is \(A\,m^{-1}\). The unit \(A\,m^2\) belongs to magnetic dipole moment, not to magnetisation.
209. A material is made of many tiny magnetic dipoles. In an unmagnetised ferromagnetic sample, the domains are usually arranged so that
ⓐ. random domains give nearly zero net magnetisation
ⓑ. all magnetic field lines disappear permanently
ⓒ. no microscopic magnetic moments exist
ⓓ. all domains are perfectly aligned in one direction
Correct Answer: random domains give nearly zero net magnetisation
Explanation: Ferromagnetic materials contain domains, which are small regions where many microscopic magnetic moments are aligned. In an unmagnetised sample, different domains point in different directions. Their contributions largely cancel, so the net magnetisation of the whole sample can be nearly zero. This does not mean microscopic magnetic moments are absent. Magnetisation appears strongly when favourable domains grow or align under an applied magnetic field.
210. A ferromagnetic specimen becomes magnetised when placed in a suitable external magnetic field mainly because
ⓐ. its volume becomes zero
ⓑ. favourable domains align with the field
ⓒ. Earth's magnetic field becomes zero inside it
ⓓ. its atoms lose all magnetic moments
Correct Answer: favourable domains align with the field
Explanation: Ferromagnetic materials have domains containing aligned microscopic magnetic moments. Without an applied field, the domain directions may be arranged so that the total magnetisation is small. When an external magnetic field is applied, domains favourably oriented with the field can grow or become more aligned. This increases the net magnetic moment per unit volume. The process changes the internal magnetic ordering, not the existence of atoms or the volume of the specimen.
211. The vector direction of magnetisation \(\vec{M}\) in a uniformly magnetised specimen is along
ⓐ. the direction of gravitational field only
ⓑ. a line perpendicular to the net magnetic moment
ⓒ. the direction opposite to every microscopic dipole
ⓓ. the direction of the net magnetic moment per unit volume
Correct Answer: the direction of the net magnetic moment per unit volume
Explanation: Magnetisation \(\vec{M}\) is a vector quantity because it is based on magnetic dipole moment, which has direction. For a uniformly magnetised specimen, \(\vec{M}\) points in the direction of the net magnetic moment per unit volume. If the microscopic dipoles align more in one direction, the magnetisation points along that resultant direction. The magnitude of \(M\) gives how much magnetic moment exists per unit volume. Direction is therefore part of magnetisation, not an extra unrelated property.
212. A specimen has volume \(5.0\times10^{-5}\,m^3\) and magnetisation \(M=400\,A\,m^{-1}\). Its net magnetic moment is
ⓐ. \(8.0\times10^6\,A\,m^2\)
ⓑ. \(4.0\times10^2\,A\,m^2\)
ⓒ. \(2.0\times10^{-2}\,A\,m^2\)
ⓓ. \(1.25\times10^{-7}\,A\,m^2\)
Correct Answer: \(2.0\times10^{-2}\,A\,m^2\)
Explanation: \( \textbf{Given:} \) \(V=5.0\times10^{-5}\,m^3\) and \(M=400\,A\,m^{-1}\).
\( \textbf{Required:} \) Net magnetic moment \(m_{net}\).
Start with:
\[
M=\frac{m_{net}}{V}
\]
Solving for \(m_{net}\):
\[
m_{net}=MV
\]
\( \textbf{Substitution:} \)
\[
m_{net}=(400\,A\,m^{-1})(5.0\times10^{-5}\,m^3)
\]
\( \textbf{Numerical calculation:} \)
\[
400\times5.0\times10^{-5}=2000\times10^{-5}
\]
\[
2000\times10^{-5}=2.0\times10^{-2}
\]
\( \textbf{Unit calculation:} \)
\[
(A\,m^{-1})(m^3)=A\,m^2
\]
\( \textbf{Final answer:} \) The net magnetic moment is \(2.0\times10^{-2}\,A\,m^2\).
This is the reverse use of the magnetisation definition.
213. The dimensional expression of magnetisation \(\vec{M}\) can be obtained from \(M=\frac{m_{net}}{V}\). If magnetic moment has unit \(A\,m^2\), the dimension of \(M\) is
ⓐ. \([A\,L^2]\)
ⓑ. \([A\,L^{-1}]\)
ⓒ. \([M\,L\,T^{-2}]\)
ⓓ. \([A\,L^{-3}]\)
Correct Answer: \([A\,L^{-1}]\)
Explanation: \( \textbf{Definition:} \) Magnetisation is magnetic moment per unit volume.
\[
M=\frac{m_{net}}{V}
\]
\( \textbf{Unit of magnetic moment:} \)
\[
[m_{net}]=A\,m^2
\]
\( \textbf{Unit of volume:} \)
\[
[V]=m^3
\]
\( \textbf{Unit of magnetisation:} \)
\[
[M]=\frac{A\,m^2}{m^3}
\]
\[
[M]=A\,m^{-1}
\]
\( \textbf{Dimension:} \)
\[
[M]=[A\,L^{-1}]
\]
\( \textbf{Final answer:} \) The dimensional expression of magnetisation is \([A\,L^{-1}]\).
The power of length becomes \(-1\) because magnetic moment is divided by volume.
214. Two specimens have the same net magnetic moment \(m_{net}\). Specimen P has volume \(V\), while specimen Q has volume \(2V\). The ratio of their magnetisations \(\frac{M_P}{M_Q}\) is
ⓐ. \(\frac{1}{2}\)
ⓑ. \(4\)
ⓒ. \(1\)
ⓓ. \(2\)
Correct Answer: \(2\)
Explanation: \( \textbf{Magnetisation definition:} \)
\[
M=\frac{m_{net}}{V}
\]
\( \textbf{For specimen P:} \)
\[
M_P=\frac{m_{net}}{V}
\]
\( \textbf{For specimen Q:} \)
\[
M_Q=\frac{m_{net}}{2V}
\]
\( \textbf{Ratio:} \)
\[
\frac{M_P}{M_Q}=\frac{\frac{m_{net}}{V}}{\frac{m_{net}}{2V}}
\]
\[
\frac{M_P}{M_Q}=2
\]
\( \textbf{Final answer:} \) \(\frac{M_P}{M_Q}=2\).
For the same net magnetic moment, spreading it through double the volume halves the magnetisation.
215. Study the table and identify the row that correctly matches a quantity with its unit.
| Row | Quantity | SI unit |
|---|
| P | Magnetic dipole moment \(m\) | \(A\,m^{-1}\) |
| Q | Magnetisation \(M\) | \(A\,m^{-1}\) |
| R | Magnetisation \(M\) | \(A\,m^2\) |
| S | Volume \(V\) | \(A\,m^{-1}\) |
ⓐ. Row S
ⓑ. Row Q
ⓒ. Row P
ⓓ. Row R
Correct Answer: Row Q
Explanation: Magnetisation is magnetic moment per unit volume, so its SI unit is \(A\,m^{-1}\). Row Q therefore matches the correct quantity and unit. Magnetic dipole moment has unit \(A\,m^2\), so Row P gives the unit of magnetisation instead of moment. Row R reverses the same distinction by assigning the unit of magnetic moment to magnetisation. Volume is measured in \(m^3\), not in \(A\,m^{-1}\). The unit difference is important because \(m\) describes a total moment, while \(M\) describes moment density.
216. A uniformly magnetised rod has net magnetic moment directed from its left end toward its right end. The direction of its magnetisation \(\vec{M}\) is
ⓐ. undefined because magnetisation is scalar
ⓑ. from left to right
ⓒ. from right to left
ⓓ. perpendicular to the rod only
Correct Answer: from left to right
Explanation: Magnetisation \(\vec{M}\) is magnetic moment per unit volume, so it has the same direction as the net magnetic moment density. If the net magnetic moment of the rod points from left to right, the magnetisation also points from left to right. The quantity is a vector, not merely a scalar magnitude. A perpendicular direction would require the net magnetic moment to be perpendicular to the rod, which is not the stated case. The arrow on \(\vec{M}\) carries physical meaning because material response has direction.
217. A graph is plotted with magnetisation \(M\) on the vertical axis and net magnetic moment \(m_{net}\) on the horizontal axis for samples of the same volume \(V\).
The graph is a straight line through the origin.
The slope of the graph is
ⓐ. \(m_{net}V\)
ⓑ. \(\frac{V}{m_{net}}\)
ⓒ. \(\frac{1}{V}\)
ⓓ. \(V\)
Correct Answer: \(\frac{1}{V}\)
Explanation: \( \textbf{Definition of magnetisation:} \)
\[
M=\frac{m_{net}}{V}
\]
For a fixed volume \(V\), rewrite this relation as:
\[
M=\left(\frac{1}{V}\right)m_{net}
\]
This is of the form \(y=(\text{slope})x\).
Here \(y\) is \(M\), and \(x\) is \(m_{net}\).
\[
\text{slope}=\frac{1}{V}
\]
\( \textbf{Final answer:} \) The slope is \(\frac{1}{V}\).
A larger volume gives a smaller slope because the same magnetic moment is distributed through more material.
218. A sample has magnetisation \(M=1.6\times10^3\,A\,m^{-1}\) and net magnetic moment \(m_{net}=0.080\,A\,m^2\). Its volume is
ⓐ. \(1.6\times10^{-1}\,m^3\)
ⓑ. \(2.0\times10^{-5}\,m^3\)
ⓒ. \(1.28\times10^{2}\,m^3\)
ⓓ. \(5.0\times10^{-5}\,m^3\)
Correct Answer: \(5.0\times10^{-5}\,m^3\)
Explanation: \( \textbf{Given:} \) \(M=1.6\times10^3\,A\,m^{-1}\) and \(m_{net}=0.080\,A\,m^2\).
\( \textbf{Required:} \) Volume \(V\).
Start with the definition:
\[
M=\frac{m_{net}}{V}
\]
Solve for \(V\):
\[
V=\frac{m_{net}}{M}
\]
\( \textbf{Substitution:} \)
\[
V=\frac{0.080\,A\,m^2}{1.6\times10^3\,A\,m^{-1}}
\]
\( \textbf{Numerical calculation:} \)
\[
\frac{0.080}{1.6\times10^3}=5.0\times10^{-5}
\]
\( \textbf{Unit calculation:} \)
\[
\frac{A\,m^2}{A\,m^{-1}}=m^3
\]
\[
V=5.0\times10^{-5}\,m^3
\]
\( \textbf{Final answer:} \) The volume is \(5.0\times10^{-5}\,m^3\).
The result has unit \(m^3\), confirming that the relation has been rearranged correctly.
219. Magnetic intensity \(\vec{H}\) is best described as
ⓐ. the applied magnetising field
ⓑ. the total magnetic dipole moment of a sample
ⓒ. magnetic moment per unit volume only
ⓓ. magnetic flux through a closed surface
Correct Answer: the applied magnetising field
Explanation: Magnetic intensity \(\vec{H}\) represents the applied magnetising field responsible for magnetising a material. It is different from magnetisation \(\vec{M}\), which describes the magnetic moment per unit volume produced in the material. The symbol \(\vec{B}\) represents magnetic field or magnetic flux density, not the same idea as \(\vec{H}\). Magnetic intensity is especially useful when separating the source field from the material response. This distinction becomes necessary when studying susceptibility and permeability.
220. The SI unit of magnetic intensity \(\vec{H}\) is
ⓐ. \(A\,m^2\)
ⓑ. \(T\,m^2\)
ⓒ. \(J\,T^{-1}\)
ⓓ. \(A\,m^{-1}\)
Correct Answer: \(A\,m^{-1}\)
Explanation: Magnetic intensity \(\vec{H}\) has SI unit \(A\,m^{-1}\). Magnetisation \(\vec{M}\) also has the same SI unit, but the two quantities do not mean the same thing. \(\vec{H}\) represents the magnetising field, while \(\vec{M}\) represents magnetic moment per unit volume of the material. The unit \(A\,m^2\) belongs to magnetic dipole moment. Sharing a unit does not make \(\vec{H}\) and \(\vec{M}\) physically identical.
