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Magnetism and Matter — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Magnetism and Matter MCQs with step-by-step solutions covering Bar Magnet and Magnetic Dipole, Earth Magnetism, Magnetic Materials. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Bar Magnet and Magnetic Dipole · medium · theory
A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It
A. will become rigid showing no movement
B. will stay in any position ✓ Correct
C. will stay in North-South direction only
D. will stay in East-West direction only
Solution: At a geomagnetic pole the horizontal component of the earth's magnetic field is zero, so there is no restoring torque on the needle and it can stay in any position.
Q2 — Bar Magnet and Magnetic Dipole · medium · numerical
A bar magnet having a magnetic moment of $2\times10^{4}\,\text{J T}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B=6\times10^{-4}\,\text{T}$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^\circ$ from the field is
A. $0.6\,\text{J}$
B. $12\,\text{J}$
C. $6\,\text{J}$ ✓ Correct
D. $2\,\text{J}$
Solution: $W=MB(\cos0^\circ-\cos60^\circ)=2\times10^{4}\times6\times10^{-4}\times\left(1-\dfrac12\right)=6\,\text{J}$.
Q3 — Bar Magnet and Magnetic Dipole · medium · theory
A bar magnet of magnetic moment $\vec{M}$ is placed in a magnetic field of induction $\vec{B}$. The torque exerted on it is
A. $\vec{M}\cdot\vec{B}$
B. $-\vec{M}\cdot\vec{B}$
C. $\vec{M}\times\vec{B}$ ✓ Correct
D. $-\vec{M}\times\vec{B}$
Solution: The torque on a magnetic dipole placed in an external field is $\vec\tau=\vec{M}\times\vec{B}$.
Q4 — Earth Magnetism · medium · theory
The relations amongst the three elements of earth's magnetic field, namely horizontal component $H$, vertical component $V$ and dip $\delta$, are ($B_E=$ total magnetic field)
A. $V=B_E\tan\delta,\,H=B_E$
B. $V=B_E\sin\delta,\,H=B_E\cos\delta$ ✓ Correct
C. $V=B_E\cos\delta,\,H=B_E\sin\delta$
D. $V=B_E,\,H=B_E\tan\delta$
Solution: From the vector triangle of the earth's magnetic field, the horizontal component is $H=B_E\cos\delta$ and the vertical component is $V=B_E\sin\delta$.
Q5 — Magnetic Materials · medium · numerical
An iron rod of susceptibility 599 is subjected to a magnetising field of $1200\,\text{A m}^{-1}$. The permeability of the material of the rod is (Take, $\mu_0=4\pi\times10^{-7}\,\text{T m A}^{-1}$)
A. $8.0\pi\times10^{-5}\,\text{T m A}^{-1}$
B. $2.4\pi\times10^{-5}\,\text{T m A}^{-1}$
C. $2.4\pi\times10^{-7}\,\text{T m A}^{-1}$
D. $2.4\pi\times10^{-4}\,\text{T m A}^{-1}$ ✓ Correct
Solution: Permeability $\mu=\mu_0(1+\chi_m)=4\pi\times10^{-7}\times(1+599)=4\pi\times10^{-7}\times600=2.4\pi\times10^{-4}\,\text{T m A}^{-1}$.
Q6 — Magnetic Materials · medium · theory
The magnetic susceptibility is negative for
A. paramagnetic material only
B. ferromagnetic material only
C. paramagnetic and ferromagnetic materials
D. diamagnetic material only ✓ Correct
Solution: Relative permeability $\mu_r=1+\chi_m$. For diamagnetic substances $\mu_r<1$, so $\chi_m$ is negative. For paramagnetic and ferromagnetic substances $\chi_m$ is positive.
Q7 — Magnetic Materials · medium · theory
There are four light weight rod samples $A$, $B$, $C$ and $D$ separately suspended by thread. A bar magnet is slowly brought near each sample and the following observations are noted: (i) $A$ is feebly repelled (ii) $B$ is feebly attracted (iii) $C$ is strongly attracted (iv) $D$ remains unaffected. Which one of the following is true?
A. $C$ is of a diamagnetic material
B. $D$ is of a ferromagnetic material
C. $A$ is of a non-magnetic material
D. $B$ is of a paramagnetic material ✓ Correct
Solution: Diamagnetic materials are feebly repelled ($A$), paramagnetic materials are feebly attracted ($B$), ferromagnetic materials are strongly attracted ($C$), and materials with negligible susceptibility remain unaffected ($D$). So $B$ is paramagnetic.
Q8 — Magnetic Materials · medium · theory
If a diamagnetic substance is brought near the North or the South pole of a bar magnet, it is
A. repelled by both the poles ✓ Correct
B. repelled by the North pole and attracted by the South pole
C. attracted by the North pole and repelled by the South pole
D. attracted by both the poles
Solution: A diamagnetic substance gets weakly magnetised opposite to the applied field, so it tends to move from a region of stronger field to weaker field, i.e. it is repelled by both the North and South poles of a bar magnet.
Q9 — Magnetic Materials · medium · theory
Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond the Curie temperature, then it will show
A. paramagnetism ✓ Correct
B. anti-ferromagnetism
C. no magnetic property
D. diamagnetism
Solution: Above the Curie temperature, thermal agitation destroys the exchange coupling responsible for ferromagnetism, and the material becomes simply paramagnetic.
Q10 — Magnetic Materials · medium · theory
Above the Curie temperature,
A. a ferromagnetic substance becomes paramagnetic ✓ Correct
B. a paramagnetic substance becomes diamagnetic
C. a diamagnetic substance becomes paramagnetic
D. a paramagnetic substance becomes ferromagnetic
Solution: Ferromagnetism decreases with rising temperature. Above a definite (Curie) temperature, the persistent alignment of atomic dipoles breaks down and the ferromagnetic substance becomes simply paramagnetic.
Q11 — Magnetic Materials · medium · theory
According to Curie's law, the magnetic susceptibility of a paramagnetic substance at an absolute temperature $T$ is proportional to
A. $\dfrac{1}{T^2}$
B. $T^2$
C. $\dfrac{1}{T}$ ✓ Correct
D. $T$
Solution: Curie's law states that the magnetic susceptibility of a paramagnetic substance is inversely proportional to its absolute temperature, i.e. $\chi_m\propto \dfrac{1}{T}$.
Q12 — Magnetic Materials · medium · theory
A diamagnetic material in a magnetic field moves
A. perpendicular to the field
B. from weaker to the stronger parts of the field
C. from stronger to the weaker parts of the field ✓ Correct
D. None of the above
Solution: A diamagnetic material is magnetised opposite to the external field, so it experiences a net force pushing it from stronger to weaker parts of a non-uniform magnetic field.
Q13 — Magnetic Materials · medium · theory
In which type of material does the magnetic susceptibility not depend on temperature?
A. Diamagnetic ✓ Correct
B. Paramagnetic
C. Ferromagnetic
D. Ferrite
Solution: Diamagnetism arises from induced currents in electron orbits and does not depend on thermal agitation, so the magnetic susceptibility of a diamagnetic substance is independent of temperature.
Q14 — Magnetic Materials · medium · theory
A diamagnetic substance is brought near a strong magnet, then it is
A. attracted by the magnet
B. repelled by the magnet ✓ Correct
C. repelled by the North pole and attracted by the South pole
D. attracted by the North pole and repelled by the South pole
Solution: A diamagnetic substance is feebly magnetised opposite to the applied field, so near a strong magnet it is repelled by the magnet (by both its poles).
Q15 — Magnetic Materials · medium · theory
For protecting a sensitive equipment from the external magnetic field, it should be
A. placed inside an aluminium can
B. placed inside an iron can ✓ Correct
C. wrapped with insulation around it when passing current through it
D. surrounded with fine copper sheet
Solution: Iron is a ferromagnetic (high permeability) material, so magnetic field lines get concentrated within it and effectively excluded from the enclosed region. Placing the equipment inside an iron can shields it from the external magnetic field.
Q16 — Bar Magnet and Magnetic Dipole · medium · numerical
A 250-turn rectangular coil of length $2.1\,\text{cm}$ and width $1.25\,\text{cm}$ carries a current of $85\,\mu\text{A}$ and is subjected to a magnetic field of strength $0.85\,\text{T}$. The work done for rotating the coil by $180^\circ$ against the torque is
A. $9.1\,\mu\text{J}$ ✓ Correct
B. $4.55\,\mu\text{J}$
C. $2.3\,\mu\text{J}$
D. $1.5\,\mu\text{J}$
Solution: $W=2MB=2NIAB=2\times250\times(85\times10^{-6})\times(2.1\times1.25\times10^{-4})\times0.85\approx9.5\times10^{-6}\,\text{J}$, closest to $9.1\,\mu\text{J}$.
Q17 — Bar Magnet and Magnetic Dipole · medium · theory
A uniform conducting wire of length $12a$ and resistance $R$ is wound up as a current ($I$) carrying coil in the shape of (1) an equilateral triangle of side $a$, and (2) a square of side $a$. The magnetic dipole moments of the coil in each case respectively are
A. $\sqrt{3}\,Ia^2$ and $3Ia^2$ ✓ Correct
B. $3Ia^2$ and $Ia^2$
C. $3Ia^2$ and $4Ia^2$
D. $4Ia^2$ and $3Ia^2$
Solution: Triangle: perimeter $3a\Rightarrow n=4$ turns, area $=\dfrac{\sqrt3}{4}a^2$, so $M=nIA=4I\cdot\dfrac{\sqrt3}{4}a^2=\sqrt3\,Ia^2$. Square: perimeter $4a\Rightarrow n=3$ turns, area $=a^2$, so $M=nIA=3Ia^2$.
Q18 — Bar Magnet and Magnetic Dipole · medium · theory
A wire of length $L$ metre carrying a current of $I$ ampere is bent in the form of a circle. Its magnetic moment is
A. $\dfrac{IL^2}{4}\,\text{Am}^2$
B. $\dfrac{\pi IL^2}{4}\,\text{Am}^2$
C. $\dfrac{2IL^2}{\pi}\,\text{Am}^2$
D. $\dfrac{IL^2}{4\pi}\,\text{Am}^2$ ✓ Correct
Solution: Circumference $L=2\pi r\Rightarrow r=\dfrac{L}{2\pi}$. Area $A=\pi r^2=\dfrac{L^2}{4\pi}$. Magnetic moment $M=IA=\dfrac{IL^2}{4\pi}\,\text{Am}^2$.
Q19 — Bar Magnet and Magnetic Dipole · medium · theory
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^\circ$ is $W$. Now the torque required to keep the magnet in this new position is
A. $\dfrac{W}{\sqrt3}$
B. $\sqrt3\,W$ ✓ Correct
C. $\dfrac{\sqrt3\,W}{2}$
D. $\dfrac{2W}{\sqrt3}$
Solution: $W=MB(\cos0^\circ-\cos60^\circ)=MB\left(1-\dfrac12\right)=\dfrac{MB}{2}\Rightarrow MB=2W$. Torque at $60^\circ$ is $\tau=MB\sin60^\circ=2W\cdot\dfrac{\sqrt3}{2}=\sqrt3\,W$.
Q20 — Bar Magnet and Magnetic Dipole · medium · theory
A bar magnet of length $l$ and magnetic dipole moment $M$ is bent in the form of a circular arc that subtends an angle of $60^\circ$ at the centre of the circle (i.e. the magnet's length equals one-sixth of the circle's circumference). The new magnetic dipole moment will be
A. $M$
B. $\dfrac{3}{\pi}M$ ✓ Correct
C. $\dfrac{2}{\pi}M$
D. $\dfrac{M}{2}$
Solution: Since the arc subtends $60^\circ$, $l=\dfrac{2\pi r}{6}=\dfrac{\pi r}{3}\Rightarrow r=\dfrac{3l}{\pi}$. Pole strength $m=\dfrac{M}{l}$, and the new moment (pole strength times the straight-line distance between poles) is $M'=m\cdot r=\dfrac{M}{l}\cdot\dfrac{3l}{\pi}=\dfrac{3M}{\pi}$.
Q21 — Bar Magnet and Magnetic Dipole · medium · theory
A charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$. The associated magnetic moment $\mu$ is given by
A. $\dfrac{qvR}{2}$ ✓ Correct
B. $qvR^2$
C. $\dfrac{qvR^2}{2}$
D. $qvR$
Solution: Current equivalent to the revolving charge is $i=\dfrac{qv}{2\pi R}$, so $\mu=iA=\dfrac{qv}{2\pi R}\times\pi R^2=\dfrac{qvR}{2}$.
Q22 — Earth Magnetism · medium · theory
At a point $A$ on the earth's surface the angle of dip is $\delta=+25^\circ$. At a point $B$ on the earth's surface the angle of dip is $\delta=-25^\circ$. We can interpret that
A. $A$ is located in the southern hemisphere and $B$ is located in the northern hemisphere
B. $A$ is located in the northern hemisphere and $B$ is located in the southern hemisphere ✓ Correct
C. $A$ and $B$ are both located in the southern hemisphere
D. $A$ and $B$ are both located in the northern hemisphere
Solution: The angle of dip is taken positive in the northern hemisphere and negative in the southern hemisphere, so $A$ (dip $+25^\circ$) lies in the northern hemisphere and $B$ (dip $-25^\circ$) lies in the southern hemisphere.
Q23 — Earth Magnetism · medium · theory
If $\delta_1$ and $\delta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\delta$ is given by
A. $\cot^2\delta=\cot^2\delta_1+\cot^2\delta_2$ ✓ Correct
B. $\tan^2\delta=\tan^2\delta_1+\tan^2\delta_2$
C. $\cot^2\delta=\cot^2\delta_1-\cot^2\delta_2$
D. $\tan^2\delta=\tan^2\delta_1-\tan^2\delta_2$
Solution: With $\cot\delta_1=\dfrac{B_H\cos\theta}{B_V}$ and $\cot\delta_2=\dfrac{B_H\sin\theta}{B_V}$, squaring and adding gives $\cot^2\delta_1+\cot^2\delta_2=\dfrac{B_H^2}{B_V^2}=\cot^2\delta$.
Q24 — Earth Magnetism · medium · numerical
A vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2\,\text{s}$ in the earth's horizontal magnetic field of $24\,\mu\text{T}$. When a horizontal field of $18\,\mu\text{T}$ is produced opposite to the earth's field by placing a current carrying wire, the new time period of the magnet will be
A. $1\,\text{s}$
B. $2\,\text{s}$ ✓ Correct
C. $3\,\text{s}$
D. $4\,\text{s}$
Solution: Time period of a vibration magnetometer is $T=2\pi\sqrt{\dfrac{I}{MB_H}}$, so $T\propto \dfrac{1}{\sqrt{B_H}}$. Thus $\dfrac{T_1}{T_2}=\sqrt{\dfrac{(B_H)_2}{(B_H)_1}}\Rightarrow \dfrac{2}{T_2}=\sqrt{\dfrac{18}{24}}\Rightarrow T_2\approx 2\,\text{s}$.
Q25 — Earth Magnetism · medium · theory
Two bar magnets having the same geometry with magnetic moments $M$ and $2M$ are firstly placed in such a way that their similar poles are on the same side, then its period of oscillation is $T_1$. Now, the polarity of one of the magnets is reversed and the time period of oscillations becomes $T_2$. Then,
A. $T_1<T_2$ ✓ Correct
B. $T_1=T_2$
C. $T_1>T_2$
D. $T_2=\infty$
Solution: With similar poles together, net moment $M_1=M+2M=3M$ and $T_1=2\pi\sqrt{\dfrac{I}{3MH}}$. With one polarity reversed, net moment $M_2=2M-M=M$ and $T_2=2\pi\sqrt{\dfrac{I}{MH}}$. Since $M_1>M_2$, $T_1<T_2$.
Q26 — Earth Magnetism · medium · theory
Due to the earth's magnetic field, charged cosmic ray particles
A. can never reach the poles
B. can never reach the equator
C. require less kinetic energy to reach the equator than the poles
D. require greater kinetic energy to reach the equator than the poles ✓ Correct
Solution: At the poles the earth's field is vertical and parallel to the particle's velocity, so the magnetic force $qv B\sin\theta=0$ and particles reach easily. At the equator the field is horizontal (perpendicular to $v$), so the deflecting force $qvB$ is maximum; only particles with greater kinetic energy can overcome this and reach the equator.
Q27 — Earth Magnetism · medium · theory
A bar magnet is oscillating in the earth's magnetic field with a period $T$. What happens to its period of motion if its mass is quadrupled?
A. Motion remains simple harmonic with new period $\dfrac{T}{2}$
B. Motion remains simple harmonic with new period $2T$ ✓ Correct
C. Motion remains simple harmonic with new period $4T$
D. Motion remains simple harmonic and the period stays nearly constant
Solution: $T=2\pi\sqrt{\dfrac{I}{MB}}$, so $T\propto\sqrt{I}$. Since $I\propto m$ (for the same geometry), quadrupling the mass quadruples $I$, so $T$ becomes $\sqrt{4}=2$ times, i.e. new period $=2T$.
Q28 — Magnetic Materials · medium · theory
A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence, the rod gains gravitational potential energy. The work required to do this comes from
A. the lattice structure of the material of the rod
B. the magnetic field
C. the current source ✓ Correct
D. the induced electric field due to the changing magnetic field
Solution: A diamagnetic substance is repelled from the stronger field region and moves to the weaker field region, gaining gravitational potential energy. This energy is supplied by the current source that maintains the magnetic field against the back-reaction of the moving rod.
Q29 — Magnetic Materials · medium · theory
If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material are denoted by $\mu_d$, $\mu_p$ and $\mu_f$ respectively, then
A. $\mu_d=0$ and $\mu_f\ne 0$
B. $\mu_d\ne 0$ and $\mu_f=0$
C. $\mu_d=0$ and $\mu_p\ne 0$ ✓ Correct
D. $\mu_p\ne 0$ and $\mu_f=0$
Solution: In diamagnetic substances, electron spins in each pair are opposite and cancel, so the atomic magnetic moment $\mu_d=0$. Paramagnetic and ferromagnetic substances have atoms with a net excess of unpaired spins, so $\mu_p\ne 0$ (and $\mu_f\ne 0$).