Torque on Current Loop & Magnetic Moment — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Torque on Current Loop & Magnetic Moment MCQs with step-by-step solutions (21 questions). Part of Magnetic Fields due to Electric Current. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Torque on Current Loop & Magnetic Moment · easy · theory
The magnetic dipole moment of a coil of $N$ turns, area $A$ carrying current $I$ is:
A. $NIA$ ✓ Correct
B. $\dfrac{NI}{A}$
C. $N^2IA$
D. $\dfrac{NA}{I}$
Solution: Its direction is normal to the plane of the coil, given by the right-hand rule applied to the current.
Q2 — Torque on Current Loop & Magnetic Moment · easy · theory
The torque on a current loop of magnetic moment $m$ in a uniform field $B$ at angle $\theta$ is:
A. $mB\tan\theta$
B. $mB\sin\theta$ ✓ Correct
C. $mB\cos\theta$
D. $\dfrac{mB}{\sin\theta}$
Solution: Equivalently $\tau = NIAB\sin\theta$, where $\theta$ is measured between $\vec{m}$ and $\vec{B}$.
Q3 — Torque on Current Loop & Magnetic Moment · medium · theory
The torque on a current loop in a uniform magnetic field is maximum when the plane of the loop is:
A. Perpendicular to the field
B. At $45^\circ$ to the field
C. Parallel to the field ✓ Correct
D. Any orientation
Solution: With the plane parallel to $\vec{B}$, the normal $\vec{m}$ is perpendicular to it, so $\sin\theta = 1$.
Q4 — Torque on Current Loop & Magnetic Moment · medium · theory
The torque on a current loop is zero when the plane of the loop is:
A. Perpendicular to the field ✓ Correct
B. At $60^\circ$ to the field
C. At $45^\circ$ to the field
D. Parallel to the field
Solution: Then $\vec{m}$ lies along $\vec{B}$, giving $\sin\theta = 0$ — the position of stable equilibrium.
Q5 — Torque on Current Loop & Magnetic Moment · easy · theory
The SI unit of magnetic dipole moment is:
A. $\text{J}\cdot\text{T}$
B. $\text{A}\cdot\text{m}$
C. $\text{A}\cdot\text{m}^2$ ✓ Correct
D. $\text{A}/\text{m}$
Solution: From $m = IA$, the unit is ampere times square metre, equivalently joule per tesla.
Q6 — Torque on Current Loop & Magnetic Moment · medium · theory
The magnetic dipole moment of a current loop is a:
A. Vector lying in the plane of the loop
B. Scalar quantity
C. Vector along the current direction
D. Vector directed along the normal to the plane of the loop ✓ Correct
Solution: Curling the right-hand fingers along the current makes the thumb point along $\vec{m}$.
Q7 — Torque on Current Loop & Magnetic Moment · medium · theory
The potential energy of a magnetic dipole of moment $m$ in a field $B$ at angle $\theta$ is:
A. $+mB\cos\theta$
B. $-mB\cos\theta$ ✓ Correct
C. $mB\tan\theta$
D. $-mB\sin\theta$
Solution: $U = -\vec{m}\cdot\vec{B}$, minimum when the moment is aligned with the field.
Q8 — Torque on Current Loop & Magnetic Moment · hard · theory
A moving coil galvanometer uses a radial magnetic field so that:
A. The torque becomes zero
B. The field becomes uniform
C. The coil rotates faster
D. The deflection is directly proportional to the current ✓ Correct
Solution: A radial field keeps the coil plane always parallel to $\vec{B}$, so $\sin\theta = 1$ at every position and $\tau = NIAB$.
Q9 — Torque on Current Loop & Magnetic Moment · medium · theory
The net force on a current loop placed in a uniform magnetic field is:
A. Equal to $NIAB$
B. Equal to $mB$
C. Equal to $BIL$
D. Zero ✓ Correct
Solution: The forces on opposite sides cancel in pairs, leaving only a couple that produces rotation.
Q10 — Torque on Current Loop & Magnetic Moment · medium · numerical
A coil of $100$ turns and area $0.01\text{ m}^2$ carries $2\text{ A}$. Its magnetic moment is:
A. $0.2\text{ A}\cdot\text{m}^2$
B. $200\text{ A}\cdot\text{m}^2$
C. $2\text{ A}\cdot\text{m}^2$ ✓ Correct
D. $0.02\text{ A}\cdot\text{m}^2$
Solution: $m = NIA = 100 \times 2 \times 0.01 = 2\text{ A}\cdot\text{m}^2$.
Q11 — Torque on Current Loop & Magnetic Moment · hard · numerical
A coil of $50$ turns and area $0.02\text{ m}^2$ carrying $1\text{ A}$ is placed with its plane parallel to a field of $0.5\text{ T}$. The torque on it is:
A. $0.25\text{ N}\cdot\text{m}$
B. $1\text{ N}\cdot\text{m}$
C. $0.5\text{ N}\cdot\text{m}$ ✓ Correct
D. $0.05\text{ N}\cdot\text{m}$
Solution: $\tau = NIAB\sin 90^\circ = 50 \times 1 \times 0.02 \times 0.5 = 0.5\text{ N}\cdot\text{m}$.
Q12 — Torque on Current Loop & Magnetic Moment · medium · numerical
A loop of area $4 \times 10^{-4}\text{ m}^2$ carries $5\text{ A}$. Its magnetic moment is:
A. $2 \times 10^{-3}\text{ A}\cdot\text{m}^2$ ✓ Correct
B. $1.25 \times 10^4\text{ A}\cdot\text{m}^2$
C. $8 \times 10^{-5}\text{ A}\cdot\text{m}^2$
D. $2 \times 10^{-4}\text{ A}\cdot\text{m}^2$
Solution: $m = IA = 5 \times 4 \times 10^{-4} = 2 \times 10^{-3}\text{ A}\cdot\text{m}^2$.
Q13 — Torque on Current Loop & Magnetic Moment · easy · numerical
A dipole of magnetic moment $2\text{ A}\cdot\text{m}^2$ is placed in a field of $0.3\text{ T}$. The maximum torque on it is:
A. $0.15\text{ N}\cdot\text{m}$
B. $0.3\text{ N}\cdot\text{m}$
C. $0.6\text{ N}\cdot\text{m}$ ✓ Correct
D. $6\text{ N}\cdot\text{m}$
Solution: $\tau_{max} = mB = 2 \times 0.3 = 0.6\text{ N}\cdot\text{m}$.
Q14 — Torque on Current Loop & Magnetic Moment · medium · numerical
The potential energy of a magnetic dipole aligned with the field is:
A. $\dfrac{mB}{2}$
B. $-mB$ ✓ Correct
C. Zero
D. $+mB$
Solution: $U = -mB\cos 0^\circ = -mB$, the minimum possible energy and hence stable equilibrium.
Q15 — Torque on Current Loop & Magnetic Moment · easy · numerical
If the area of a current loop is doubled at constant current, its magnetic moment:
A. Becomes four times
B. Halves
C. Remains unchanged
D. Doubles ✓ Correct
Solution: $m = IA \propto A$.
Q16 — Torque on Current Loop & Magnetic Moment · easy · numerical
If the number of turns of a coil is doubled at constant current and area, its magnetic moment:
A. Becomes four times
B. Doubles ✓ Correct
C. Remains unchanged
D. Halves
Solution: $m = NIA \propto N$.
Q17 — Torque on Current Loop & Magnetic Moment · hard · numerical
A circular loop of radius $0.1\text{ m}$ carries $2\text{ A}$. Its magnetic moment is approximately:
A. $0.2\text{ A}\cdot\text{m}^2$
B. $0.0063\text{ A}\cdot\text{m}^2$
C. $0.63\text{ A}\cdot\text{m}^2$
D. $0.063\text{ A}\cdot\text{m}^2$ ✓ Correct
Solution: Area $= \pi(0.1)^2 = 0.0314\text{ m}^2$, so $m = 2 \times 0.0314 \approx 0.063\text{ A}\cdot\text{m}^2$.
Q18 — Torque on Current Loop & Magnetic Moment · medium · numerical
A current loop is placed with its plane perpendicular to a uniform magnetic field. The torque on it is:
A. Equal to $mB$
B. Maximum
C. Equal to $\dfrac{mB}{2}$
D. Zero ✓ Correct
Solution: The magnetic moment is then parallel to the field, so $\tau = mB\sin 0^\circ = 0$.
Q19 — Torque on Current Loop & Magnetic Moment · hard · numerical
The work done in rotating a magnetic dipole from alignment with the field to a position perpendicular to it is:
A. $mB$ ✓ Correct
B. $2mB$
C. Zero
D. $\dfrac{mB}{2}$
Solution: $W = U_{90^\circ} - U_{0^\circ} = 0 - (-mB) = mB$.
Q20 — Torque on Current Loop & Magnetic Moment · medium · numerical
A coil of $200$ turns and area $10^{-3}\text{ m}^2$ carries $0.5\text{ A}$. Its magnetic moment is:
A. $0.1\text{ A}\cdot\text{m}^2$ ✓ Correct
B. $1\text{ A}\cdot\text{m}^2$
C. $100\text{ A}\cdot\text{m}^2$
D. $0.01\text{ A}\cdot\text{m}^2$
Solution: $m = NIA = 200 \times 0.5 \times 10^{-3} = 0.1\text{ A}\cdot\text{m}^2$.
Q21 — Torque on Current Loop & Magnetic Moment · hard · numerical
A dipole of moment $m$ in a field $B$ is turned from $0^\circ$ to $60^\circ$. The work done is:
A. $1.5mB$
B. $0.5mB$ ✓ Correct
C. $0.87mB$
D. $mB$
Solution: $W = -mB\cos 60^\circ - (-mB\cos 0^\circ) = -0.5mB + mB = 0.5mB$.