Force and Torque on Current Carrying Conductor — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Force and Torque on Current Carrying Conductor MCQs with step-by-step solutions (10 questions). Part of Moving Charges and Magnetism. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Force and Torque on Current Carrying Conductor · hard · numerical
A metallic rod of mass per unit length $0.5\ \text{kg m}^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of $30^\circ$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction $0.25\ \text{T}$ is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
A. $14.76\ \text{A}$
B. $5.98\ \text{A}$
C. $7.14\ \text{A}$
D. $11.32\ \text{A}$ ✓ Correct
Solution: For equilibrium along the incline, $mg\sin\theta = F\cos\theta = ILB\cos\theta$, giving $I=\dfrac{m}{L}\dfrac{g\tan\theta}{B}=\dfrac{0.5\times 9.8\times\tan 30^\circ}{0.25}=\dfrac{0.5\times 9.8}{0.25\sqrt{3}}\approx 11.32\ \text{A}$.
Q2 — Force and Torque on Current Carrying Conductor · medium · numerical
A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength $0.2\ \text{Wb/m}^2$. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of $30^\circ$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be
A. $0.15\ \text{Nm}$
B. $0.20\ \text{Nm}$ ✓ Correct
C. $0.24\ \text{Nm}$
D. $0.12\ \text{Nm}$
Solution: Since the plane makes $30^\circ$ with $B$, the angle between the normal and $B$ is $60^\circ$. $\tau=NIAB\sin\theta=50\times 2\times(0.12\times 0.1)\times 0.2\times\sin 60^\circ\approx 0.20\ \text{Nm}$.
Q3 — Force and Torque on Current Carrying Conductor · medium · theory
A current loop in a magnetic field
A. experiences a torque whether the field is uniform or non-uniform in all orientations
B. can be in equilibrium in one orientation
C. can be in equilibrium in two orientations, both the equilibrium states are unstable
D. can be in equilibrium in two orientations, one stable while other is unstable ✓ Correct
Solution: A current loop's equilibrium orientations occur when its magnetic moment $\vec{M}$ is parallel or antiparallel to $\vec{B}$. The parallel orientation is stable equilibrium, and the antiparallel orientation is unstable equilibrium.
Q4 — Force and Torque on Current Carrying Conductor · medium · theory
A current carrying closed loop in the form of a right angled isosceles $\triangle ABC$ is placed in a uniform magnetic field acting along $AB$. If the magnetic force on the arm $BC$ is $\vec{F}$, the force on the arm $AC$ is
A. $-\vec{F}$ ✓ Correct
B. $\vec{F}$
C. $\sqrt{2}\,\vec{F}$
D. $-\sqrt{2}\,\vec{F}$
Solution: For a closed current loop in a uniform field, $\vec F_{AB}+\vec F_{BC}+\vec F_{CA}=0$. Since arm $AB$ is parallel to $\vec B$, $\vec F_{AB}=0$, so $\vec F_{CA}=-\vec F_{BC}=-\vec F$.
Q5 — Force and Torque on Current Carrying Conductor · medium · theory
A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\vec F$, the net force on the remaining three arms of the loop is
A. $3\vec F$
B. $-\vec F$ ✓ Correct
C. $-3\vec F$
D. $\vec F$
Solution: The net force on a closed current loop placed in a uniform magnetic field is always zero, so the sum of forces on the remaining three arms must be $-\vec F$, equal and opposite to the force on the first arm.
Q6 — Force and Torque on Current Carrying Conductor · hard · theory
A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole pieces of a permanent magnet such that $\vec{F}$ is in the plane of the coil. If due to a current $i$ in the triangle a torque $\tau$ acts on it, the side $l$ of the triangle is
A. $\dfrac{2}{\sqrt3}\left(\dfrac{\tau}{Bi}\right)^{1/2}$
B. $\dfrac{2}{\sqrt3}\left(\dfrac{\tau}{Bi}\right)$
C. $2\left(\dfrac{\tau}{\sqrt3\,Bi}\right)^{1/2}$ ✓ Correct
D. $\dfrac{1}{\sqrt3}\dfrac{\tau}{Bi}$
Solution: Torque on the coil is $\tau = iAB\sin90^\circ = iAB$, and the area of an equilateral triangle of side $l$ is $A=\dfrac{\sqrt3}{4}l^2$. So $\tau=\dfrac{\sqrt3}{4}l^2Bi \Rightarrow l=2\left(\dfrac{\tau}{\sqrt3\,Bi}\right)^{1/2}$.
Q7 — Force and Torque on Current Carrying Conductor · easy · theory
Current is flowing in a coil of area $A$ and number of turns $N$, then magnetic moment of the coil, $M$ is equal to
A. $NiA$ ✓ Correct
B. $\dfrac{Ni}{A}$
C. $\dfrac{Ni}{\sqrt{A}}$
D. $N^2Ai$
Solution: The magnetic dipole moment (magnetic moment) of a current-carrying coil with $N$ turns, current $i$, and area $A$ is $M=NiA$.
Q8 — Force and Torque on Current Carrying Conductor · easy · numerical
Two long parallel wires are at a distance of $1\,\text{m}$. Both of them carry $1\,\text{A}$ of current. The force of attraction per unit length between the two wires is
A. $2\times10^{-7}\,\text{N/m}$ ✓ Correct
B. $2\times10^{-8}\,\text{N/m}$
C. $5\times10^{-8}\,\text{N/m}$
D. $10^{-7}\,\text{N/m}$
Solution: $\dfrac{F}{l}=\dfrac{\mu_0}{2\pi}\dfrac{i_1i_2}{r}=\dfrac{4\pi\times10^{-7}}{2\pi}\times\dfrac{1\times1}{1}=2\times10^{-7}\,\text{N/m}$.
Q9 — Force and Torque on Current Carrying Conductor · easy · numerical
A straight wire of length $0.5\,\text{m}$ and carrying a current of $1.2\,\text{A}$ is placed in uniform magnetic field of induction $2\,\text{T}$. The magnetic field is perpendicular to the length of the wire. The force on the wire is
A. $2.4\,\text{N}$
B. $1.2\,\text{N}$ ✓ Correct
C. $3.0\,\text{N}$
D. $2.0\,\text{N}$
Solution: $F=BiL\sin\theta$, with $\theta=90^\circ$ (field perpendicular to wire): $F=2\times1.2\times0.5=1.2\,\text{N}$.
Q10 — Force and Torque on Current Carrying Conductor · easy · theory
A current carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes
A. inclined at $45^\circ$ to the magnetic field
B. inclined at any arbitrary angle to the magnetic field
C. parallel to the magnetic field ✓ Correct
D. perpendicular to the magnetic field
Solution: The coil orients itself so that its magnetic moment $\vec M$ becomes parallel to $\vec B$, which means the plane of the coil becomes parallel to the field, giving zero net torque (stable equilibrium).