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MHTCET 23 Apr 2025 Shift 1 Paper

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Question : 14 of 150

Marks: +1, -0

A square coil ABCD of side ' L ' is carrying a current in clockwise direction. A straight conductor carrying current

\mathrm{I}_{2}

(upward direction) is kept parallel to side AB at a distance

\frac{\mathrm{L}}{3}

in the plane of

A B C D

. The net force on the coil ABCD is (

\mu_{0}=

magnetic permeability)

$\frac{\mu_{0} \mathrm{I}_{1} \mathrm{I}_{2}}{3 \pi}$
$\frac{3 \mu_{0} \mathrm{I}_{1} \mathrm{I}_{2}}{4 \pi}$
$\frac{3 \mu_{0} \mathrm{I}_{1} \mathrm{I}_{2}}{8 \pi}$
$\frac{9 \mu_{0} \mathrm{I}_{1} \mathrm{I}_{2}}{8 \pi}$

Solution:

Concept:

Force between a long straight current-carrying wire and a parallel current-carrying side is

F = \frac{\mu_0 I_1 I_2}{2\pi r}L

.

Parallel currents attract, while antiparallel currents repel.

Explanation:

The straight conductor is at distance

\frac{L}{3}

from side AB. For clockwise current in the square coil, side AB and side CD carry currents in opposite directions.

Force on side AB (attractive):

F_{AB}=\frac{\mu_0}{2\pi}\frac{I_1I_2}{L/3}L=\frac{3\mu_0I_1I_2}{2\pi}

Distance of side CD from the conductor is

\frac{L}{3}+L=\frac{4L}{3}

Force on side CD (repulsive):

F_{CD}=\frac{\mu_0}{2\pi}\frac{I_1I_2}{4L/3}L=\frac{3\mu_0I_1I_2}{8\pi}

Forces on sides BC and AD are equal and opposite, so they cancel each other.

Net force on the coil:

F_{net}=F_{AB}-F_{CD}=\frac{3\mu_0I_1I_2}{2\pi}-\frac{3\mu_0I_1I_2}{8\pi}

\therefore F_{net}=\frac{9\mu_0I_1I_2}{8\pi}

Answer:

\frac{9\mu_0I_1I_2}{8\pi}

Option D

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