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JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 6)

A wire X of length $$50 \mathrm{~cm}$$ carrying a current of $$2 \mathrm{~A}$$ is placed parallel to a long wire $$\mathrm{Y}$$ of length $$5 \mathrm{~m}$$. The wire $$\mathrm{Y}$$ carries a current of $$3 \mathrm{~A}$$. The distance between two wires is $$5 \mathrm{~cm}$$ and currents flow in the same direction. The force acting on the wire $$\mathrm{Y}$$ is

JEE Main 2022 (Online) 29th July Evening Shift Physics - Magnetic Effect of Current Question 74 English

$$1.2 \times 10^{-5} \mathrm{~N}$$ directed towards wire $$\mathrm{X}$$.
$$1.2 \times 10^{-4} \mathrm{~N}$$ directed away from wire $$\mathrm{X}$$.
$$1.2 \times 10^{-4} \mathrm{~N}$$ directed towards wire $$\mathrm{X}$$.
$$2.4 \times 10^{-5} \mathrm{~N}$$ directed towards wire $$\mathrm{X}$$.

Explanation

JEE Main 2022 (Online) 29th July Evening Shift Physics - Magnetic Effect of Current Question 74 English Explanation

Length of wire Y is very high compared to length of wire X. So we can assume length of wire Y is infinite compare to wire X.

Magnetic field due to wire y from 5 cm from wire towards wire X is,

$$B = {{{\mu _0}{I_1}} \over {2\pi d}} = {{{\mu _0} \times 3} \over {2\pi \times 0.05}}$$

And direction of B is outward.

Now force on wire X due to wire Y,

$${F_{YX}} = {I_2}B\,l$$

$$ = 2 \times {{{\mu _0} \times 3} \over {2\pi \times 0.05}} \times 5$$

$$ = 1.2 \times {10^{ - 5}}\,N$$

Current flowing in both the wire in same direction so both wire will attract each other with equal force.

$$\therefore$$ $${F_{YX}} = {F_{XY}}$$

$$\therefore$$ $${F_{XY}} = 1.2 \times {10^{ - 5}}\,N$$, attractive force towards wire X.

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