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JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 15)

A square loop of side 2$$a$$, and carrying current I, is kept in XZ plane with its centre at origin. A long wire carrying the same current I is placed parallel to the z-axis and passing through the point (0, b, 0), (b >> a). The magnitude of the torque on the loop about zaxis is given by :
$${{2{\mu _0}{I^2}{a^2}} \over {\pi b}}$$
$${{{\mu _0}{I^2}{a^2}} \over {2\pi b}}$$
$${{{\mu _0}{I^2}{a^3}} \over {2\pi {b^2}}}$$
$${{2{\mu _0}{I^2}{a^3}} \over {\pi {b^2}}}$$

Explanation

JEE Main 2020 (Online) 5th September Morning Slot Physics - Magnetic Effect of Current Question 129 English Explanation

We know, $$\tau = MB\sin \theta $$

Here M = IA = I(2a)2 = 4a2I

B = $${{{\mu _0}I} \over {2\pi b}}$$

Angle between B and M = 90o

$$ \therefore $$ $$\tau = \left( {4{a^2}I} \right)\left( {{{{\mu _0}I} \over {2\pi b}}} \right)\sin 90^\circ $$

= $${{2{\mu _0}{I^2}{a^2}} \over {\pi b}}$$

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