Verified Solution • 2025–26 Session moderate

QJEEM20Ph2113: JEE Main 2020 Physics - Class 12 JEEM20Ph2113 Solution: A square loop of side 2 , and carrying current I, is kept in XZ …

A square loop of side 2 aa , 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 :

Multiple Choice Options

  • A 2μ0I2a2πb{{2{\mu _0}{I^2}{a^2}} \over {\pi b}}
  • B μ0I2a22πb{{{\mu _0}{I^2}{a^2}} \over {2\pi b}}
  • C μ0I2a32πb2{{{\mu _0}{I^2}{a^3}} \over {2\pi {b^2}}}
  • D 2μ0I2a3πb2{{2{\mu _0}{I^2}{a^3}} \over {\pi {b^2}}}

Step-by-Step Verified Solution

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

We know, τ=MBsin⁡θ\tau = MB\sin \theta

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

B = μ0I2πb{{{\mu _0}I} \over {2\pi b}}

Angle between B and M = 90o

∴\therefore τ=(4a2I)(μ0I2πb)sin⁡90∘\tau = \left( {4{a^2}I} \right)\left( {{{{\mu _0}I} \over {2\pi b}}} \right)\sin 90^\circ

= 2μ0I2a2πb{{2{\mu _0}{I^2}{a^2}} \over {\pi b}}

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