A square loop of side 2a, and… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Electromagnetism

2020mcqmedium

A square loop of side $2a,$ 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 $z$ -axis is given by.

Official previous-year question

Held on 5 Sept 2020 · Verified 6 Jul 2026.

Options

A
$\frac{{\mu }_{0}{I}^{2}{a}^{2}}{2\pi b}$
B
$\frac{{\mu }_{0}{I}^{2}{a}^{3}}{2\pi {b}^{2}}$
C
$\frac{2{\mu }_{0}Ia22}{\pi b}$
D
$\frac{2{\mu }_{0}Ia32}{\pi {b}^{2}}$

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Solution

JEE Main 2020 Physics, Magnetic Effects of Current — solution figure 1

Net force acting on wire $\mathrm{AB}$ & $\mathrm{CD}$ are zero

JEE Main 2020 Physics, Magnetic Effects of Current — solution figure 2

${F}_{1}=ILB=I\times (2a)\times (\frac{{\mu }_{0}I}{2\pi \sqrt{{a}^{2}+{b}^{2}}})=\frac{{\mu }_{0}{I}^{2}a}{\pi \sqrt{{a}^{2}+{b}^{2}}}$

$\tau ={F}_{1}cos\theta \times 2a=\frac{{\mu }_{0}{I}^{2}a}{\pi \sqrt{{a}^{2}+{b}^{2}}}\times \frac{b}{\sqrt{{a}^{2}+{b}^{2}}}\times 2a$

$\tau =\frac{2{\mu }_{0}{I}^{2}{a}^{2}b}{\pi ({a}^{2}+{b}^{2})}\approx \frac{2{\mu }_{0}{I}^{2}{a}^{2}}{\pi b},(a\ll b)$

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Solution

JEE Main 2020 Physics, Magnetic Effects of Current — solution figure 1

Net force acting on wire $\mathrm{AB}$ & $\mathrm{CD}$ are zero

JEE Main 2020 Physics, Magnetic Effects of Current — solution figure 2

${F}_{1}=ILB=I\times (2a)\times (\frac{{\mu }_{0}I}{2\pi \sqrt{{a}^{2}+{b}^{2}}})=\frac{{\mu }_{0}{I}^{2}a}{\pi \sqrt{{a}^{2}+{b}^{2}}}$

$\tau ={F}_{1}cos\theta \times 2a=\frac{{\mu }_{0}{I}^{2}a}{\pi \sqrt{{a}^{2}+{b}^{2}}}\times \frac{b}{\sqrt{{a}^{2}+{b}^{2}}}\times 2a$

$\tau =\frac{2{\mu }_{0}{I}^{2}{a}^{2}b}{\pi ({a}^{2}+{b}^{2})}\approx \frac{2{\mu }_{0}{I}^{2}{a}^{2}}{\pi b},(a\ll b)$

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Solution

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Mechanics

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