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JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2020
- Hard
- MCQ
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 \(z-\)axis and passing through point \((0, b , 0),( b >> a ) .\) The magnitude of torque on the loop about \(z-\)ax is will be
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{2 \mu_{0} I^{2} a^{2} b}{\pi\left(a^{2}+b^{2}\right)}\)
Step-by-step Solution
Detailed explanation
\(F = BI 2 a =\frac{\mu_{0} I }{2 \pi r } I \times 2 a\)
\(F =\frac{\mu_{0} I ^{2} a }{\pi \sqrt{ b ^{2}+ a ^{2}}}\)
\(\tau= F \cos \theta \times 2 a\)
\(=\frac{\mu_{0} I^{2} a}{\pi \sqrt{b^{2}+a^{2}}} \times \frac{b}{\sqrt{b^{2}+a^{2}}} \times 2 a\)
\(\tau=\frac{2 \mu_{0} I^{2} a^{2} b}{\pi\left(a^{2}+b^{2}\right)}\)
If \(b \gg a\) then \(\tau=\frac{2 \mu_{0} I ^{2} a ^{2}}{\pi b }\)
But among the given options \((1)\) is most appropriate
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