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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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
- A \(\frac{2 \mu_{0} I^{2} a^{2} b}{\pi\left(a^{2}+b^{2}\right)}\)
- B \(\frac{\mu_{0} I^{2} a^{2} b}{2 \pi\left(a^{2}+b^{2}\right)}\)
- C \(\frac{\mu_{0} I^{2} a^{2}}{2 \pi b}\)
- D \(\frac{2 \mu_{0} I^{2} a^{2}}{\pi b}\)
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
\(F = BI 2 a =\frac{\mu_{0} I }{2 \pi r } I \times 2 a\) \(F =\frac{\mu_{0} I ^{2} a }{\pi \sqrt{ b ^{2}+ a ^{2}}}\) \(\tau= F \cos \theta \times 2 a\) \(=\frac{\mu_{0} I^{2} a}{\pi \sqrt{b^{2}+a^{2}}} \times \frac{b}{\sqrt{b^{2}+a^{2}}} \times 2 a\)…
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