JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A current carrying circular loop of radius \(2\) cm with unit normal \(\hat{n} = \dfrac{\hat{k}+\hat{i}}{\sqrt{2}}\) is placed in a magnetic field, \(\vec{B} = B_0(3\hat{i}+2\hat{k})\). If \(B_0 = 4\times 10^{-3}\) T and current \(I=100\sqrt{2}\) A, the torque experienced by the loop is ________ Wb·A. (\(\pi=3.14\))
- A \(16\times 10^{-5}\,\hat{k}\)
- B \(5024\times 10^{-7}\,\hat{k}\)
- C \(5024\times 10^{-7}\,\hat{i}\)
- D \(5024\times 10^{-7}\,\hat{j}\)
Answer & Solution
Correct Answer
(D) \(5024\times 10^{-7}\,\hat{j}\)
Step-by-step Solution
Detailed explanation
The magnetic dipole moment of the current carrying loop is given by \(\vec{M} = I A \hat{n}\). Given radius \(r = 2\) cm \(= 2 \times 10^{-2}\) m, the area of the loop is: \(A = \pi r^2 = \pi (2 \times 10^{-2})^2 = 4\pi \times 10^{-4}\) m\(^2\) Substituting the given values…
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