JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A circular coil of radius \(2\) cm and \(125\) turns carries a current of \(1\) A. The coil is placed in a uniform magnetic field of magnitude \(0.4\) T. The axis of the coil makes an angle of \(30°\) with the direction of the magnetic field. The torque acting on the coil is \(\alpha \times 10^{-4}\) N.m. The value of \(\alpha\) is ______.
(\(\pi=3.14\))
- A 3.14
- B 314
- C 0.0314
- D 0.003
Answer & Solution
Correct Answer
(B) 314
Step-by-step Solution
Detailed explanation
The magnetic moment of the coil is given by \(M = NIA\). The torque acting on the coil is \(\tau = MB \sin\theta = NIAB \sin\theta\). Substituting the given values: \(\tau = 125 \times 1 \times (\pi \times (0.02)^2) \times 0.4 \times \sin 30^\circ\)…
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