Solution:
The torque is given by \(\tau = MBsin\theta\), where \(M = I A = I(\pi R^2)\) and \(\theta = 60^\circ\). Thus, \(tau = I(\pi R^2)Bsin 60^\circ = \frac{\sqrt{3}\pi R^2 I B}{2}\).
A circular coil of radius \(R\) having current \(I\) is placed in a uniform magnetic field \(B\). If the angle between the area vector of the coil and the magnetic field is \(60^circ\), then the torque on the coil will be:
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