Angular acceleration of conducting ring in magnetic field

Question

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Attempt:

$\tau = \mu \times B$
$\tau = I\alpha = MR^2 \alpha$

$\mu = iA= i\pi R^2$

$\implies \alpha = \dfrac{\pi i B}{M} = 20 \pi\text{rad/s^2}$

But answer given is $a) 40 \pi$

Answer 1

Your mistake is in the expression for moment of inertia.

You have used the moment of inertia for rotation about the horizontal axis perpendicular to the plane of the ring (going into the page). Call this direction $x$. But the ring rotates about the vertical axis $y$ in its own plane. Since the disk is a planar object its moments of inertia are related by the Perpendicular Axes Theorem : $I_x=I_y+I_z$ where $z$ is the horizontal axis in the plane of the page (and the disk).

Perhaps you are confused between the rotation axis and the direction of the magnetic moment $\mu$ which is perpendicular to the plane (direction $x$) and looks like a rotation axis.