# Angular acceleration of conducting ring in magnetic field

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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 $a) 40 \pi$

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.