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Angular acceleration of conducting ring in magnetic field

1 vote



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

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

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

But answer given is $a) 40 \pi$

asked Aug 3, 2018 in Physics Problems by Reststack (422 points)

1 Answer

1 vote
Best answer

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.

answered Aug 4, 2018 by sammy gerbil (28,896 points)
edited Aug 5, 2018 by sammy gerbil