Question

A plastic circular loop has radius R, and a positive charge q is distributed uniformly around the circumference of the loop. The loop is then rotated around its central axis, perpendicular to the plane of the loop, with angular speed v. If the loop is in a region where there is a uniform magnetic field B S directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.

Answer 1

Answer:

$\tau = \frac{q\omega R^2 B}{2}$

Explanation:

As we know that charge is uniformly distributed on the circumference of the circular loop

so here we have

total charge = q

also we know that loop is moving with uniform angular speed

angular speed = $\omega$

now we know that current in the loop due to motion of the ring is given as

$i = \frac{q}{T} = \frac{q\omega}{2\pi}$

now the magnetic moment of the loop is given as

$M = i A$

$M = (\frac{q\omega}{2\pi})\pi R^2$

$M = \frac{q\omega R^2}{2}$

now we know that the torque on the loop is given as

$\tau = \vec M \times \vec B$

$\tau = MBsin90$

$\tau = \frac{q\omega R^2 B}{2}$