Question

A thin wire of the length L is made of an insulaing material . The wire is bent to from a circular loop, and a positive charge q is distributed uniformly around the circumference of the loop . The loop is then set into rotation with angular speed $\omega$ around an axis through its centre . If the loop is in the region with angular speed $\omega$ around an axis through its centre . If the loop is in the region where there is a uniform magnetic field B directed parallel to the plane of the loop , calculate the magnitude of the magnetic torque on the loop.

• A. $\frac{q\omega L^2B}{8\pi^2}$
• B. $\frac{q\omega L^2B}{4\pi^2}$
• C. $\frac{q\omega L^2B}{2\pi^2}$
• D. $\frac{q\omega L^2B}{\pi^2}$

Answer 1

Answer: (A)

$\tau=mB\,\sin\,\theta =I A B \,\sin\,\theta =L AB (as\theta =90^{\circ})$
Further ,as $I=\frac{q}{T}=\frac{q}{2\pi/\omega}=\frac{q \omega}{2\pi}$
$A=\pi\,r^2=\pi \left(\frac{L}{2\pi}\right)^2=\frac{L^2}{4 \pi}$
$\tau =\left(\frac{q \omega}{2 \pi}\right)\left(\frac{L^2}{4 \pi} \right)B=\frac{q\omega\,L^2B}{8\pi^2}$