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

Answer 1

Answer: (A)

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