Q. When a rod of length $l$ is rotated with angular velocity of $\omega$ in a perpendicular field of induction $B$, about one end, the emf across its ends is

A) $Bl^{2}\omega$
B) $\frac{Bl^{2}\omega}{2}$
C) $Bl\omega$
D) $\frac{Bl\omega}{2}$
Explanation

(b) A conducting rod of length$\ l$ whose one end is fixed, is rotated about the axis passing through its fixed end and perpendicular to its length with constant angular velocity$\ \omega$. Magnetic field $(B)$ is perpendicular to the plane of the paper.
Emf induced across the ends of the rod is $e = BAn$
[image:6_image72.emf]
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = B\pi l^{2}n$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{Bl^{2}\pi}{T}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}Bl^{2}\omega$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.