Home › MHT-CET › Physics › Electromagnetic Induction › A conducting rod of length $\ell$ rotates about …
A conducting rod of length $\ell$ rotates about one end with angular velocity $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of rotation. The induced EMF between the two ends of the rod is:
A$B\omega \ell$
B$\tfrac{1}{2} B\omega^2 \ell$
C$B\omega \ell^2$
D$\tfrac{1}{2} B\omega \ell^2$
Answer & Solution
Correct answer: D. $\tfrac{1}{2} B\omega \ell^2$
Each element $dr$ at distance $r$ from the pivot moves with velocity $\omega r$ and contributes $dE = B(\omega r)\,dr$. Integrating from 0 to $\ell$: $e = \int_0^\ell B\omega r\,dr = \tfrac{1}{2}B\omega \ell^2$.
Related questions
When a rod moves perpendicular to a magnetic field at constant velocity, the FREE charges An LC circuit oscillates with angular frequency ω equal to:Maxwell's correction to Ampère's law introduced:A bar magnet is dropped through a vertical conducting ring. The magnet:A coil of inductance 0.5 H carries a current changing at 4 A s⁻¹. The induced EMF is:A magnetic flux through a coil changes from 4 Wb to 1 Wb in 0.3 s. The average EMF inducedEddy currents are reduced in transformer cores by:A transformer works on the principle of: