ExamBro
ExamBro
TS EAMCET · Physics · Electromagnetic Induction

A conducting rod of length \(L\) rotates with angular speed \(\omega\) in a uniform magnetic field of induction \(B\) which is perpendicular to its motion. The induced emf developed between the two ends of the rod is

  1. A \(\frac{B L^2 \omega}{4}\)
  2. B \(\frac{B L^2 \omega}{2}\)
  3. C \(B L^2 \omega\)
  4. D \(2 B L^2 \omega\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{B L^2 \omega}{2}\)

Step-by-step Solution

Detailed explanation

Linear velocity of the rod \(v=r \omega=\frac{L \omega}{2}\) \(\therefore \quad\) Induced emf \( e=B v L=B \times \frac{L \omega}{2} L=\frac{1}{2} B \omega L^2 \)
Same subject
Explore more questions on app