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
- A \(\frac{B L^2 \omega}{4}\)
- B \(\frac{B L^2 \omega}{2}\)
- C \(B L^2 \omega\)
- D \(2 B L^2 \omega\)
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 \( e=B v L=B \times \frac{L \omega}{2} L=\frac{1}{2} B \omega L^2 \)
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