Question

A copper rod of length ℓ rotates about its end with angular velocity ω in a uniform magnetic field B. The emf developed between the ends of the rod if the field is normal to the plane of rotation is:

• A. \( B\omega \ell^2 \)
• B. \( \dfrac{1}{2}B\omega \ell^2 \)
• C. \( 2B\omega \ell^2 \)
• D. (1)/(4)Bω ℓ²

Hint:

For rotating conductors, integrate motional emf over length.

Answer 1

The Correct Option is B

Step 1: Motional emf in a rotating rod: E = int₀^ℓ Bω rdr
Step 2: E = Bω [(r²)/(2)]₀^ℓ = (1)/(2)Bω ℓ²