JEE Main20265 April 2026Evening ShiftPhysicsElectromagnetic InductionActual
A metal rod of length L rotates about one end at origin with a uniform angular velocity . The magnetic field radially falls off as B(r) = B₀ e^ - r ; being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :
Options
- AB₀ [ 1 ^2 - e^ - L ( 1 ^2 + L ) ]
- BB₀ [ 1 ^2 + e^ - L ( 1 ^2 + L ) ]
- CB₀ [ 4 ^2 - e^ -2 L ( 1 ^2 + 2L ) ]
- DB₀ [ 3 ^2 - e^ -3 L ( 3 ^2 + L ) ]
Correct answer
A. B₀ [ 1 ^2 - e^ - L ( 1 ^2 + L ) ]
Step-by-step solution
The motional emf dE induced in a small element of length dr at a distance r from the origin is given by: dE = B(r) v(r) dr Since the rod is rotating with angular velocity , the velocity of the element is v(r) = r . dE = B₀ e^ - r ( r) dr The total emf induced E is the integral of dE from r = 0 to r = L : E = ₀^L B₀ r e^ - r dr E = B₀ ₀^L r e^ - r dr Using integration by parts: r e^ - r dr = r ( e^ - r - ) - 1 ( e^ - r - ) dr = - r e^ - r - 1 ^2 e^ - r Evaluating the definite integral from 0 to L : ₀^L r e^ - r dr =