A simple pendulum with a bob of mass m and a conducting wire of length L, swings under gravity with an angular amplitude `2theta`. If the horizontal component of the earth's magnetic field perpendicular to the plane of motion of the pendulum is B, then the maximum emf induced across the pendulum is

Options

A. `2BLsin(theta/2)(gL)^(1//2)`
B. `BLsin(theta/2)(gL)`
C. `BLsin(theta/2)(gL)^(3//2)`
D. `BLsin(theta/2)(gL)^(2)`

Answer

A

Explanation

`h=L(1-costheta)`
`v^(2)=2gh`
`rArrv^(2)=2gL(1-costheta)`
`rArrv^(2)=2gL(2"sin"^(2)theta/2)`
`thereforev=2sqrt(gL)sintheta//2`
emf e = BvL = `2B(sqrt(gL)sintheta//2)L`
= `2BLsin(theta//2)(gL)^(1/2)`.