A wire loop enclosing a semicircle of radius `R` islocated on the boundary of a uniform magnetic field `B`. At the moment `t = 0`, the loop is set into rotation with constant angular acceleration `alpha` about an axis `O`. The clockwise emf direction is taken to be positive.

The variation of emf as function of time is

Options

A. (a) `(1)/(2)BR^(2)alpha t`
B. (b) `(3)/(2)BR^(2)alpha t A`
C. ( c) `sqrt(3)BR^(2)alpha t`
D. (d) `(BR^(2)alpha t)/(sqrt(2)`

Answer

A

Explanation

(a)emf is induced in the loop because area inside the magnetic field is continually changing.
From `theta = 0 to pi`, `2pi to 3pi`, `4pi to 5pi`, the loop beings to enter the magnetic field. Thus the magnetic field passing through the loop is increasing.Hence, current in the loop is anticlockwise, and for `theta = pi` to` 2pi`, `3pi` to `4pi`, `5pi` to `6pi`, etc. magnetic field passing through the loop is decreasing. Hence current in the loop is clockwise. Let at any time, angle rotated is `theta`, then `theta = (1)/(2) alphat^(2)`
Area inside magnetic field. `A = (1)/(2)R^(2)theta = (1)/(2)R^(2)((1)/(2) alphat^(2)) = (1)/(4)R^(2)alphat^(2)`
Flux in the loop: `phi = BA = (B)/(4)R^(2)alphat^(2)`
emf: `e = -(dphi)/(dt) = -(B)/(2)R^(2)alphat` `rarr` `e prop t`
Time taken to complete first half circle: `t_(1) = sqrt((2pi)/(alpha))`
When within magnetic field:
`A = (1)/(2)R^(2)(2pi - theta) = piR^(2) - (R^(2)theta)/(2)`
emf `e = -(dphi)/(dt) = (B)/(2)R^(2)alphat`
`e = -(BR^(2))/(2) = alphat` `rarr` e `prop t`
Time taken to complete second half revolution
`t_(2) = sqrt((4pi)/(alpha)) = sqrt((2pi)/(alpha))`
We see that `t_(2) lt t_(1)`
We can write induced emf as `e = (-1)^(n)[(1)/(2)BR^(2)alphat]`
Where `n = 1, 2, 3,......` is the number of half revolution completed by loop Smaller time will be taken to complete the second half revolution as compared to the previous half revolution.