A wire loop enclosing as semicircle of radius `R` is located on the boudary of uniform magnetic field `B`. At the moment `t=0`, the loop is set into rotation with a costant angular acceleration `alpha` about an axis `O` coinciding with a line of vector B on the boundary. Find the emf induced in the loop as a function of time. Draw the approximate plot of this function.The arrow in the figure shows the emf direction taken to be positive.

Explanation

`theta=1/2at^2`
`:. t=sqrt((2theta)/alpha)=` time taken to rotate angle` theta`
where `theta=0`, to `pi,2pi` to `3pi, 4pi` to `5pi` etc.
`ox` magnetic field passing through the loop is increasing. Hence, current in the loop is anti clockwise or induced emf is negative. And for `theta=pi` to `2pi, 3pi to 3pi, 5pi to 6pi` etc.
`ox` magnetic field passing through the loop is decresing.Hence, current in the loop is clockwise or emf is positive.
So,
`t_(1) =` time taken to rotate an angle `pi = sqrt((2pi)/(alpha))`
`t_(2) =` time taken to rotate an angle `2pi = sqrt((4pi)/(alpha))`
`" ..." " ..."" ..." " ..."" ..."`
`t_(n) =`time taken to rotate an angle `npi = sqrt((2npi)/(alpha))`
Now, from 0 to `t_(1)` emf is negative
`t_(1)` to `t_(2)` is positive
`t_(2)` to `t_(3)` emf is again negative
and so on.
Now, at time t, angle rotated us `theta = (1)/(2) alpha t^(2)`
Area inside the filed is `S = (piR)^(2) ((theta)/(2pi)) = (1)/(2) R^(2) theta`
or `S = (1)/(4) R^(2) alpha t^(2)`
So, flux passing through the loop, `phi = BS = (1)/(4) BR^(2) alpha t^(2)`
`e = |(d phi)/(dt)| = (1)/(2) BR^(2) alphat`
`e prop t`
i.e. `e-t` graph is a staright line passing through origin. `e-t` equation wih sign can be written as
`e = (-1)^(n) ((1)/(2)BR^(2)alphat)`
Here, `n = 1,2,3`.. is the number of half revolutions that the loop performs at the given moment `t`.
The `e-t` graph is as shown in figure.