A standing wave `y = 2A sin kx cos omegat` is set up in the wire `AB` fixed at both ends by two vertical walls (see Fig. 3.197). The region between the walls contains a constant magnetic field `B`. The wire is found to vibrate in the third harmonic. Now answer the following question:

In the above qestion, the time when the emf becomes maximum for the time is

Options

A. (a) `(2pi)/(omega)`
B. (b) `(pi)/(omega)`
C. ( c) `(pi)/(2omega)`
D. (d) `(pi)/(4omega)`

Answer

C

Explanation

( c) `y = 2A sin kx cos omegat`
`v = (dy)/(dt) = -2A sin kx omega sin omegat`
`v_(max) = -2A omega sin kx, k = (3pi)/(AB)`
`e = int_(0)^(l = AB) bv_(max)dx = -2aomegaBint_(0)^(AB) sin kx dx = + (2omegaAB)/(k)[cos(3pi)/(AB)AB - cos theta] = (-4(AB)omega)/(k)`
`omegat = (pi)/(2)`
`t = (pi)/(2omega)`
For second harmonic `k = (2pi)/(AB)`