ABCDA is a closed loop of conducting wire consisting of two semicircular sections, the part ABCD lying in the XY plane and the part CDA lying in the YZ plane, both the parts having the centre at origin O (see the diagram). This loop is placed in a uniform field `vec(B)` which varies with time.
Radius of the semicircular section is 0.5 m

If `vec(B)` be directed along the vector `(vec(i) - vec(j))` and decreases at the rate of `10^(-2) Tesla/second`. The magnitude and sense of the induced emf in the loop as seen along the positive x-axis will be

Options

A. zero
B. `(5pi)/(4sqrt(2))` mV in the counterclockwise sense
C. `(5 pi)/(4sqrt(2))` mV in the clockwise sense
D. `(5pi)/(2sqrt(2))`mV in the clockwise sense.

Answer

C

Explanation

It is given that `vec(B)=(hat(i)=hat(j))/(sqrt(2)) |B|`
This direction is perpendicular to z-axis.
Hence there will be no flux in the semicular section ABC. The induced emf in the other semiculaer section ADC will be
`e=-(d phi)/(dt)=(d)/(dt)[vec(B)*vec(A)]`
`=-(d)/(dt)[B(pia^(2))/(dt)*cos(pi)/(4)]`
Since the field makes an angle of `pi//4` with the yz plane
`e=(pi a^(2))/(2sqrt(2))*(dB)/(dt)=(pia^(2))/(2sqrt(2))*10^(-2)V`
Putting `a=0.5m e, (pi(0.5)^(2))/(2sqrt(2))10^(-2)V=(5 pi)/(4sqrt(2))mV`
This will be in the clockwise sence looking along positive x-axis ie,`vec(CDA)`.