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 `(+ vec(i) - vec(k))` and decreases at the rate of `10^(-2) Tesla//second`, the magnitude and the sense of the induced emf in the loop, as seen along the magnetic field direction 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

D

Explanation

it is given `vec(B)=((vec(i)-vec(k))/(sqrt(2))) |B| and |(dB)/(dt)| = 10^(-2)T//s`
It is obvious that B is directed perpendicular to y-axis. Components `B_(x) and B_(z)` of the field B are `B_(x)=(B)/(sqrt(2)) and B_(z)=-(B)/(sqrt(2))`.
Then the flux (at instant t) through `ABC=phi_(1) = ((pi a^(2))/(2))B_(z)`

Hence emf `e_(1)` induced in loop ABC is
`e_(1)=-(d phi_(1))/(dt) = - (pi a^(2))/(2) * (dB_(z))/(dt) = + (pi a^(2))/(2 sqrt(2)) ((dB)/(dt))`
This emf will be directed clockwise (looking along negative z axis) `vec(ABC)`.
Similarly the flux (at instant t) throght `ADC =(phi_(2)=)=(pi a^(2))/(2) Bx`.
Hence emf `e_(2)` induced in loop ADC is
`e_(2)=-(d phi_(2))/(dt) = - (pi a^(2))/(2) (dB_(x))/(dt) = - (pi a^(2))/(2 sqrt(2)) ((dB)/(dt))`
this emf will be directed clockwise (looking along positive x axis ) `vec(CDA)`.
It is easy to see that these two emfs along the two semicircles will add up.
Hence induced emf in the closed loop will be
`e=|e_(1)|+|e_(2)|=(pi a^(2))/(2 sqrt(2)) (dB)/(dt)+(pi a^(2))/(2sqrt(2))(dB)/(dt)`
It is given `(dB)/(dt)= 10^(-2)T//s` and `a=0.5m`.
Hence `e=(pi(0.5)^(2))/(sqrt(2))*10^(-2)V=(5pi)/(2sqrt(2)) xx 10^(-3)V=(5pi)/(2 sqrt(2))mV`
This will be in clokwise sence, when looking along B.