The wire loop, shown in , is made by taking a flat rectangular loop of sides `10 cm` and `20 cm` bending at the long sides a their midpoints to produce two mutually perpendicular square parts. The loop is laced in an oscillating magnetic field `B = B_(0) sin2pivt`, with `B_(0) = 1.2 xx 10^(-3) T` and `v = 60 Hz`. The magnetic field is induced at angle `theta` with `x-z` plane.
(a) Express the emf around the loop as a function of time and the angle `theta`.
(b) For what angle `theta` does the induced emf have the largest amplitude ?

Answer

(a) `(1.44pi xx 10^(-3)) cos (120pit)(cos theta + sin theta)V`; (b) `(pi)/(4)`

Explanation

(a)
Now flux through loop `(1)`, `phi_(1) = Ba^(2) cos theta`
Flux through loop `(2)`, `phi_(2) = Ba^(2) sin theta`
emf induced in loop `(1)`,
`E_(1) = (d phi_(1))/(dt) = (2piv)B_(0) cos (2pivt)a^(2) cos theta`
emf induced in loop `(2)`,
`E_(2) = (d phi_(2))/(dt) = (2piv)B_(0) cos (2pivt)a^(2) sin theta`
`:.` net induced emf, `E_(net) = E_(1) + E_(2)`
`= (2piv)B_(0) cos(2pivt)a^(2) (cos theta + sin theta)`
`= (120pi)(1.2 xx 10^(-3)) cos (120pit)(0.01) xx (cos theta + sin theta)`
`= (1.44pi xx 10^(-3)) cos(120pit)(cos theta + sin theta)V`
(b) emf will be maximum when `cos theta + sin theta` has maximum value.
Let `y = cos theta + sin theta`
`(dy)/(d theta) = - sin theta + cos theta = 0` `rarr` `theta = (pi)/(4)`
`(d^(2)y)/(dtheta^(2)) = -cos theta - sintheta = negative` at `theta = (pi)/(4)`, hence at `theta = (pi)/(4)`,
induced emf is maximum.