Question

I just need part B answered, thanks!

In a model AC generator, a 504 turn rectangular coil 8.0 cm by 26 cm rotates at 120 rev/min in a uniform magnetic field of 0.61 T. What is the maximum emf induced in the coil? v What is the instantaneous value of the emf in the coil at t = (pi/32) s? Assume that the emf is zero at t = 0. What is the smallest value of t for which the emf will have its maximum value? s

Answer 1

Ans

$Emf= -N\frac{\partial \phi }{\partial t}$

Where N is no of turn.

$\phi = B.A =BA cos\theta$

( $\theta$ is the angle between magnetic field B and area vector A )

is the magnetic flux.

t is time.

Also using equation of circular motion

$\theta = \theta ^{_{_{_{0}}}} + \omega t$

therefore

$Emf =-N\frac{\partial BAcos(\omega t +\theta _{0})}{\partial t}$

            = $Emf =NBA\omega sin(\omega t+\theta _{0})$

At t=0, Emf=0;

i.e    $sin(\theta _{0}) = 0$

or   $\theta _{0} = 0$

therefore $Emf =NBA\omega sin(\omega t)$

                            ( $\omega = 2\pi *120/60 =4\pi$ in rad per sec)

                            = 504*0.61*8*26*10^-4*4*3.14*sin(4* $\pi$ ²/32)

                              =75.78 volt