Question

Suppose that a 0.375 m radius 500 turn coil produces an average emf of 12000 V when rotated one-fourth of a revolution in 4.27 ms, starting from its plane being perpendicular to the magnetic field. what is the peak emf generated by this coil, in volts?

Answer 1

`\epsilon_0=12000\ V`

Step-by-step explanation:

Given:

no. of turns in the coil,
`n=500`

area of the coil,
`A=\pi* 0.375^2=0.442\ m^2`

average emf induced,
`\epsilon=12000\ V`

angle turned by the coil,
`\psi=(\pi)/(4) \ radians`

time taken to sweep the given angle,
`t=4.27* 10^(-3)\ s`

peak emf,
`\epsilon_0=?`

We have the relation between peak emf and average emf as:

`\epsilon=\epsilon_0. sin (\omega t)` .............................(1)

where:

`\rm \omega= angular\ velocity`

we already know the value:

`\omega t=\psi=90^(\circ)`

From eq. (1) we have:

`\epsilon=\epsilon_0. sin((\pi)/(4))^c`

`\epsilon=\epsilon_0`

`\epsilon_0=12000\ V`

Answer 2

epsilon_{peak}= 18833 V

Step-by-step explanation:

Average emf

`\epsilon_(avg) =(NAB)/(\Delta t)`

⇒
`B=(\epsilon_(avg)\Delta t)/(NA)`

`=(12000*4.27*10^(-3))/(500*\pi 0.375^2)`

solving this we get

B=0.2320 T

No. of revolutions are 1/4 rev

Δθ=
`(1)/(4)2\pi =1.57 rads`

angular velocity ω =
`(\Delta\theta)/(\Delta t)`

=
`(1.57)/(4.27*10^(-3))`

=367.68 rad/sec

The peak emf

`\epsilon_(peak) =NAB\omega`

putting values we get
`\epsilon_(peak)=500*\pi*0.375^2*0.232*367.68`

solving we get

epsilon_{peak}= 18833 V