Answer:
epsilon_{peak}= 18833 V
Explanation:
Average emf
[tex]\epsilon_{avg} =\frac{NAB}{\Delta t}[/tex]
⇒ [tex]B=\frac{\epsilon_{avg}\Delta t}{NA}[/tex]
[tex]=\frac{12000\times4.27\times10^{-3}}{500\times\pi 0.375^2}[/tex]
solving this we get
B=0.2320 T
No. of revolutions are 1/4 rev
Δθ= [tex]\frac{1}{4}2\pi =1.57 rads[/tex]
angular velocity ω = [tex]\frac{\Delta\theta}{\Delta t}[/tex]
= [tex]\frac{1.57}{4.27\times10^{-3}}[/tex]
=367.68 rad/sec
The peak emf
[tex]\epsilon_{peak} =NAB\omega[/tex]
putting values we get [tex]\epsilon_{peak}=500\times\pi\times0.375^2\times0.232\times367.68[/tex]
solving we get
epsilon_{peak}= 18833 V
Answer:
[tex]\epsilon_0=12000\ V[/tex]
Explanation:
Given:
no. of turns in the coil, [tex]n=500[/tex]
area of the coil, [tex]A=\pi\times 0.375^2=0.442\ m^2[/tex]
average emf induced, [tex]\epsilon=12000\ V[/tex]
angle turned by the coil, [tex]\psi=\frac{\pi}{4} \ radians[/tex]
time taken to sweep the given angle, [tex]t=4.27\times 10^{-3}\ s[/tex]
peak emf, [tex]\epsilon_0=? [/tex]
We have the relation between peak emf and average emf as:
[tex]\epsilon=\epsilon_0. sin (\omega t)[/tex] .............................(1)
where:
[tex]\rm \omega= angular\ velocity[/tex]
we already know the value:
[tex]\omega t=\psi=90^{\circ}[/tex]
From eq. (1) we have:
[tex]\epsilon=\epsilon_0. sin(\frac{\pi}{4})^c [/tex]
[tex]\epsilon=\epsilon_0[/tex]
[tex]\epsilon_0=12000\ V[/tex]
