Question

A circular coil of radius 8 cm and 20 turns rotates about its vertical diameter with

an angular speed of 50 s --1in a uniform horizontal magnetic field of magnitude3×102 T. Find the maximum and average value of the emf induced in the coil.

Answer 1

Here the magnetic flux through the circular coil at any instant = BAcos$\theta$
where $\theta =\omega t$
Therefore induced emf is $E=-\frac{d\phi }{dt}=-\frac{dBA\cos \left(\omega t\right)}{dt}=BA\omega \sin \left(\omega t\right)$
for N turns of coil ,​$E=NBA\omega \sin \left(\omega t\right)$
$E=E_o\sin \left(\omega t\right);Where{E}_{o}isthemaximuninducedemf.$
For one time period average emf is ​
$E_{avrage}=\frac{{\int }_0^{T}Edt}{{\int }_0^{T}dt}=\frac{{\int }_0^T{E}_o\cos \left(\omega t\right)dt}{{\int }_0^{T}dt}=0.$
Here $A={\mathrm{\pi r}}^2=\mathrm{\pi }\times 0.08\times 0.08=0.020096{\mathrm{m}}^2$
Maximum induced emf ​$E_o=NBA\omega =75360V$$E_o=NBA\omega =20\times 3.0\times {10}^2\times 0.020096\times 50=6034.28V$