A circular coil of radius 8.0 cm and 20 turns rotates about its vertical diameter with an angular speed of 50 `s^(-1)` in a uniform horizontal magnetic field of magnitude `3 xx 10^(-2) T`. Obtain the maximum and average e.m.f. induced in the coil . If the coil forms a closed loop of resistance `10 Omega`, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating.

Explanation

Given, radius of the circular coil, `r = 8 cm=8 xx 10^(-2) m`
Number of turns in coil, N = 20
Angular speed of rotation, `omega = "50 rad"s^(-1)`
Magnetic field, `B = 3.0 xx 10^(-2) T`
Area enclosed by the each turn of the coil, `A=pir^(2)`
If `theta=omegat` is the angle between the normal to the plane of the loop and magnetic field, then flux through each turn of the loop,
`phi="NBA" cos theta = N pir^(2) B cos omegat`
The induced emf in circular coil on account of its rotation in magnetic field is given by
`epsilon=-(dphi)/(dt)=N omega pir^(2) B sin omegat` ...(i)
Maximum induced emf in the coil will be
`epsilon_("max") = N omega pir^(2) B`,
`= 20 xx 50 xx 7 xx 64 xx 10^(-4) xx 3.0 xx 10^(-2) = 0.603 V`
From (i), we can see that the induced emf is a sinusoidal function. The average value of sinusoidal function over a complete cycle is zero.
Maximum induced current will be
`I_("max") = (epsilon_("max"))/(R)=(0.603 V)/(10 Omega)=0.0603 A`
Average power loss due to Joule heating in the resistance R will be
`P=(epsilon_("max")I_("max"))/(2)=(0.603 xx 0.0603)/(2)=0.018 W`
The induced current produces a torque opposing the rotation of the coil. An external agent must supply torque (and do work) to cancel this torque in order to keep the coil rotating uniformly. Thus the source of the power dissipated as heat in the coil is the external agent (a rotor).