Question

40-turn circular coil (radius = 4.0 cm, total resistance = 0.20 ) is placed in a uniform magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies with time as given by B = 50 sin(10 t) mT where t is measured in s. What is the magnitude of the induced current in the coil at 0.10 s?

Answer 1

Answer:

$EMF = 316 Volts$

Explanation:

As we know that magnetic flux through the coil is given by

$\phi = NBA$

now by Faraday law we know that rate of change in magnetic flux is equal to the EMF induced in the coil

so we have

$-\frac{d\phi}{dt} = EMF$

$EMF = -NA\frac{dB}{dt}$

now we have

$B = 50 sin(10\pi t)$

$A = \pi r^2 = \pi(0.04)^2 = 5.03 \times 10^{-3} m^2$

now we have

$EMF = -(40)(5.03 \times 10^{-3})\frac{d(50 sin(10\pi t))}{dt}$

$EMF = -0.2012(50 \times 10\pi)cos(10\pi t)$

$EMF = -316 cos(10\pi t)$

now at t = 0.10 s

$EMF = 316 Volts$