Question

A coil 5.00 cm in radius, containing 250 turns, is placed in a uniform magnetic field that varies with time according to:
B = (0.230 T/s)t + (4.00 x 10^-5 T/s⁵)t⁵.  The coil is connected to a 450 ohm resistor, and its plane is perpendicular to the magnetic field.  The resistance of the coil can be neglected.  Find the induced emf in the coil as a function of time.

Answer 1

Induced EMF is given by:

EMF = -N*d(phi)/dt

phi = Magnetic flux = B*A*cos theta

So, EMF = -N*A*cos theta*dB/dt

N = number of turns = 250.0 turns

A = area = pi*0.05^2 = 7.85*10^-3 m^2

B = magnetic field = (0.230 T/s)t + (4.00 x 10^-5 T/s⁵)t⁵

then, dB/dt = (0.230 T/s)*1 + (5*4.00 x 10^-5 T/s⁵)t^5-1

dB/dt = (0.230) + (2.00 x 10^-4)t⁴

since, plane of coil is perpendicular to the magnetic field

So, theta = 0 deg, cos theta = 1

then, EMF = -250*7.85*10^-3*cos 0 deg*[(0.230) + (2.00 x 10^-4)t⁴]

EMF = -0.451375 - 3.925*10^-4*t^4

EMF = -(0.451375 - 3.925*10^-4*t⁴)

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