Question

A circular coil of radius 0.11 m contains a single turn and is located in a constant magnetic field of magnitude 0.25 T. The magnetic field has the same direction as the normal to the plane of the coil. The radius increases to 0.31 m in a time of 0.080 s.

a) Determine the magnitude of the emf induced in the coil.
b) The coil has a resistance of 0.71 Ω. Find the magnitude of the induced current.

Answer 1

Average induced EMF is given by:

EMF = -N*d(phi)/dt

phi = Magnetic flux = B*A  

So,

EMF = -N*d(B*A)/dt

Since Area is changing w.r.t time, So

EMF = -N*B*(dA/dt)

Given that

N = 1 turn

B = magnetic field = 0.25 T

dA = Af - Ai = pi*rf^2 - pi*ri^2 = pi*0.31^2 - pi*0.11^2 =

dt = 0.080 sec

Using given values:

EMF = -1*0.25*(pi*0.31^2 - pi*0.11^2)/0.080

EMF = -0.82467 V

Magnitude of EMF will be

|EMF| = 0.825 V

Part B.

Induced current will be

i = EMF/R (from Ohm's law)

i = 0.825/0.71

i = 1.16 Amp

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