Question

A circular coil (830 turns, radius = 0.087 m) is rotating in a uniform magnetic field. At t = 0 s, the normal to the coil is perpendicular to the magnetic field. At t = 0.021 s, the normal makes an angle of 45o with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.075 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

Answer 1

According to lenz law which states that it always opposes the cause producing it hence

V =-d(B×A)/dt, B is magnetic field and A is area of coil

B is constant hence V= -B.dA/dt( dot product, B and A are vectors with normal vector to the plane of the coil times the area of the coil making the vector A)

As above is dot product of two vectors so angle will be equal to cosθ between them hence in scalar terms

V=- |B| d( |A| cos(θ))/dt

Let coil have N turns so A= N  π r²

N=830, r=.087 hence A = 830×3.14×.087×.087

=19.72m²

Now we know θ=wt where w is angular frequency of coil and t is time.

since it takes 0.021 s to turn π/4 radians(1/8 revolutions), T/8=. 021 ,T=.168 sec

w=2π/T=37.3 rad/sec

So d(|A| cosθ )/dt = |A| d(cos(wt))/dt = -|A| w sin(wt)

Substituting value of Area

=- Nπ a²w sin(wt)

So V = -|B| d|A| cos(wt)/dt = |B| w N π a²sin(wt)

Now you are given V = 0.075V at T= 0.168s so solve for |B|

|B| = V/(w Nπ a²sin(wt))= 0.075V/(37.3×19.72×sin 45 degree)

=144 micro tesla

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