Question

. A coil is wrapped with 400 turns of wire on the perimeter of a circular frame of radius 8.5cm. Each turn has the same area, equal to that of the frame. A uniform magnetic field is turned on perpendicular to the plane of the coil. This field changes at a constant rate from 10.0 to 80.0mT in a time of 20.0ms. What is the magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50.0mT

Answer 1

The emf in the coil when the magnetic field is 50.0 mT is 31.78 V.

What is emf?

According to Faraday's law of electromagnetism, the emf of a coil  

is the negative of the rate of change of magnetic flux with respect to time.

The emf of a coil is given by the formula,

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

where E is the emf, dΦ is the change in flux and dt is the change in time.

Magnetic flux: The scalar product of the magnetic field B and the total area vector A perpendicular to it is called the magnetic flux. The magnetic flux for a circular coil with n number of turns is given by,

$\phi=n\pi r^2B$

where r is the radius of the coil.

Substitute $\phi=n\pi r^2B$ in the above equation and solve it.

$E=-\frac{d(n\pi r^2 B)}{dt}\\E=-n\pi r^2 \frac{dB}{dt}$

Where dB/dt is the rate of change in the magnetic field with respect to time.

Note: 1mT/ms= 1 T/s, 1 cm= 0.01 m and 1mT=0.001 T

Given the magnetic field changes from 10.0mT to 80.0mT, the change in magnetic field dB = 80-10=70mT. The time interval dt= 20.0ms. Therefore,

dB/dt = 70/ 20 =3.5mT/(ms)

dB/dt = 3.5 T/s

Substitute dB/dt= 3.5T/s,  n=400, r=8.5 cm or r=0.085 m in the formula of E and solve it. Since the value of magnitude is positive, ignore the negative sign.

E=400*π*(0.085)^2*(3.5)

E=31.78 V

Since the rate of change of magnetic field is constant, so the magnitude of the emf of the coil will remain the same at all instants. Hence E=31.75 V when the magnitude of the magnetic field is 50.0 T

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