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The magnetic field inside a 4.0cm diameter superconducting solenoid varies sinusoidally between 8.0T and 12.0T at a freqency of 10Hz.

a) what is the maximum electric field strength at a point 1.5cm from the solenoid axis?
b) What is the value of B at the instant E reaches its maximum value

1 Answer

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E max = 0.942 V/m is the maximum electrical field amplitude at a position 1.5 cm away from the solenoid axis.

An electric field is what?

The physical force that surrounds ionised objects and exerts force on every other energetic particles with in field, either attracting of repel them, is known as an electric field (or E-field). It can also refer to a collection containing charge particles' physical field.

Briefing

Diameter, D, equals 4 cm = 400 parts per million m

R = d/2 ≈ 0.04/2 = 0.02 metres for the circumference.

The electric potential is provided by Faraday's law;

Frequency; f = 10 Hz

From Faraday’s law, the electric field is given by;

E = -(r/2)(dB/dt)

We can formulate the magnetic field's equation as follows based on the magnetic field's variation:

B(t) = B_c + B_o*sin (2πft)

B(t) = 2sin(2*10t) + 10

Thus;

dB/dt = 40π cos 20πt

When cos 20t = -1, the electric field is at its maximum value.

(dB/dt)_max = -40π

Thus, with respect to the solenoid axis, at r = 1.5 centimeter = 0.015 m, we obtain;

E = -(0.015/2) × -40π

E = 0.942 V/m

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