5. A copper disc of radius 0.10 m rotates 1200 times min-¹ with its plane perpendicular to the magnetic field. If the induced e.m.f. between the centre and the edge of the disc is 28 mV, calculate the intensity of the field.
Question
- A copper disc of radius 0.10 m rotates 1200 times min-¹ with its plane perpendicular to the magnetic field. If the induced e.m.f. between the centre and the edge of the disc is 28 mV, calculate the intensity of the field.
Solution
To solve this problem, we need to use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.
The formula for the induced emf in a rotating disc is given by:
E = ½ * B * ω * r²
where: E is the induced emf, B is the magnetic field strength, ω is the angular velocity, and r is the radius of the disc.
We are given: E = 28 mV = 28 * 10^-3 V, ω = 1200 rotations per minute = 1200 * 2π / 60 rad/s (since 1 rotation = 2π rad), and r = 0.10 m.
We need to find B.
Substituting the given values into the formula, we get:
28 * 10^-3 = ½ * B * (1200 * 2π / 60) * (0.10)²
Solving for B, we get:
B = (28 * 10^-3) / (½ * (1200 * 2π / 60) * (0.10)²)
After calculating the above expression, we get the intensity of the magnetic field B.
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