A metallic rod of 1 m length is rotated with a frequency of 50 revolution per second, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field of 1 T parallel to the axis is present everywhere. The emf between the centre and the metallic ring is:
Question
A metallic rod of 1 m length is rotated with a frequency of 50 revolution per second, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field of 1 T parallel to the axis is present everywhere. The emf between the centre and the metallic ring is:
Solution
The problem is asking for the emf generated in a rotating rod in a magnetic field. This is a classic problem in electromagnetism and can be solved using Faraday's law of electromagnetic induction.
Here are the steps to solve the problem:
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First, we need to understand that the emf (ε) generated in a rotating rod of length (l) in a magnetic field (B) is given by the formula: ε = 1/2 * B * l^2 * ω, where ω is the angular velocity.
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In this problem, the length of the rod (l) is 1 m, the magnetic field (B) is 1 T, and the frequency of rotation (f) is 50 revolutions per second.
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The angular velocity (ω) is related to the frequency of rotation by the formula: ω = 2πf. So, ω = 2π * 50 = 100π rad/s.
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Substituting these values into the formula for emf, we get: ε = 1/2 * 1 T * (1 m)^2 * 100π rad/s = 50π V.
So, the emf between the centre and the metallic ring is 50π volts.
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