A Wheel with 12 metal spokes each 0.6m long is rotated with a speed of 180 r.p.m in a plane normal to earth’s magnetic field at a place. If the magnitude of the field is 0.6 G, what is the magnitude of induced e.m.f. between the axle and rime of the wheel?
Question
A Wheel with 12 metal spokes each 0.6m long is rotated with a speed of 180 r.p.m in a plane normal to earth’s magnetic field at a place. If the magnitude of the field is 0.6 G, what is the magnitude of induced e.m.f. between the axle and rime of the wheel?
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 (E) in a rotating wheel is given by:
E = 1/2 * B * N * ω * r^2
where:
- B is the magnetic field strength,
- N is the number of turns (or in this case, spokes),
- ω is the angular velocity, and
- r is the radius of the wheel.
Given in the problem:
- B = 0.6 G = 0.6 * 10^-4 T (since 1 G = 10^-4 T),
- N = 12,
- ω = 180 rpm = 180 * 2π / 60 rad/s (since 1 rpm = 2π / 60 rad/s), and
- r = 0.6 m.
Substituting these values into the formula, we get:
E = 1/2 * 0.6 * 10^-4 T * 12 * 180 * 2π / 60 rad/s * (0.6 m)^2
After calculating, we find that the magnitude of the induced emf between the axle and rim of the wheel is approximately 0.0389 V or 38.9 mV.
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