a metal rod of length 'l' is rotated about one end with a constant angular velocity 'w' in a uniform transverse magnetic field B . change in magnetic flux is given by the formula ? 1. d'fi =B*1/2*l square*d teeta...2.d 'fi'=B*1/2*l*d teeta ...
plz explain the derivation....
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A₀ = area covered by the rod = π L² where L = length of the rod. This is the area swept in one rotation.
B = magnetic field.
number of complete rotations per second = f = ω/2π = number of turns
Area swept by the rod in time dt = dA = A₀ * f * dt
The angle between the plane of rotation of the rod and the magnetic field is 90 deg. The angle between the normal to the plane and the magnetic field is 0.
Total flux change in a second = dФ
= B * dA * dt * Cos0 = B * π L² * ω/2π * dt
= B (dθ/dt) L² /2 * dt
here ω = dθ / dt where θ = angle of rotation of the rod about its fixed end.
so dΦ = B L² dθ /2
B = magnetic field.
number of complete rotations per second = f = ω/2π = number of turns
Area swept by the rod in time dt = dA = A₀ * f * dt
The angle between the plane of rotation of the rod and the magnetic field is 90 deg. The angle between the normal to the plane and the magnetic field is 0.
Total flux change in a second = dФ
= B * dA * dt * Cos0 = B * π L² * ω/2π * dt
= B (dθ/dt) L² /2 * dt
here ω = dθ / dt where θ = angle of rotation of the rod about its fixed end.
so dΦ = B L² dθ /2
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