Question

A. Draw a labeled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.


B. A circular coil of cross-sectional area and 20 turns is rotated about the vertical diameter with angular speed of in uniform magnetic field of magnitude Calculate the maximum value of the current in the coil.

Answer 1

A. a labeled diagram of AC generator. The expression for the instantaneous value of the emf induced in the coil.

• The magnetic flux passing through the coil is given by,
• φ  = N (B . A)
• φ  = Bcos θ. AN
• and θ = ωt
• [ω is angular velocity of the coil]
• So, φ  = N A B cos ωt  
• where, t = perpendicular vector to the plane of the coil  
• θ = angle with the direction of magnetic field B.
• If e is the instantaneous induced emf produced in the coil, then
• e = -dφ/dt = -d/dt (NAB cos ωt ) = -NωAB(sin ωt )
• Maximum value or peak value of e is attained when  sin ωt = ± 1
• ∴ emax = Nω AB  
• So,   e =   emax sin ωt

B. A circular coil of cross-section area 200 centimetre square  and 20 turns is rotated about the vertical diameter with an angular speed of 50 rad/s in a uniform magnetic field of magnitude 3*10-2T(10 to the power -2). The maximum value of current in the coil is  0.6/R ampere.

• Given, N = 50 turns
• A = 200 cm^2
• B = 3 × 10^-2 T
• w = 50 rad/s
• emf = NABw
• = 20 × 200 × 10^-4 × 3 × 10^-2 × 50
• = 0.6 V
• I = e/R
• ∴ I = 0.6/R ampere.