Final answer:
The self-inductance of coil A is 3 H. The flux linking with coil B is 0.008 Wb. The average emf induced in coil B when the flux with it changes from zero to full value in 0.02 s is 200 V. The mutual inductance is 0.5 H. The average emf in coil B when the current in A changes from 0 to 8 A in 0.05 s is 80 V.
Step-by-step explanation:
(a) The formula to calculate self-inductance is L = N²R/I, where L is the self-inductance, N is the number of turns, R is the reluctance, and I is the current. Using this formula, the self-inductance of coil A can be calculated as:
LA = (600² * 0.04) / 8 = 3 H
(b) Since the coefficient of coupling is 0.2, the flux linking with coil B can be calculated as:
Flux linking with B = coefficient of coupling * flux in A = 0.2 * 0.04 = 0.008 Wb
(c) To calculate the average emf induced in coil B, we can use the formula E = -N (dΦ/dt), where E is the emf, N is the number of turns, and dΦ/dt is the rate of change of flux. In this case, the flux changes from zero to full value in 0.02 s, so the rate of change of flux is (0.008 Wb - 0) / (0.02 s) = 0.4 Wb/s. Therefore, the average emf induced in coil B is:
EB = -500 * 0.4 = -200 V (negative sign indicates the direction of the induced emf)
(d) The formula to calculate mutual inductance is M = k * √(LA * LB), where M is the mutual inductance, k is the coefficient of coupling, and LA and LB are the self-inductances of coils A and B, respectively. Using this formula, the mutual inductance can be calculated as:
M = 0.2 * √(3 * LB) = 0.2 * √(3 * 0.5) = 0.5 H
(e) To calculate the average emf in coil B when the current in A changes from 0 to 8 A in 0.05 s, we can use the same formula as in (c) with the rate of change of flux equal to (0.008 Wb - 0) / (0.05 s) = 0.16 Wb/s. Therefore, the average emf induced in coil B is:
EB = -500 * 0.16 = -80 V (negative sign indicates the direction of the induced emf)