In a magnetic field of 0.05 T , area of a coil changes from 101 cm 2 to 100 cm 2 without changing the resistance which is 2 Ω . The amount of charge (in μ C ) that flows during this period is

Step-by-step solution

B = 0.05   T A 1 = 101     cm 2 = 101 × 10 − 4     m 2 A 2 = 100     cm 2 = 100 × 10 − 4     m 2 R = 2    Ω Amount of charge flown is given by, q = B ( Δ A ) R = 0.05 × 1 0 1 × 1 0 - 4 - 1 0 0 × 1 0 - 4 2 = 0.05 × 1 × 1 0 - 4 2 = 2.5 × 1 0 - 6   C