An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area 24 cm². The two ends of the wire are connected to a resistor. The total resistance in the circuit is 12 Ω. If an externally applied uniform magnetic field in the core along its axis changes from 1.5 T in one direction to 1.5 T in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be ___________ mC.

Solution

Physics sir · 4 steps

The charge depends only on the total change in flux, not on how fast the field flips.

1. Charge from flux. The induced current is I = ε/R = N/RdΦ/dt. Since I = dq/dt, the time cancels, and the total charge depends only on the total flux change.

   q = (N ΔΦ)/R

2. Change in flux. The field goes from +1.5 T to -1.5 T, so Δ B = 3 T. With A = 24 × 10⁻⁴ m², the flux change per turn is ΔΦ = 3 × 24 × 10⁻⁴ = 7.2 × 10⁻³ Wb.
3. Put in numbers. q = (100 × 7.2 × 10⁻³)/12 = 0.72/12 = 0.06 C.
4. So. The charge that flows is 0.06 C = 60 mC.

The common slip: 30. Taking the change in field as 1.5 T instead of 3 T ignores the reversal and gives 30, half the correct charge.

The idea to keep. Charge through a coil is q = NΔΦ/R, whatever the rate of change. When a field reverses, the flux change is twice the initial flux, not once.

Answer from the NTA final answer key. Solution by PrepWiser.