Question

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 129. 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.

Answer 1

Correct Answer: 60

The induced emf (\( \mathcal{E} \)) in the circuit is given by Faraday's law: \[ \mathcal{E} = -N \frac{\Delta \Phi_B}{\Delta t}, \] where \( N = 100 \) is the number of turns, and \( \Delta \Phi_B \) is the change in magnetic flux. The total change in magnetic flux (\( \Delta \Phi_B \)) is: \[ \Delta \Phi_B = A \Delta B, \] where \( A = 24 \, \text{cm}^2 = 24 \times 10^{-4} \, \text{m}^2 \) is the cross-sectional area, and \( \Delta B = 2 \times 1.5 = 3 \, \text{T} \) is the change in magnetic field (from \( 1.5 \, \text{T} \) in one direction to \( 1.5 \, \text{T} \) in the opposite direction). Substitute values: \[ \Delta \Phi_B = 24 \times 10^{-4} \times 3 = 7.2 \times 10^{-3} \, \text{Wb}. \] The total emf induced is: \[ \mathcal{E} = \frac{N \Delta \Phi_B}{R}, \] where \( R = 12 \, \Omega \). Substitute \( N = 100 \): \[ \mathcal{E} = \frac{100 \cdot 7.2 \times 10^{-3}}{12} = 6 \times 10^{-3} \, \text{C}. \] Thus, the charge flowing through the circuit is \( \boxed{60 \, \text{mC}} \).