Question

The magnetic field B crossing normally a square metallic plate of area \( 4 \, \text{m}^2 \) is changing with time as shown in the figure. The magnitude of induced emf in the plate during \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \) is _______ mV.

Hint:

The induced emf in a conductor can be found using Faraday's law, where the emf is proportional to the rate of change of the magnetic field through the area.

Answer 1

Correct Answer: 8

The induced emf in the plate is given by Faraday's Law of Induction: \[ \text{emf} = - \frac{d\Phi}{dt} \] Where \( \Phi \) is the magnetic flux given by: \[ \Phi = B \times A \] Here, \( A = 4 \, \text{m}^2 \), and \( \frac{dB}{dt} \) is the slope of the \( B \)-time graph from \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \). From the graph, the change in \( B \) is \( B_2 - B_1 = 8 - 4 = 4 \, \text{T} \) and the change in time is \( \Delta t = 4 - 2 = 2 \, \text{s} \). Now, the induced emf is: \[ \text{emf} = \frac{dB}{dt} \times A = \frac{4 \, \text{T}}{2 \, \text{s}} \times 4 \, \text{m}^2 = 8 \, \text{V} \] Thus, the magnitude of the induced emf is \( 8 \, \text{V} \).