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.
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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.
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 \( 8 \, \text{V} \).
