The magnetic field B crossing normally a square metallic plate of area 4 m² is changing with time as shown in figure. The magnitude of induced emf in the plate during t=2 s to t=4 s, is __________ mV.

Solution

Physics sir · 4 steps

Read the slope off the graph and multiply by the area; that is the whole problem.

1. Faraday's law. The field is normal to the plate and the area is fixed, so the flux is φ = BA. Only B changes, so the emf is A times the rate of change of B.

   |ε| = dφ/dt = A dB/dt

2. Slope from graph. At t = 2 s, B = 4 mT. At t = 4 s, B = 8 mT. So dB/dt = (8 - 4)/(4 - 2) = 2 mT s⁻¹. The graph is a straight line, so this rate is constant over the interval.
3. Put in numbers. |ε| = 4 m² × 2 × 10⁻³ T s⁻¹ = 8 × 10⁻³ V.
4. So. The induced emf is 8 mV, so the answer is 8.

The common slip: 16. Multiplying the area by the change Δ B = 4 mT and not dividing by Δ t = 2 s gives 16. That is the change in flux in mWb, not the emf.

The idea to keep. For a B versus t graph, the induced emf is area times slope. A straight line means a constant slope, and so a constant emf over the whole interval.

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