Magnetic flux (in weber) in a closed circuit of resistance 20 Ω varies with time t(s) at φ = 8t² - 9t + 5. The magnitude of the induced current at t = 0.25 s will be ____________ mA.

Solution

Physics sir · 4 steps

The emf comes from how fast the flux changes, so differentiate first and only then put in t = 0.25 s.

1. Faraday's law. The induced emf equals the rate of change of flux, e = -dφ/dt. The value of φ at an instant does not decide the emf. Its slope does.

   dφ/dt = d/dt(8t² - 9t + 5) = 16t - 9

2. Put in t. At t = 0.25 s, dφ/dt = 16 × 0.25 - 9 = 4 - 9 = -5 Wb s⁻¹. So the magnitude of the emf is |e| = 5 V.
3. Current. By Ohm's law, I = |e|/R = 5/20 = 0.25 A.
4. So. 0.25 A = 0.25 × 1000 mA, so the induced current is 250 mA.

The common slip: 0.25. Stopping at the current in amperes gives 0.25. The question asks for milliamperes, so multiply by 1000 to get 250.

The idea to keep. Induced emf is the time derivative of flux, not the flux itself. Differentiate, put in the instant, take the magnitude, divide by R, and watch the units asked for.

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