The magnetic flux φ (in weber) linked with a closed circuit of resistance 8 Ω varies with time (in seconds) as φ=5 t²-36 t+1. The induced current in the circuit at t=2 s is __________ A.

Solution

Physics sir · 4 steps

Current comes from how fast the flux changes, so differentiate first and only then put in t = 2 s.

1. Faraday's law. The induced emf equals the rate of change of flux. Its size is |ε| = |dφ/dt|.

   dφ/dt = d/dt(5t² - 36t + 1) = 10t - 36

2. Put in the time. At t = 2 s, dφ/dt = 10 × 2 - 36 = -16 Wb s⁻¹. The minus sign only gives the direction of the emf (Lenz's law), so |ε| = 16 V.
3. Ohm's law. I = |ε|/R = (16 V)/(8 Ω) = 2 A.
4. So. The induced current at t = 2 s is 2 A.

The common slip: 6.375. Dividing the flux itself by R instead of its derivative gives |φ(2)|/8 = 51/8 = 6.375, not 2.

The idea to keep. Flux never drives current by itself. Only its rate of change does: differentiate φ(t), put in the time, then divide by R.

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