In the given figure the magnetic flux through the loop increases according to the relation φ B (t) = 10t² + 20t, where φ B is in milliwebers and t is in seconds.
The magnitude of current through R = 2Ω resistor at t = 5 s is ___________ mA.

Solution

Physics sir · 4 steps

Faraday cares how fast the flux changes, not how much there is: differentiate, then divide by R.

1. Rate of change. By Faraday's law, the induced emf equals the rate of change of flux. Differentiate the given flux with respect to time.

   |ε| = dφ_B/dt = d/dt(10t² + 20t) = 20t + 20

2. At five seconds. |ε| = 20 × 5 + 20 = 120 mV. The emf comes out in millivolts because the flux is in milliwebers.
3. Ohm's law. I = |ε|/R = (120 mV)/(2 Ω) = 60 mA.
4. So. The current through the resistor at t = 5 s is 60 mA.

The common slip: 175. Using the flux value φ_B(5) = 350 mWb in place of its rate of change gives 350/2 = 175, not 60.

The idea to keep. Induced emf comes from dφ/dt, never from φ itself. Differentiate first, put in the time second, then divide by the resistance to get the current.

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