A coil has an inductance of 2 H and resistance of 4 Ω. A 10 V is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ___________ × 10⁻² J.

Solution

Physics sir · 3 steps

At steady state the inductor stops opposing, so only the resistance sets the current. Find it, then use ½LI².

1. Steady current. Once the current stops changing, dI/dt = 0, so the induced emf LdI/dt is zero. The coil then acts like a plain 4 Ω resistor, so I₀ = V/R = 10/4 = 2.5 A.

   U = 1/2 L I₀²

2. Put in numbers. U = 1/2 × 2 × (2.5)² = 6.25 J.
3. So. 6.25 J = 625 × 10⁻² J, so the answer is 625.

The common slip: 1250. Dropping the 1/2 in 1/2LI² gives 12.5 J, which reads as 1250 instead of 625.

The idea to keep. For an RL coil on a DC supply, the final current is just V/R. The inductance decides how fast the current gets there, and then it sets the stored energy 1/2LI².

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