A coil with an inductance of 2.0 H and a resistance of 20 is connected to a battery having an emf of 4.0 V . Determine the magnetic field energy at the instant 0.20 s after the connection is made.

Step-by-step solution

The growth of current in an LR circuit is given by i = i₀ (1 - e^ -t/ ) where the steady state current i₀ = E R and the time constant = L R . Substituting the given values: i₀ = 4.0 20 = 0.2 A = 2.0 20 = 0.1 s At t = 0.20 s , the current is: i = 0.2 (1 - e^ -0.20/0.1 ) = 0.2 (1 - e⁻²) Using e⁻² 0.135 : i 0.2 (1 - 0.135) = 0.2 0.865 = 0.173 A The magnetic field energy stored in the inductor is: U = 1 2 L i^2 U = 1 2 2.0 (0.173)^2 0.0299 J 0.03 J