If current of 4 A produces magnetic flux of 3 10⁻³ Wb through a coil of 400 turns, the energy stored in the coil will be

Step-by-step solution

Total magnetic flux linked with the coil is given by = N Substituting the given values: = 400 3 10⁻³ = 1.2 Wb The self-inductance of the coil is L = I L = 1.2 4 = 0.3 H The energy stored in the coil is U = 1 2 L I^2 U = 1 2 0.3 (4)^2 = 2.4 J Answer: 2.4 J