Electromagnetic Induction - Result Question 34
34. A 100 millihenry coil carries a current of $1 A$.
Energy stored in its magnetic field is [1991]
(a) $0.5 J$
(b) $1 A$
(c) $0.05 J$
(d) $0.1 J$
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Answer:
Correct Answer: 34. (c)
Solution:
(c) $E=\frac{1}{2} L i^{2}=\frac{1}{2} \times(100 \times 10^{-3}) \times 1^{2}=0.05 J$ In building a steady current in the circuit, the source emf has to do work against self inductance of coil and whatever energy consumed for tis work stored in magnetic field of coil-called magnetic potential energy (u) of coil $u=\frac{1}{2} L I^{2}=\frac{N \phi i}{2}$
