Electromagnetic Induction Ques 28
28. 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: 28.(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 $u=\frac{1}{2} L I^{2}=\frac{N \phi i}{2}$
BETA
