A) 250 J
B) 250 erg
C) 125 J
D) 125 erg
Correct Answer: A
Solution :
Final current, \[I=\frac{E}{R}\] \[I=10A\] Energy stored in the magnetic field \[U=\frac{1}{2}Li\] \[\frac{1}{2}\times 5\times {{(10)}^{2}}=250J\]
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