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A 100mH coil carries a current of 1 ampe...

A `100mH` coil carries a current of `1` ampere. Energy stored in its magnetic field is

A

`0.5J`

B

`0.05J`

C

1J

D

`0.1J`

Text Solution

Verified by Experts

The correct Answer is:
B
Here, `L=100 mH=100xx10^(-3)H`
`I-=1A`
`U=(1)/(2)LI^(2)=(1)/(2)xx(10xx10^(-3))xx1^(2)=0.05J`
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