CBSE Class 12 Physics 2013 Paper

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Question : 18
Total: 29
Starting from the expression for the energy W=‌
1
2
L
I2
, stored in a solenoid of self-inductance L to build up the current I, obtain the expression for the magnetic energy in terms of the magnetic field B, area A and length l of the solenoid having n number of turns per unit length. Hence show that the energy density is given by ‌
B2
2µ0
.
Solution:  
‌B=µ0nI
Or, ‌B2=µ02I2n2
Or, ‌I2=‌
B2
µ02n2

And ‌L=µ0n2lA
Putting in the given expression,
‌W=‌
1
2
L
I2

Or, ‌W=‌
1
2
(µ0n2lA)
I2
[substituting L]
Or, ‌W=‌
1
2
(µ0n2V)
I2
‌‌‌[‌ Putting Volume ‌=V=l]
Or, W‌=‌
1
2
(µ0n2V)
×
‌
B2
µ02n2
‌ [substituting ‌l2]‌
Or, W‌=‌
B2V
2µ0


∴‌‌ Energy density ‌‌=‌
W
V
=‌
B2
2µ0
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