CBSE Class 12 Physics 2014 Outside Delhi Set 2 Paper

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Question : 8
Total: 9
Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance I to build up a current I through it.
Solution:  
The ratio of magnetic flux through the solenoid to the current passing through it is called selfinductance of a solenoid. It is given by
L=
ϕ
I


Energy stored in an inductor: When a current grows through an inductor, a back e.m.f. is set up which opposes the growth of current. So work needs to be done against back e.m.f. (e) in building up the current. This work done is stored as magnetic potential energy.
Let I be the current through the inductor L at any instant t. The current rises at the rate
dI
dt
. So the induced e.m.f. is
e=
LdI
dt

The work done against induced e.m.f. in dt is
dW=Pdt
=eIdt[P=VI]
=
LdI
dt
I
dt

=LIdI
For total work from 0 to I0 current
=
1
2
L
I02

Hence, this work done is stored as the magnetic potential energy U in the inductor
U=
1
2
L
I2
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