Starting from the expression for the energy ...

CBSE Class 12 Physics 2013 Paper

Question : 18 of 29

Marks: +1, -0

Starting from the expression for the energy

W = \frac{1}{2} L I^2

, 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

\frac{B^2}{2 \mu_0}

Solution:

B = \mu_0 n I

Or,

B^2 = \mu_0^2 I^2 n^2

Or,

I^2 = \frac{B^2}{\mu_0^2 n^2}

And

L = \mu_0 n^2 l A

Putting in the given expression,

W = \frac{1}{2} L I^2

Or,

W = \frac{1}{2} (\mu_0 n^2 l A) I^2

[substituting L]

Or,

W = \frac{1}{2} (\mu_0 n^2 V) I^2

[ \text{Putting Volume} = V = l ]

Or,

W = \frac{1}{2} (\mu_0 n^2 V) \times \frac{B^2}{\mu_0^2 n^2}

\text{[substituting } l^2]

Or,

W = \frac{B^2 V}{2 \mu_0}

\therefore \text{Energy density} = \frac{W}{V} = \frac{B^2}{2 \mu_0}

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