Question

The magnetic energy stored in a long solenoid of area of cross section a

Answer 1

If there is magnetic field of solenoid of specific A and l.

Then,

U = 1 / 2 εo E ^2 - - - - - - -> i

As,

c^2 = 1 / εo μo

εo = 1 / c^2 μo

E = B c

Substituting the values of E and εo in eq i

U = 1 / 2 B ^2 / μo x A x l

Answer 2

Answer:

The self inductance of the solenoid is $3.01\times10^{-6}\ H$

Explanation:

Suppose, The magnetic energy stored in a long solenoid of length 40 cm, cross section area 24 cm²  and 20 turns per centimeter carries a current of 8.0 A. Find  the self inductance of the solenoid,

We need to calculate the self inductance of the solenoid

Using formula of self inductance

$L=\dfrac{\mu N^2 A}{l}$

Where, A = cross section area

N = number of turns

I = current

Put the value into the formula

$L=\dfrac{4\pi\times10^{-7}\times20\times20\times24\times10^{-4}}{40\times10^{-2}}$

$L=0.00000301\ H$

$L=3.01\times10^{-6}\ H$

Hence, The self inductance of the solenoid is $3.01\times10^{-6}\ H$