SATHEE: electromagnetic-induction Question 29

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Question: Q. 11. Derive the expression for the magnetic energy stored in a solenoid in terms of magnetic field $B$ , area $A$ and length $l$ of the solenoid carrying a steady current $I$ . How does this magnetic energy per unit volume compare with the electrostatic energy density stored in a parallel plate capacitor?

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Solution:

Ans. Rate of work done,

Total amount of work done,

$\begin{aligned} \int d W & = \int L I d I W & = \frac{1}{2} L I^{2} \end{aligned}$

For the solenoid :

Inductance, $L = μ_{0} n^{2} A l$ ; also $B = μ_{0} n I$

$\begin{aligned} ∴ W = U_{B} = \frac{1}{2} L I^{2} & = \frac{1}{2} (μ_{0} n^{2} A l) {(\frac{B}{μ_{0} n})}^{2} & = \frac{B^{2} A l}{2 μ_{0}} \end{aligned}$

$1 / 2$

Magnetic energy per unit volume $= \frac{B^{2}}{2 μ_{0}}$

Also, Electrostatic Energy Stored Per Unit Volume

$= \frac{1}{2} ε_{0} E^{2}$

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