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Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field \(B,\) area \(A\) and length \(l\) of the solenoid.
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Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field \(B,\) area \(A\) and length \(l\) of the solenoid.

Answer

\(U_m=\frac{B^2\mathit{Al}}{2\mu _0}\)
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Solution

Magnetic energy stored in the solenoid is given by:
 \(U_m=\frac 1 2LI^2\) 
where\(L\)is the self-inductance and\(I\)is the current flowing
But, magnetic field inside the solenoid:
 \(B=\mu _0nI\) 
where\(n\)is the number of turns per unit length
 \(⟹I=\frac B{\mu _0n}\)
Substitute the value of \(I\) in \(U_m\) , we get
 \(U_m=\frac 1 2L(\frac B{\mu _0n})^2\)  ...(i)
And, the self-inductance for the solenoid:
 \(L=\mu _0n^2\mathit{Al}\) 
where\(A\)is the area of cross-section of the solenoid
Substitute the value of\(L\)in (i).
Hence, the magnetic energy stored in the solenoid is given by:
 \(U_m=\frac 1 2(\mu _0n^2\mathit{Al})\left(\frac B{\mu _0n}\right)^2\) 
 \(⟹U_m=\frac{B^2\mathit{Al}}{2\mu _0}\)  
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