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Obtain and expression for magnetic energy …

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Obtain and expression for magnetic energy stored in a a solenoid in terms of of magnetic field b area a a and length l of the solenoid . How does this magnetic energy compare with the electrostatic energy stored in a capacitor
  • 1 answers

Sia ? 4 years, 5 months ago

  1. The work done against the induced emf in building up the current from 0 to I is
    W = {tex}\int d W=\int \mathfrak{E} I d t=\int L \frac{d I}{d t} I d t=\int_{0}^{I} L I d I=\frac{1}{2} L I^{2}{/tex}
    The magnetic energy stored in the solenoid is
    UB = W = {tex}\frac{1}{2} L I^{2}=\frac{1}{2} L\left(\frac{B}{\mu_{0} n}\right)^{2}{/tex} {tex}\left[\because B=\mu_{0} n I\right]{/tex}
    {tex}\frac{1}{2}\left(\mu_{0} n^{2} A l\right)\left(\frac{B}{\mu_{0} n}\right)^{2}{/tex} {tex}\left[\because L=\mu_{0} n^{2} A l\right]{/tex}
    {tex}U_{B}=\frac{1}{2 \mu_{0}} B^{2} A l{/tex}
  2. The magnetic energy stored per unit volume of the solenoid is
    {tex}u_{B}=\frac{U_{B}}{V}=\frac{U_{B}}{A l}{/tex} [Here V is the volume that contains flux]
    [Here V is volume that contains flux]
    or {tex} u_{B}=\frac{B^{2}}{2 \mu_{0}}{/tex} ....(i)
    We know that the electrostatic energy stored per unit volume in a parallel plate capacitor is
    {tex}u_{E}=\frac{1}{2} \varepsilon_{0} E^{2}{/tex} .....(ii)
    In both the cases energy is proportional to the square of the field strengths. Eqs. (i) and (ii) are general and valid for any region of space in which a magnetic field or/and an electric field exist.
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