Question

Define the self-inductance of the coil. Show that magnetic energy required to build up the current I in the coil of self-inductance L is given by $\dfrac{1}{2}L{I^2}$.

Answer 1

Hint The above problem is based on the magnetic effect of the current and electromagnetic induction in the coil. If a charge moves inside the conductor then it creates a region in which other particles in this region experience the effect of the moving charge. The variation in the magnetic flux induces a voltage in the conductor called induced emf.

Complete step by step answer
The induced emf in the conductor by the phenomenon of electromagnetic induction always resists the variation in the current that causes the induced emf. The property by which the self-induced emf opposes the change in the current in the conductor is called the self-inductance of the coil. The self-inductance is also the same as the ratio of the induced emf to variation in the current with time. The coil of higher self-inductance allows very small variation in the current.
Let us consider initially the current in the coil is zero. As the current increases from zero to some value the self-induced emf opposes the change in the current in the coil. If the current $dI$ varies in time $dt$ then the self-induced emf of the coil is given as,
$\varepsilon = - L\left( {\dfrac{{dI}}{{dt}}} \right)$
Negative sign indicates the oppose of change in current by the self-induced emf.
The work done by the induced emf to oppose the change in current in the coil is given as,
$dW = \left| \varepsilon \right|Idt......\left( 1 \right)$
Substitute $ - L\left( {\dfrac{{dI}}{{dt}}} \right)$ for $\varepsilon $ in the above expression to calculate the work done.
\[dW = \left| { - L\left( {\dfrac{{dI}}{{dt}}} \right)} \right|Idt\]
\[dW = \left( {L\left( {\dfrac{{dI}}{{dt}}} \right)} \right)Idt\]
\[dW = \left( {LI} \right)dI......\left( 2 \right)\]
Integrate the above expression (2) for current 0 to I and work 0 to W to find the total work done by the induced emf to oppose the change in the current.
$\int\limits_0^W {dW} = \int\limits_0^I {\left( {LI} \right)dI} $
$\left( W \right)_0^W = L\left( {\dfrac{{{I^2}}}{2}} \right)_0^I$
$W - 0 = L\left( {\dfrac{{{I^2}}}{2} - \dfrac{0}{2}} \right)$
$W = \dfrac{1}{2}L{I^2}......\left( 3 \right)$
The work given by the expression (3) is the same as the magnetic energy stored in the magnetic field of the coil due to change in the current.
Thus, the above expression (3) shows that magnetic energy required to build up the current I in the coil of self-inductance L is given by $\dfrac{1}{2}L{I^2}$.

Note The work done by the self-induced emf to oppose the change in current in the coil in the magnetic field is the same as the magnetic energy of the coil in the magnetic field. The self-inductance of the coil depends on the number of turns of the coil, length of the coil and cross sectional area of the coil.