How much energy is stored in a 3.0-cm-diameter, 12-cm-long solenoid that has 200 turns of wire and carries a current of 0.80 A?

We have given that that, the 12 cm long solenoid has 3.0 cm diameter and 200 turns of wire and carries a current of 0.80 A.

We need to calculate the value of self inductance, to calculate the value of stored energy.

Formula to calculate the self inductance :

$L=\frac{{\mu }_0{N}^{2}A}{l}$

Here $L$= Self inductance

${\mu }_0$= permeability constant

$N$=No. of turns of wire

$A$= Cross section area

$l$= Length of wire

Substituting the values in the formula :

localid="1651029502538" $L=\frac{(4\pi \times {10}^{-7}H/m)\times {\left(200\right)}^2\times \pi \times {(0.015m)}^2}{0.12m}$

localid="1651314937109" $L=29.57\times {10}^{-5}H$

The formula of stored energy is :

$U=\frac{1}{2}L{I}^2$

Substituting the values in formula :

localid="1651314977368" $U=\frac{1}{2}\times (29.57\times {10}^{-5}H)\times {(0.80A)}^2$

localid="1651315065815" $U=9.4624\times {10}^{-5}J$