91Ó°ÊÓ

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 :

$L=\frac{{\mathrm{μ}}_0{N}^{2}A}{l}$

Here $L$= Self inductance

${\mathrm{μ}}_0$= permeability constant

$N$=No. of turns of wire

$A$= Cross section area

$l$= Length of wire

Substituting the values in the formula :

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

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

The formula of stored energy is :

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

Substituting the values in formula :

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

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