Question

Calculate the total energy UB and the energy density Up of the magnetic field stored in a solenoid $0.5\mathrm{m}$ long, having $5000$turns and a current $10\mathrm{A}$. (radius of the solenoid is $4\mathrm{cm}$).

Answer 1

Step 1: Given data

Length of the solenoid$\left(\mathrm{L}\right)$= $0.5\mathrm{m}$

Turns$\left(\mathrm{N}\right)$ = $5000$

Current$\left(\mathrm{I}\right)$ = $10\mathrm{A}$

radius$\left(\mathrm{r}\right)$ = $4\mathrm{cm}$

Step 2: Formula used

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{NI}}{\mathrm{l}}$

Here,

$\mathrm{B}$ is the magnetic field.

${\mathrm{\mu }}_0$ is the permeability of free space$(4\mathrm{\pi }\times {10}^{-7})$.

$\mathrm{N}$ is the number of turns in the given solenoid.

$\mathrm{I}$ is the current through the solenoid.

$\mathrm{E}=\frac{{\mathrm{B}}^2}{2{\mathrm{\mu }}_0}{\mathrm{A}}_{\mathrm{L}}$

where,

$\mathrm{E}$ is the electric field.

$\mathrm{B}$ is the magnetic field.

${\mathrm{\mu }}_0$ is the permeability of free space$(4\mathrm{\pi }\times {10}^{-7})$.

${\text{A}}_L$is the area of the solenoid.

Step 3: Calculation of total energy

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{NI}}{\mathrm{l}}$

$\mathrm{B}=\frac{4\times \mathrm{\pi }\times {10}^{-7}\times 5000\times 10}{0.5}$

$\mathrm{B}=0.04\mathrm{\pi }$.

$\mathrm{E}=\frac{{\mathrm{B}}^2}{2{\mathrm{\mu }}_0}{\mathrm{A}}_{\mathrm{L}}$

$\mathrm{E}=\frac{(0.04\mathrm{\pi }{)}^2}{2\times 4\mathrm{\pi }\times {10}^{-7}}\times \mathrm{\pi }(0.04{)}^2\times 0.5$

$\mathrm{E}=15.7\mathrm{joules}$

Therefore, the total energy is $15.7\mathrm{joules}$.