Question

A toroidal inductor with an inductance of 9.0.mH encloses a volume of $\mathbf{0}\mathbf{.}\mathbf{0200}{\mathbf{m}}^{\mathbf{3}}$. If the average energy density in the toroid is$\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{J}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$, what is the current through the inductor?

Short answer

5.58 A

Step-by-step solution

Given

i) Inductance of inductor$L=90.0\mathrm{mH}=90.0\times {10}^{-3}\mathrm{H}$

ii) Volume of toroid$V=0.0200{\mathrm{m}}^3$

ii) Average density$u_B=70.0\mathrm{J}/{\mathrm{m}}^3$l

Understanding the concept

We use the formula of total energy stored for volumein the magnetic field into the formula of energy stored in the inductor’s magnetic field to find the current through the inductor.

Formula:

$$\begin{gathered} U_B=\frac{1}{2}L{i}^2 \\ {U}_B=u_{B}V \end{gathered}$$

Calculate the current through the inductor

The magnetic field stored in the toroid is given by

$U_B=\frac{1}{2}L{i}^2$

Butthe total energy stored for volume V in the magnetic field is given by

$U_B=u_{B}V$

Therefore,

$$\begin{gathered} \frac{1}{2}L{i}^2=u_{B}V \\ i=\sqrt{\frac{2u_{B}V}{L}} \\ i=\sqrt{\frac{2\left(70.0{\displaystyle \frac{J}{m^3}}\right)\left(0.0200m^3\right)}{\left(90.0\times {10}^{-3}H\right)}} \\ i=5.58A \end{gathered}$$

Therefore, the current through the inductor is 5.58 A.