Question

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T magnetic field?

Short answer

$1.5\times {10}^8\mathrm{V}/\mathrm{m}$

Step-by-step solution

Given

Magnetic field$b=0.50\mathrm{T}$

Understanding the concept

We use the formula of energy density in the magnetic field and energy density in an electric field to find the magnitude of a uniform electric field if it is to have the same energy density as that possessed by amagnetic field.

Formula:

$$\begin{gathered} u_E=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}^2 \\ {\mathrm{u}}_{\mathrm{B}}=\frac{1}{2}\frac{B^2}{u_0} \end{gathered}$$

Calculate the magnitude of a uniform electric field required

The electric field density is given by

$u_E=\frac{1}{2}{\epsilon }_0{E}^2$

The magnetic field density is given by

$u_B=\frac{1}{2}\frac{B^2}{u_0}$

According to the given condition,

$$\begin{gathered} u_E=u_B \\ \frac{1}{2}{\epsilon }_0{E}^2=\frac{1}{2}\frac{B^2}{u_0} \\ {E}^2=\frac{B^2}{{\mu }_0{\epsilon }_0} \\ E=\frac{B}{\sqrt{{\mu }_0{\epsilon }_0}} \\ E=\frac{\left(0.50\mathrm{T}\right)}{\sqrt{\left(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\right)\left(4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m}\right)}} \\ E=1.5\times {10}^8\mathrm{V}/\mathrm{m} \end{gathered}$$

Therefore, the magnitude of a uniform electric field is $1.5\times {10}^8\mathrm{V}/\mathrm{m}$.