The energies required to set up in a cube of side $10 \mathrm{~cm}$ (i) a uniform electric field of $10^7…
The energies required to set up in a cube of side $10 \mathrm{~cm}$
(i) a uniform electric field of $10^7 \mathrm{Vm}^{-1}$ and
(ii) a uniform magnetic field of $0.25 \mathrm{Wbm}^{-2}$ are respectively about
$
\left(\mu_0=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}, \varepsilon_0=8.9 \times 10^{-12} \mathrm{Fm}^{-1}\right)
$
0.445 J, 25 J
4.45 J, 2.5 J
44.5 J, 25 J
0.44 J, 2.5 J
Solution
Energy densities are $u_E=\frac{1}{2} \varepsilon_0 E^2$
$
u_B=\frac{1}{2} \frac{B^2}{\mu_0}
$
So, energy required to setup a uniform electric field in cube of side $10 \mathrm{~cm}$ is
$
\begin{aligned}
U_E & =u_E \times \text { Volume of cube }=\frac{1}{2} \varepsilon_0 E^2 \times l^3 \\
& =\frac{1}{2} \times 8.9 \times 10^{-12} \times\left(10^7\right)^2 \times(0.1)^3 \\
& =4.45 \times 10^{-12+14-3} \\
& =4.45 \times 10^{-1}=0.445 \mathrm{~J}
\end{aligned}
$
and energy required to setup a uniform magnetic field in cube of side $10 \mathrm{~cm}$ is
$
\begin{aligned}
U_B & =u_B \times \text { Volume of cube } \\
& =\frac{B^2 l^3}{2 \mu_0}=\frac{0.25 \times 0.25 \times(0.1)^3}{4 \pi \times 10^{-7} \times 2}=25 \mathrm{~J}
\end{aligned}
$