The energies required to set up in a cube of | vedclass

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Question

(A) The energy density of an electric field is $u_E = \frac{1}{2} \varepsilon_0 E^2$ and the energy density of a magnetic field is $u_B = \frac{1}{2}

Vedclass · Accepted Answer

$0.445 \,J, 25 \,J$

Vedclass · Answer

(A) The energy density of an electric field is $u_E = \frac{1}{2} \varepsilon_0 E^2$ and the energy density of a magnetic field is $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.
Volume of the cube $V = l^3 = (0.1 \,m)^3 = 10^{-3} \,m^3$.
For the electric field:
$U_E = u_E 	imes V = \frac{1}{2} \varepsilon_0 E^2 	imes V = \frac{1}{2} 	imes 8.9 	imes 10^{-12} 	imes (10^7)^2 	imes 10^{-3} = 4.45 	imes 10^{-12+14-3} = 4.45 	imes 10^{-1} = 0.445 \,J$.
For the magnetic field:
$U_B = u_B 	imes V = \frac{B^2}{2 \mu_0} 	imes V = \frac{(0.25)^2 	imes 10^{-3}}{2 	imes 4 \pi 	imes 10^{-7}} = \frac{0.0625 	imes 10^{-3}}{8 \pi 	imes 10^{-7}} \approx \frac{0.0625 	imes 10^4}{25.13} \approx 24.87 \,J \approx 25 \,J$.

The energies required to set up in a cube of side $10 \,cm$ $(i)$ a uniform electric field of $10^7 \,Vm^{-1}$ and (ii) a uniform magnetic field of $0.25 \,Wbm^{-2}$ are respectively about $(\mu_0=4 \pi \times 10^{-7} \,Hm^{-1}, \varepsilon_0=8.9 \times 10^{-12} \,Fm^{-1})$

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