The energies required to set up in a cube of side $10$ cm,~ $( a )$ A uniform electric field of ${10}^{7}Vm^{- 1}$ and~ $( b )$ ~A~uniform magnetic field if $0 . 25Wbm^{- 2}$ , are respectively about
\left({μ}_{0} = 4\pi \times {10}^{- 7}Hm^{- 1}, {ϵ}_{0} = 8 . 9 \times {10}^{- 12}Fm^{- 1}\right)

Question

The energies required to set up in a cube of side

10

cm,~

( a )

A uniform electric field of

{10}^{7}Vm^{- 1}

and~

( b )

~A~uniform magnetic field if

0 . 25Wbm^{- 2}

, are respectively about
\left({μ}_{0} = 4\pi \times {10}^{- 7}Hm^{- 1}, {ϵ}_{0} = 8 . 9 \times {10}^{- 12}Fm^{- 1}\right)

Quantrex Academy · Accepted Answer

Correct answer: A. 0 . 445J, 25J

Recall the formula of energy in terms of electric field and magnetic field strength,

${E}_{E} = \dfrac{1}{2}{ε}_{0}\left(E\right)2V \text{ and } {E}_{B} = \dfrac{1}{2}{ε}_{0}\left(Bc\right)2V ⇒ {E}_{E} = \dfrac{1}{2}{ε}_{0}\left(E\right)2V ⇒ {E}_{E} = \dfrac{1}{2} \times 8 . 9 \times {10}^{- 12} \times \left({10}^{7}\right)2 \times {10}^{- 3} = 0 . 445J ⇒ {E}_{B} = \dfrac{1}{2}{ε}_{0}\left(Bc\right)2V ⇒ {E}_{B} = \dfrac{1}{2} \times 8 . 9 \times {10}^{- 12} \times \left(0 . 25\right)2 \times \left(3 \times {10}^{8}\right)2 \times {10}^{- 3} = 25J$

The energies required to set up in a cube of side 10 cm,~ ( a ) A...

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