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) = 4pi xx 10^(-7) Hm^(-1), epsi_(0) = 8.9 xx 10^(-12) Fm^(-1))`

Options

A. 0.445J, 25J
B. 4.45J, 2.5J
C. 44.5J, 25J
D. 0.44J, 2.5J

Answer

A

Explanation

Energy densities are `u_(E) = (1)/(2) epsi_(0)E^(2)`
`u_(B) = (1)/(2) (B^(2))/(mu_(0))`
so, energy required to setup a uniform electric field in cube of side 10 cm is
`U_(E ) = u_(E) xx` Volume of cube `=(1)/(2) epsi_(0) E^(2) xx l^(3)`
`=(1)/(2) xx 8.9 xx 10^(-12) xx (10^(7))^(2) xx (0.1)^(3)`
`=4.45 xx 10^(-12 + 14-3)`
`=4.45 xx 10^(-1) = 0.445J`
and energy required to setup a uniform magnetic field in cube of side 10 cm is
`U_(B) = u_(B) xx` Volume of cube
`=(B^(2)l^(3))/(2mu_(0)) = (0.25 xx 0.25 xx (0.1)^(3))/(4pi xx 10^(-7) xx 2) = 25J`