Units & Measurements

Dimension of $\varepsilon $₀ = M^$-$1L^$-$3T⁴A²

Dimension of electric field E = MLT^$-$3A^$-$1

Now, dimension of

${1 \over 2}$$\varepsilon $₀E² = (M^$-$1L^$-$3T⁴A²)(MLT^$-$3A^$-$1)²

= M^$-$1L^$-$3T⁴A²M²L²T^$-$6A^$-$2 = ML^$-$1T^$-$2

Answer (C)

From coulombs law, we know that

$F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} \Rightarrow {\varepsilon _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {{r^2}F}}$

Therefore, ${\varepsilon _0} = {{[{I^2}{T^2}]} \over {[{L^2}][ML{T^{ - 2}}]}} = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$

Now, $c = {1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} \Rightarrow {c^2} = {1 \over {{\varepsilon _0}{\mu _0}}}$

Therefore, ${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}} = {1 \over {{{[L{T^{ - 1}}]}^2}[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}]}} = [ML{T^{ - 2}}{I^{ - 2}}]$

Answer (B), (C)

Induction $L = {\phi \over i} = $ weber/Ampere

Also, $e = - L\left( {{{di} \over {dt}}} \right) \Rightarrow L = {{ - e} \over {(di/dt)}} = {{volt - {\mathop{\rm second}\nolimits} } \over {Ampere}}$

Also, $U = {1 \over 2}L{i^2} \Rightarrow L = {{2U} \over {{i^2}}} = {{Joule} \over {{{(Ampere)}^2}}}$

and $U = {1 \over 2}L{i^2} = {i^2}RT \Rightarrow L = RT = $ Ohm-second

Answer (A), (B), (C), (D)