JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 7)
In SI units, the dimensions of $$\sqrt {{{{ \in _0}} \over {{\mu _0}}}} $$ is :
A–1 TML3
A2T3M–1L–2
AT–3ML3/2
AT2M–1L–1
Explanation
$$\sqrt {{{{ \in _0}} \over {{\mu _0}}}} $$ = $${{{ \in _0}} \over {\sqrt {{\mu _0}{ \in _0}} }}$$ = c $$ \times $$ $${{ \in _0}}$$
$$ \therefore $$ $$\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$$ = $$\left[ {L{T^{ - 1}}} \right] \times \left[ {{ \in _0}} \right]$$
We know, F = $${1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}$$
$$ \therefore $$ $${ \in _0} = {{{q^2}} \over {4\pi {r^2}F}}$$
$$ \Rightarrow $$ $$\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right] \times \left[ {{L^2}} \right]}}$$ = $$\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$$
$$ \therefore $$ $$\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$$ = $$\left[ {L{T^{ - 1}}} \right]$$ $$ \times $$ $$\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$$
= [A2T3M–1L–2]
$$ \therefore $$ $$\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$$ = $$\left[ {L{T^{ - 1}}} \right] \times \left[ {{ \in _0}} \right]$$
We know, F = $${1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}$$
$$ \therefore $$ $${ \in _0} = {{{q^2}} \over {4\pi {r^2}F}}$$
$$ \Rightarrow $$ $$\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right] \times \left[ {{L^2}} \right]}}$$ = $$\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$$
$$ \therefore $$ $$\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$$ = $$\left[ {L{T^{ - 1}}} \right]$$ $$ \times $$ $$\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$$
= [A2T3M–1L–2]
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