Question:
Let $ [ \varepsilon_0 ] $ denote the dimensional formula of the permittivity of the vacuum and $ [ \mu_0 ] $ that of the permeability of the vacuum. If $M$ = mass, $L$ = length, $T$ = time and $I$ = electric current.
Let $ [ \varepsilon_0 ] $ denote the dimensional formula of the permittivity of the vacuum and $ [ \mu_0 ] $ that of the permeability of the vacuum. If $M$ = mass, $L$ = length, $T$ = time and $I$ = electric current.
Updated On: Apr 2, 2024
- $ [ \varepsilon_0 ] = [ M^{ - 1} L^{ - 3} T^2 I ] $
- $ [ \varepsilon_0 ] = [ M^{ - 1} \, L^{ - 3} \, T^4 \, I^2 ] $
- $ [ \mu_0 ] = [ ML \, T^{ - 2} I^{ - 2} ] $
- $ [ \mu_0 ] = [ ML^2 \, T^{ - 1} \, I ] $
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The Correct Option is C
Solution and Explanation
F = $ \frac{1}{ 4 \pi \varepsilon_0} . \frac{ q_1 q_2 }{ r^2 } $
$ [ \varepsilon_0 ] = \frac{ [ q_1 ] [ q_2 ] }{ [ F] [ r^2 ] } = \frac{ [ IT^2 ] }{ [ MLT^{ - 2} ] \, [ L^2 ] } = [ M^{ - 1} \, L^{ - 3} \, T^4 \, I^2 ] $
Speed of light, c = $ \frac{1}{ \sqrt{ \varepsilon_0 \, \mu_0 }} $
$\therefore [ \mu_0 ] = \frac{1}{ [ \varepsilon_0 ] \, [ c]^2 } = \frac{ 1}{ [ M^{ - 1} \, L^{ - 3} \, T^4 I^2 ] \, [ LT^{ - 1} ]^2 } $
= $ [ MLT^{ - 2} \, I^{ - 2} ] $
$ [ \varepsilon_0 ] = \frac{ [ q_1 ] [ q_2 ] }{ [ F] [ r^2 ] } = \frac{ [ IT^2 ] }{ [ MLT^{ - 2} ] \, [ L^2 ] } = [ M^{ - 1} \, L^{ - 3} \, T^4 \, I^2 ] $
Speed of light, c = $ \frac{1}{ \sqrt{ \varepsilon_0 \, \mu_0 }} $
$\therefore [ \mu_0 ] = \frac{1}{ [ \varepsilon_0 ] \, [ c]^2 } = \frac{ 1}{ [ M^{ - 1} \, L^{ - 3} \, T^4 I^2 ] \, [ LT^{ - 1} ]^2 } $
= $ [ MLT^{ - 2} \, I^{ - 2} ] $
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