JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 6)
The dimension of $${{{B^2}} \over {2{\mu _0}}}$$, where B is magnetic field and $${{\mu _0}}$$
is the magnetic permeability of vacuum,
is :
ML2T–2
MLT–2
ML-1T–2
ML2T–1
Explanation
As $${{{B^2}} \over {2{\mu _0}}}$$ = Energy per unit volume
$$ \therefore $$ Dimension = $${{M{L^2}{T^{ - 2}}} \over {{L^3}}}$$ = ML-1T–2
$$ \therefore $$ Dimension = $${{M{L^2}{T^{ - 2}}} \over {{L^3}}}$$ = ML-1T–2
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