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Physics - Unit & Dimension Question with Solution | TestHub

PhysicsUnit & DimensionMiscellaneousEasy2 minQB

PhysicsEasysingle choice

Using mass $(M)$ , length $(L)$ , time $(T)$ and current $(A)$ as fundamental quantities, the dimension of permeability is

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Answer:

Solution:

From $B = \frac{μ _{0}}{2 π} \frac{i}{r}$ and
$F = i lB : [μ_{0}]$
$= [\frac{F}{i ^{2}}]$
$= \frac{[ M L T ^{- 2} ]}{[ A ^{2} ]}$
$= [M L T^{- 2} A^{- 2}]$ .

Stream:JEESubject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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