Physics

# The dimensions of emf in MKS is:

${ML^{2}}{T^{-2}}{Q^{-1}}$

##### SOLUTION
The emf across an inductor is given by:
$e = -L\dfrac {di}{dt}$

$e = \big[ {ML^{2}}{T^{-2}}{A^{-2}} \big] \big[ \dfrac {A}{T} \big]$

$e = \dfrac {{ML^{2}}{T^{-2}}}{A^{-1}{T}}$

$e = {ML^{2}}{T^{-2}}{Q^{-1}}$

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Single Correct Medium Published on 18th 08, 2020
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