Physics

# The ratio of dimensions electric field and magnetic field is,

$LT^{-1}$

##### SOLUTION
Electric field $=$ force/charge or $E=F/q$
So, $[E]=\dfrac{[MLT^{-2}]}{[IT]}=[MLT^{-3}I^{-1}]$ where $[I]=$ dimension of current.
We know that magnetic force $F=qvB$ or $B=\dfrac{F}{qv}$
So, $[B]=\dfrac{[MLT^{-2}]}{[IT][LT^{-1}]}=[MT^{-2}I^{-1}]$

Thus, $\dfrac{[E]}{[B]}=\dfrac{[MLT^{-3}I^{-1}]}{MT^{-2}I^{-1}}=[LT^{-1}]$

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Single Correct Medium Published on 18th 08, 2020
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