A) \[M{{T}^{-3}}\]
B) \[{{M}^{2}}L{{T}^{-5}}{{A}^{-2}}\]
C) \[{{M}^{2}}L{{T}^{-3}}{{A}^{-1}}\]
D) \[ML{{T}^{-2}}{{A}^{-2}}\]
Correct Answer: B
Solution :
\[F=qvB\] \[B=\frac{F}{qv}=\frac{F}{It\,v}\] \[=\frac{[ML\,{{T}^{-2}}]}{[AT][L{{T}^{-1}}]}=[M{{T}^{-2}}{{A}^{-1}}]\] \[B=\frac{F}{q}=[ML{{T}^{-3}}{{A}^{-1}}]\] \[E\times B=[ML{{T}^{-3}}{{A}^{-1}}]\times [M{{T}^{-2}}{{A}^{-1}}]\] \[=[{{M}^{2}}L{{T}^{-5}}{{A}^{-2}}]\]
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