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1.Units, Dimensions and Measurement
medium
The dimensions of $\left(\frac{ B ^{2}}{\mu_{0}}\right)$ will be.
(if $\mu_{0}$ : permeability of free space and $B$ : magnetic field)
A$\left[ ML ^{2} T ^{-2}\right]$
B$\left[ ML T ^{-2}\right]$
C$\left[ ML ^{-1} T ^{-2}\right]$
D$\left[ ML ^{2} T ^{-2} A ^{-1}\right]$
(JEE MAIN-2022)
Solution
$u =\frac{ B ^{2}}{2 \mu_{0}}$
$u \rightarrow$ Energy per unit volume
$\left[\frac{ B ^{2}}{\mu_{0}}\right]=[ u ]=\frac{\left[ ML ^{2} T ^{-2}\right]}{\left[ L ^{3}\right]}=\left[ ML ^{-1} T ^{-2}\right]$
$u \rightarrow$ Energy per unit volume
$\left[\frac{ B ^{2}}{\mu_{0}}\right]=[ u ]=\frac{\left[ ML ^{2} T ^{-2}\right]}{\left[ L ^{3}\right]}=\left[ ML ^{-1} T ^{-2}\right]$
Standard 11
Physics
