The dimensions of (( B 2/μ0)) will be: (if μ0: permeability of free space and B: magnetic field)

Question

The dimensions of (( B 2/μ0)) will be: (if μ0: permeability of free space and B: magnetic field) (A) [ M L 2 T -2] (B) [ M L T -2] (C) [ M L -1 T -2] (D) [ M L 2 T -2 A -1]

Tardigrade Admin · Accepted Answer

u =( B 2/2 μ0) u arrow Energy per unit volume [( B 2/μ0)]=[ u ]=([ ML 2 T -2]/[ L 3])=[ ML -1 T -2]

Q. The dimensions of $(\frac{B ^{2}}{μ _{0}})$ will be : (if $μ_{0}$ : permeability of free space and B : magnetic field)

$[M L^{2} T^{- 2}]$

$[M L T^{- 2}]$

$[M L^{- 1} T^{- 2}]$

$[M L^{2} T^{- 2} A^{- 1}]$

$u = \frac{B ^{2}}{2 μ _{0}}$
$u \to$ Energy per unit volume
$[\frac{B ^{2}}{μ _{0}}] = [u] = \frac{[ M L ^{2} T ^{- 2} ]}{[ L ^{3} ]} = [M L^{- 1} T^{- 2}]$

Q. The dimensions of (μ0​B2​) will be : (if μ0​ : permeability of free space and B : magnetic field)

[ML2T−2]

[MLT−2]

[ML−1T−2]

[ML2T−2A−1]