The dimension of P = frac{{{B^2}{l^2}}}{ | vedclass

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Question

$F = BIL$
$	herefore \,\,{m{Dimension}}\,{m{of}}\,{m{[}}B{m{]}} = \frac{{{m{[}}F{m{]}}}}{{{m{[}}I{m{]}}\,{m{[}}L{m{]}}}} = \frac

Vedclass · Accepted Answer

$M{L^2}{T^{ - 4}}I^{-2}$

Vedclass · Answer

$F = BIL$
$	herefore \,\,{m{Dimension}}\,{m{of}}\,{m{[}}B{m{]}} = \frac{{{m{[}}F{m{]}}}}{{{m{[}}I{m{]}}\,{m{[}}L{m{]}}}} = \frac{{[ML{T^{ - 2}}]}}{{[I]\,[L]}}$=$[M{T^{ - 2}}{I^{ - 1}}]$
$[P] = \frac{{{B^2}{l^2}}}{m} = \frac{{{{[M{T^{ - 2}}{I^{ - 1}}]}^2} 	imes [{L^2}]}}{{[M]}}$ $ = [M{L^2}{T^{ - 4}}{I^{ - 2}}]$

List$-I$	List$-II$
$(a)$ Capacitance, $C$	$(i)$ ${M}^{1} {L}^{1} {T}^{-3} {A}^{-1}$
$(b)$ Permittivity of free space, $\varepsilon_{0}$	$(ii)$ ${M}^{-1} {L}^{-3} {T}^{4} {A}^{2}$
$(c)$ Permeability of free space, $\mu_{0}$	$(iii)$ ${M}^{-1} L^{-2} T^{4} A^{2}$
$(d)$ Electric field, $E$	$(iv)$ ${M}^{1} {L}^{1} {T}^{-2} {A}^{-2}$

The dimension of $P = \frac{{{B^2}{l^2}}}{m}$ is
where $B=$ magnetic field, $l=$ length, $m =$ mass

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