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Q.
A straight wire of length $50\,cm$ carrying a current of $2.5\, A$ is suspended in mid-air by a uniform magnetic field of $0.5\, T$ (as shown in figure). The mass of the wire is $(g = 10 \, ms^{-2} )$
Given, length of wire $(I)=50\, cm =50 \times 10^{-2} \,m$
current $(I)=2.5 \,A$
magnetic field $(B)=0.5 \,T$
$g=10 \,ms ^{-2}$
For the balance condition,
$F_{B} =mg $
$BIl =m g $
$m =\frac{B I l}{g} $
$=\frac{0.5 \times 2.5 \times 50 \times 10^{-2}}{10}$
$=62.5\, g$
