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(C) The magnetic force on a current-carrying conductor is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$,or in magnitude,$F = ILB \sin 	heta$,where $

Vedclass · Accepted Answer

$9$

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(C) The magnetic force on a current-carrying conductor is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$,or in magnitude,$F = ILB \sin 	heta$,where $	heta$ is the angle between the length vector $\vec{L}$ and the magnetic field vector $\vec{B}$.For the $5\,cm$ side,the length $L = 5 	imes 10^{-2}\,m$. The magnetic field $B = 0.75\,T = \frac{3}{4}\,T$.From the geometry of the right-angled triangle,the angle $	heta$ between the $5\,cm$ side and the $13\,cm$ side (which is parallel to $\vec{B}$) is given by $\cos 	heta = \frac{5}{13}$ and $\sin 	heta = \frac{12}{13}$.Substituting these values into the force formula:$F = (2\,A) 	imes (5 	imes 10^{-2}\,m) 	imes (0.75\,T) 	imes \sin 	heta$$F = 2 	imes 0.05 	imes 0.75 	imes \frac{12}{13}$$F = 0.1 	imes 0.75 	imes \frac{12}{13} = 0.075 	imes \frac{12}{13} = \frac{3}{40} 	imes \frac{12}{13} = \frac{3 	imes 3}{10 	imes 13} = \frac{9}{130}\,N$.Comparing this with $\frac{x}{130}\,N$,we get $x = 9$.

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