A straight wire AB of mass 40,g and length 50 | vedclass

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(B) Given:Mass of the wire,$m = 40\,g = 40 	imes 10^{-3}\,kg$Length of the wire,$\ell = 50\,cm = 50 	imes 10^{-2}\,m = 0.5\,m$Magnetic field,$B = 0.

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$2$

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(B) Given:Mass of the wire,$m = 40\,g = 40 	imes 10^{-3}\,kg$Length of the wire,$\ell = 50\,cm = 50 	imes 10^{-2}\,m = 0.5\,m$Magnetic field,$B = 0.40\,T$Acceleration due to gravity,$g = 10\,ms^{-2}$To remove the tension in the supporting leads,the upward magnetic force on the wire must balance the downward gravitational force (weight) of the wire.Magnetic force,$F_m = I\ell B$Weight of the wire,$W = mg$For equilibrium,$F_m = W$$I\ell B = mg$Substituting the values:$I 	imes 0.5 	imes 0.4 = 40 	imes 10^{-3} 	imes 10$$I 	imes 0.2 = 0.4$$I = \frac{0.4}{0.2} = 2\,A$Thus,the magnitude of the current required is $2\,A$.

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