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Question

(B) For a closed current-carrying loop placed in a uniform magnetic field,the net magnetic force on the entire loop is given by $\overrightarrow{F}_{n

Vedclass · Accepted Answer

$-\overrightarrow{F}$

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(B) For a closed current-carrying loop placed in a uniform magnetic field,the net magnetic force on the entire loop is given by $\overrightarrow{F}_{net} = I(\oint d\overrightarrow{l}) 	imes \overrightarrow{B}$.Since the loop is closed,the vector sum of the length elements $\oint d\overrightarrow{l} = 0$,therefore $\overrightarrow{F}_{net} = 0$.Let the forces on the four arms of the square loop be $\overrightarrow{F}_1, \overrightarrow{F}_2, \overrightarrow{F}_3,$ and $\overrightarrow{F}_4$.The net force on the loop is $\overrightarrow{F}_1 + \overrightarrow{F}_2 + \overrightarrow{F}_3 + \overrightarrow{F}_4 = 0$.Given that the force on one arm is $\overrightarrow{F}_1 = \overrightarrow{F}$,the sum of the forces on the remaining three arms is $\overrightarrow{F}_2 + \overrightarrow{F}_3 + \overrightarrow{F}_4 = -\overrightarrow{F}_1 = -\overrightarrow{F}$.

$A$ square current-carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\overrightarrow{F}$,the net force on the remaining three arms of the loop is

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