A rectangular loop of wire,supporting a mass | vedclass

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Question

(A) The magnetic force on the top segment of the loop (length $a$) is given by $\overrightarrow{F} = i(\overrightarrow{a} 	imes \overrightarrow{B})$.

Vedclass · Accepted Answer

The weight rises due to a vertical force caused by the magnetic field and work is done on the system.

Vedclass · Answer

(A) The magnetic force on the top segment of the loop (length $a$) is given by $\overrightarrow{F} = i(\overrightarrow{a} 	imes \overrightarrow{B})$. Since the current $i$ flows from left to right in the top segment and the magnetic field $\overrightarrow{B}$ is directed into the plane,the force $\overrightarrow{F}$ acts vertically upwards.Magnitude of the magnetic force is $F = iBa$.Given that $i > mg/Ba$,we have $iBa > mg$,which means the upward magnetic force is greater than the downward gravitational force $mg$.As a result,the net force on the loop is upward,causing the mass $m$ to rise.Since the loop moves against the gravitational force,work is done on the system by the external source (the battery driving the current $i$).Therefore,the weight rises and work is done on the system.

$A$ rectangular loop of wire,supporting a mass $m$,hangs with one end in a uniform magnetic field $\overrightarrow{B}$ pointing into the plane of the paper. $A$ clockwise current $i$ is set up such that $i > mg/Ba$,where $a$ is the width of the loop. Then:

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