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Question: A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the l...

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 isF→\overrightarrow{F}, the net force on the remaining three arms of the loop is
A. 3F→3\overrightarrow{F}
B. −F→-\overrightarrow{F}
C. −3F→-3\overrightarrow{F}
D. F→\overrightarrow{F}

Explanation

Solution

You could make a rough sketch of the given situation so as to have better clarity. Then, you could use the formula as per requirement. You may recall that the force for a closed loop will be zero. We are given force on one of the four arms. We could easily find the net force due to the other three arms.
Formula used:
Magnetic force,
F→=i(∮dl→×B→)\overrightarrow{F}=i\left( \oint{\overrightarrow{dl}\times \overrightarrow{B}} \right)

Complete answer:
In the question, we are given a square current carrying loop that is suspended in a uniform magnetic field that is acting in the plane of the loop. The following figure will give you a clear picture of the situation.

Let us recall that magnetic force could be given by,
dF→=i(dl→×B→)\overrightarrow{dF}=i\left( \overrightarrow{dl}\times \overrightarrow{B} \right)
This expression is for a small element. Now for the whole loop it would be given by,
F→=i(∮dl→×B→)\overrightarrow{F}=i\left( \oint{\overrightarrow{dl}\times \overrightarrow{B}} \right)
We also know that, for a closed loop,
∮dl→=0\oint{\overrightarrow{dl}}=0
Therefore, we could conclude that the net magnetic force would be zero for the given closed loop. We are told that the magnetic force on one of the arms is F→\overrightarrow{F} and the net force due to the other three arms should be −F→-\overrightarrow{F} for the net force due to the loop to be zero.

Hence, option B is the correct answer.

Note:
We could define magnetic force as the attraction or repulsion that is caused by flow of current or the electrically charged particles. So, basically magnetic force is indeed a consequence of electromagnetic force. Another way to find this magnetic force is by using the magnetic Lorentz force.
F→=qv→×B→\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}