Question

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 $F$, the net force on the remaining three arms of the loop is

• A. $3\overrightarrow{F}$
• B. $-\overrightarrow{F}$
• C. $-3\overrightarrow{F}$
• D. $\overrightarrow{F}$

Answer 1

The correct option is B

$-\overrightarrow{F}$

Explanation for the correct answer:

In the case of option B.

Step 1. Given data:

Let us consider ABCD as a square loop, which is suspended in a uniform magnetic field $B$ as shown above.

Force on one arm of the loop = $F$,

Step 2. Formula used:

1. As shown in the figure, arms AB and CD are parallel to the direction of magnetic field.
2. The force on these two arms parallel to the field is zero.
3. Because $\overrightarrow{F}=i\oint \left(\overrightarrow{dl}\times \overrightarrow{B}\right)$, where $\overrightarrow{F}$=Force, $i$=current,
4. Since $\oint \left(dl\times B=0\right)$, where $\overrightarrow{dl}$=displacement, $\overrightarrow{B}$=Magnetic field,

Step 3. Calculations:

1. Hence, for parallel lines CD and AB, $F_{CD}=F_{AB}=0$
2. This is because the magnitude of the force depends on the sin of the angle between the direction of the current and the direction of the magnetic field.
3. So if the current-carrying wire is held parallel to the magnetic field, the force will be zero.
4. Now, the net force acting on the closed loop is zero. Hence

$F_{AB}+F_{BC}+F_{CD}+F_{DA}=0$

Or $$\begin{gathered} 0+F_{BC}+0+F_{DA}=0 \\ Or{F}_{DA}=-F_{BC} \end{gathered}$$

If we consider, force in arm BC is given as $F$

Force on remaining arms $=-F$

Thus the correct option is option B.