Question

A square of side $2\text{m}$ is placed in a uniform magnetic field $B=2\text{T}$ in a direction perpendicular to the plane of square and pointing inwards Equal current $i=3\text{A}$ is flowing in the direction shown in figure. The magnitude of magnetic force acting on the loop will be :

• A. $36\sqrt{2}\text{N}$
• B. $20\text{N}$
• C. $12\sqrt{2}\text{N}$
• D. $45\text{N}$

Answer 1

The correct option is A $36\sqrt{2}\text{N}$


If we look at the section $\text{ACD}$ of the current carrying wire, then $\text{AD}$ is the length of effective length vector.

Thus, force on section $\text{ACD}$ is given by,

${\overrightarrow{F}}_{ACD}=i(\overrightarrow{L}\times \overrightarrow{B})$

$|{\overrightarrow{F}}_{ACD}|=\left(Bi{L}_{AD}\right)\sin {90}^{\circ }=Bi{L}_{AD}$

Again, the magnetic force on section $\text{AED}$ will be given by :

${\overrightarrow{F}}_{AED}=i(\overrightarrow{L}\times \overrightarrow{B})$

Here $\overrightarrow{L}={\overrightarrow{L}}_{AD}$

$|{\overrightarrow{F}}_{AED}|=\left(Bi{L}_{AD}\right)\sin {90}^{\circ }=Bi{L}_{AD}$

Similarly on wire segment $\text{AD}$.

${\overrightarrow{F}}_{AD}=i(\overrightarrow{L}\times \overrightarrow{B})$

$|{\overrightarrow{F}}_{AD}|=\left(Bi{L}_{AD}\right)\sin {90}^{\circ }=Bi{L}_{AD}$

The direction of magnetic force on each section $\text{ACD}$, $\text{AED}$ & $\text{AD}$ will be in same direction.

$\Rightarrow |{\overrightarrow{F}}_{net}|=3\left(Bi{L}_{AD}\right)$

or, $|{\overrightarrow{F}}_{net}|=3\times 2\times 3\times (\sqrt{2}\times 2)$

$\therefore |{\overrightarrow{F}}_{net}|=36\sqrt{2}\text{N}$

Why this question ? Caution: The loop $\text{ACDEA}$ cannot be considered as a closed loop because the direction of current in the loop does not have same sense, Hence $F_{ACDEA}\ne 0$