Question

A square of side of $2.0\text{m}$ is placed in a uniform magnetic field $\overrightarrow{B}=2.0\text{T}$ in a direction perpendicular to the plane of the square inwards. Equal current $i=3.0\text{A}$ is flowing in the directions shown in figure. Find the magnitude of magnetic force acting on the loop $ABCD$.

• A. $18\sqrt{2}\text{N}$
• B. $27\sqrt{2}\text{N}$
• C. $36\sqrt{2}\text{N}$
• D. $16\sqrt{2}\text{N}$

Answer 1

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


​​​​​​From the above figure, it is clear that , it is not an example of closed loop. Rather it can be treated as combination of the following parts,


As, $\overrightarrow{F}=I(\overrightarrow{L}\times \overrightarrow{B})$

where, $L$ is effective length.

All three segments have same effective length as $AD=2\sqrt{2}\text{m}\because [AD=\sqrt{A{C}^2+C{D}^2}]$
with some direction from $A\to D$

Since, $\text{Force on ACD=Force on AD=Force on AED}$
$\therefore$Net force on loop $=3{\overrightarrow{F}}_{AD}$

$\Rightarrow F_{net}=3i[\overrightarrow{L}\times \overrightarrow{B}]$

$=3i[\overrightarrow{AD}\times \overrightarrow{B}]$

$=3\times 3\left[\right(AD\left)B\sin {90}^{\circ }\right]$

$=3\times 3[2\sqrt{2}\times 2]$

$\therefore F_{net}=36\sqrt{2}\text{N}$

Hence, option (c) is the correct answer.