Question

A wire ABCDEF (with each side of length L) bent as shown in figure and carrying a current I is placed in a magnetic field. Force on the wire is :

• A. BiL along z-axis
• B. BiL along -z axis
• C. BiL along +x-axis
• D. BiL along -x-axis

Answer 1

The correct option is A BiL along z-axis

Concider figure given in problems

The expression of force is $\overrightarrow{F}=I(\overrightarrow{l}\times \overrightarrow{B})$

The angle between $\overrightarrow{FE}$ and $\overrightarrow{B}$ is $180°$

And betwwen $\overrightarrow{BA}$ and $\overrightarrow{B}$ is $0°$

For both of these $\overrightarrow{L}\times \overrightarrow{B}=0$

The angle between $\overrightarrow{ED}$ and $\overrightarrow{B}$ is $90°$

Force on $ED$ is $(\overrightarrow{L}\times \overrightarrow{B})$ $=ILB\sin 90°=ILB$

The angle between $\overrightarrow{CB}$ and $\overrightarrow{B}$ is $270°$

Force on $CB$ is $(\overrightarrow{L}\times \overrightarrow{B})$ $=ILB\sin 270°=-ILB$

The force on $ED$ or $CB$ acts perpendicular to both $\overrightarrow{ED}$ and $\overrightarrow{B}$ or $\overrightarrow{CB}$ or $\overrightarrow{B}$ towards $x$-axis.

These two factors cancel each other as they act in opposite direction.

The angle between $\overrightarrow{DC}$ and $\overrightarrow{B}$ is $90°$

Force on $DC$ is $(\overrightarrow{L}\times \overrightarrow{B})$ $=ILB\sin 90°=ILB$

This force will act along $Z$ axis

So, $BIL$ is along $Z$ axis