Question

Same current $I=2A$ is flowing in a wire frame as shown in the figure. The frame is combination of two equilateral triangles ACD and CDE of side $1m$. It is placed in a uniform magnetic field $B=4T$ acting perpendicular to the plane of frame. The magnitude of magnetic force acting on the frame is

• A. $24N$
• B. Zero
• C. $16N$
• D. $8N$

Answer 1

The correct option is A $24N$

Let the wires lie in the x-y plane with the magnetic field in the -z direction as shown in the figure.


Force acting on a current carrying wire in a magnetic field is given by
$\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B}$)
$=ilB\sin \theta \hat{n}$ where $\theta$ is the angle between the current element vector $\overrightarrow{l}$ and the magnetic field vector $\overrightarrow{B}$ and $\hat{n}$ represents the direction of force acting on the wire.

Here, current $i$, length $l$ and magnetic field intensity $B$ are the same for all five wires. The angle $\theta$ is also ${90}^{\circ }$ for each of the wires since magnetic field is in the -z direction. Hence, the magnitude of magnetic force acting on each of the wires is going to be the same.

The direction of the force acting on the wires is different and is given by the direction of $\overrightarrow{l}\times \overrightarrow{B}$. The forces acting on each of the individual wires are shown in the diagram below.


Adding all the forces vectorially,


$F_{net}=2\left(2F\sin {30}^{\circ }\right)+F$
$=4F\frac{1}{2}+F=3F$
$=3ilB\sin {90}^{\circ }$
$=3\times 2\times 1\times 4$
$=24N$