Question

A wire is bent in the form of an equilateral triangle $PQR$ of side $20\text{cm}$ and carries a current of $2.5\text{A}$. It is placed in a magnetic field $B$ of magnitude $2.0\text{T}$ directed perpendicularly to the plane of the loop. Find the forces on the three sides of the triangle (Given an integer).

Answer 1

Suppose the field and the current have directions as shown in figure. The force on $PQ$ is$${\overrightarrow{F}}_1=i(\overrightarrow{l}\times \overrightarrow{B})$$$\Rightarrow F_1=2.5\times 20\times \times (10{)}^{-2}\times 2(\sin {90}^{\circ })=1.0\text{N}$

The rule of vector product shows that the force $F_1$ is perpendicular to $PQ$ and is directed towards the inside of the triangle.

Similarly, the forces ${\overrightarrow{F}}_2$ and ${\overrightarrow{F}}_3$ on $QR$ and $RP$ can also be obtained.

Resultant of ${\overrightarrow{F}}_2$ and ${\overrightarrow{F}}_3$

$F_{23}=\sqrt{F_2^2+F_3^2+2F_2{F}_3\cos {120}^{\circ }}$

$=\sqrt{1^2+1^2-2\times 1\times 1\times \frac{1}{2}}=1\text{N}$(Downwards)

Now, resultant of $F_{23}$ and $F_1$

$F_R=F_1-F_{23}=1-1=0$

$\text{Alternative solution:}$

This result can be generalized. Any closed current carrying loop, placed in a homogeneous magnetic field, does not experience a net magnetic force.

Accepted answer : $0$