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

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Solution

Suppose the field and the current have directions as shown in figure. The force on $P Q$ is ${\to F}_{1} = i (\to l \times \to B)$ $\Rightarrow F_{1} = 2.5 \times 20 \times \times (10)^{- 2} \times 2 (sin 90^{\circ}) = 1.0 N$

The rule of vector product shows that the force $F_{1}$ is perpendicular to $P Q$ and is directed towards the inside of the triangle.

Similarly, the forces ${\to F}_{2}$ and ${\to F}_{3}$ on $Q R$ and $R P$ can also be obtained.

Resultant of ${\to F}_{2}$ and ${\to F}_{3}$

$F_{23} = \sqrt{F_{2}^{2} + F_{3}^{2} + 2 F_{2} F_{3} cos 120^{\circ}}$

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

Now, resultant of $F_{23}$ and $F_{1}$

$F_{R} = F_{1} - F_{23} = 1 - 1 = 0$

$Alternative solution:$

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 : $0$

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

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