Question

A rectangular loop of sides 15cm and 10cm ,carrying a current of 1a ,is placed with its longer side parallel to long straight wire carrying a current of 2a placed at a distance of 2cm . The net force experienced by the loop is?

Answer 1

Since the 15 cm sides are placed parallel only these two sides will have forces  acting on them , and since current in both the longer sides will be opposite , hence force acting on them are also opposite in nature(attractive and repulsive).

There will be no force due to the 10 cm sides as they will cancel out each other.

We know that force on a conductor , F = μ₀I₁I₂L/2πr

For the first side placed at 2 cm distance ,

Force, F₁ = μ₀I₁I₂L/2πr

= 4π × 10⁻⁷ ×1×2×0.15/2π×0.02

= 30×10⁻⁷

Similarly force acting on the other side which is at 12cm distance ,

F₂ = 4π × 10⁻⁷ ×1×2×0.15/2π×0.12

= 5 × 10⁻⁷

Since both these forces act in opposite direction,

Hence resultant force F =  F₁-F₂

= 30×10⁻⁷- 5 × 10⁻⁷

= 25 ×10⁻⁷