Question

Two isolated point poles of strength 30 am and 60 am are placed at a distance of 0.3m the force of repulsion between them is

Answer 1

Given ⇒

Mass of the First pole(m₁) = 30 A-m.
Mass of the Second Pole(m₂) = 60 A-m.
Distance between them (r) = 0.3 m.

We know, μ₀/4π = 10⁻⁷ N/A²

Now, Using the Formula,

 Force of Repulsion (F) = μ₀/4π × m₁m₂/r²
⇒ F = 10⁻⁷ × 30 × 60/(0.3)²
⇒ F = 10⁻⁷ × 1800/0.09
⇒ F = 20000 × 10⁻⁷
∴ F = 2 × 10⁻³ N.


Hence, the Force of Repulsion between the Isolated point poles is 2 × 10⁻³ N.


Hope it helps.