A current carrying rectangular loop PQRS is made of uniform wire. The length P R=Q S=5 cm and P Q=R S=100 cm. If ammeter current reading changes from I to 2 I, the ratio of magnetic forces per unit length on the wire P Q due to wire R S in the two cases respectively (f_(P Q)^I: f_(P Q)²ᵗ) is:

Solution

Physics sir · 4 steps

One supply current feeds both wires, so doubling I doubles both currents and the force rises four times.

1. Split the current. The current enters at the middle of side PR and leaves at the middle of side QS. The upper path (through PQ) and the lower path (through RS) are identical uniform wire, so each carries I/2, in the same direction.
2. Force per length. PQ and RS are parallel wires a distance d = 5 cm apart. Their length of 100 cm is much larger than d, so the long-wire formula applies. Parallel currents attract, with force per unit length:

   f = μ₀/(2π d)(I/2)(I/2) = (μ₀ I²)/(8π d) ∝ I²

3. Double the current. With 2I, each wire carries I, so f becomes (μ₀ I²)/(2π d), which is 4 times the first value.
4. So. f_(PQ)^I : f_(PQ)^(2I) = I² : (2I)² = 1 : 4, option (a).

The idea to keep. Force between two wires goes as the product I₁ I₂. When one supply current feeds both wires, each share scales with it, so the force scales as its square.

Answer from the NTA final answer key. Solution by PrepWiser.