Question

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

पर्याय

• 1 : 2

• 1 : 4

• 1 : 5

• 1 : 3

Answer 1

1 : 4

Explanation:

Magnetic force per unit length on wire PQ due to wire RS is given by:

`F/l = mu_0/(4 pi) (2I_1 I_2)/r`

∴ F₁ ∝ I₁ I₂ ∝ (I)²

When the ammeter current reading is changed to 2I.

F₂ ∝ (2I)²

∴ F₁ : F₂ = 1 : 4