A current-carrying rectangular loop PQRS is made of uniform wire. The length PR = QS = \( 5 \, \text{cm} \) and PQ = RS = \( 100 \, \text{cm} \). If the 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 \( F^{I}_{PQ} : F^{2I}_{PQ} \) is:
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The magnetic force between parallel wires varies with the product of the currents. If a current doubles, the force increases by the square of the change factor.
The magnetic force per unit length \( F/\ell \) between two parallel current-carrying wires is given by: \[F/\ell = \frac{\mu_0 I_1 I_2}{4\pi r},\] where \( \mu_0 \) is the permeability of free space, \( I_1 \) and \( I_2 \) are the currents in the wires, and \( r \) is the separation distance. Step 1: Analyze the Proportionality of \( F/\ell \) The magnetic force is directly proportional to the product of the currents. If the current in a single wire is \( I \), the force is proportional to \( I^2 \): \[F_1 \propto I^2 \quad \text{(for current \( I \))}.\] If the current is doubled to \( 2I \), the force becomes proportional to: \[F_2 \propto (2I)^2 = 4I^2.\] Step 2: Calculate the Force Ratio The ratio of the forces is: \[\frac{F_1}{F_2} = \frac{I^2}{4I^2} = \frac{1}{4}.\] Thus, the ratio of the forces when the current changes from \( I \) to \( 2I \) is: \[F^{I}_{PQ} : F^{2I}_{PQ} = 1:4.\] Final Answer: \[\boxed{1:4}\]