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:
Show Hint
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 expressed as:
\[
F/\ell = \frac{\mu_0 I_1 I_2}{4\pi r},
\]
where:
- \( \mu_0 \) is the permeability of free space,
- \( I_1 \) and \( I_2 \) are the currents in the two wires,
- \( r \) is the separation between the wires.
Step 1: Analyze Proportionality for \( F/\ell \)
The magnetic force is proportional to the product of the currents:
\[
F_1 \propto I^2 \quad \text{(when current is \( I \))}.
\]
When the current is increased to \( 2I \):
\[
F_2 \propto (2I)^2 = 4I^2.
\]
Step 2: Calculate the Ratio of Forces
The ratio of the magnetic forces is:
\[
\frac{F_1}{F_2} = \frac{I^2}{4I^2} = \frac{1}{4}.
\]
Therefore, the ratio of the forces is:
\[
F^{I}_{PQ} : F^{2I}_{PQ} = 1:4.
\]
Final Answer:
\[
\boxed{1:4}
\]
