Question

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: \begin{center} \includegraphics[width=5cm]{21.png} \end{center}

• A. \( 1:2 \)
• B. \( 1:4 \)
• C. \( 1:5 \)
• D. \( 1:3 \)

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.

Answer 1

The Correct Option is B

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: \[ \boxed{1:4} \]