Question

A current carrying rectangular loop $P Q R S$ 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 $\left(f_{P Q}^I: f_{P Q}^{2 I}\right)$ is:

• A. $1: 5$
• B. $1: 3$
• C. $1: 2$
• D. $1: 4$

Answer 1

The Correct Option is D

To solve this problem, we need to calculate the magnetic force per unit length between two parallel current-carrying conductors, specifically wires \(PQ\) and \(RS\) of the rectangular loop.

The formula for the magnetic force per unit length \(f\) between two parallel wires carrying currents \(I_1\) and \(I_2\) and separated by distance \(d\) is given by:

\(f = \frac{\mu_0 I_1 I_2}{2\pi d}\)

In this scenario, initially, both wires \(PQ\) and \(RS\) carry the same current \(I\). When the current in the circuit changes from \(I\) to \(2I\), we need to find the ratio of the forces per unit length in these two conditions.

1. The initial force per unit length \(f_{PQ}^I\) when the current is \(I\) can be expressed as: 
   \(f_{PQ}^I = \frac{\mu_0 I \cdot I}{2\pi d} = \frac{\mu_0 I^2}{2\pi d}\)
2. When the current changes to \(2I\), the force per unit length \(f_{PQ}^{2I}\) becomes: 
   \(f_{PQ}^{2I} = \frac{\mu_0 (2I) \cdot (2I)}{2\pi d} = \frac{\mu_0 (2I)^2}{2\pi d} = \frac{\mu_0 \cdot 4I^2}{2\pi d}\)

Now, calculate the ratio: 
\(\frac{f_{PQ}^I}{f_{PQ}^{2I}} = \frac{\frac{\mu_0 I^2}{2\pi d}}{\frac{\mu_0 \cdot 4I^2}{2\pi d}} = \frac{I^2}{4I^2} = \frac{1}{4}\)

Thus, the ratio of magnetic forces per unit length when the current is \(I\) and \(2I\) is \(1:4\).

Therefore, the correct answer is: \(1:4\)