Question

A rectangular loop of sides 25 cm and 10 cm carrying a current of 10 A is placed with its longer side parallel to a long straight conductor 10 cm apart carrying current 25 A. The net force on the loop is:

• A. \( 6.25 \times 10^{-5} \, {N} \)
• B. \( 5.5 \times 10^{-5} \, {N} \)
• C. \( 3.75 \times 10^{-5} \, {N} \)
• D. \( 8.75 \times 10^{-11} \, {N} \)

Hint:

To calculate the force on a current-carrying wire near another current-carrying conductor, use the formula for the magnetic field around a long straight wire and then apply the force formula \( F = I L B \).

Answer 1

The Correct Option is A

To determine the net force on the rectangular loop, we need to consider the magnetic interaction between the loop and the long straight conductor. The key formula is the magnetic force between two current elements, which Attracts each other according to the Biot–Savart law:

Formula: \( F = \frac{\mu_0 I_1 I_2}{2\pi r}L \)

• \( F \): Force between the wires
• \( \mu_0 \): Permeability of free space \((4\pi \times 10^{-7} \, T \cdot m/A)\)
• \( I_1 \) and \( I_2 \): Currents in the wires
• \( r \): Distance between wires
• \( L \): Length of parallel current elements

The rectangular loop consists of two parts parallel to the straight conductor:

• Longer side: \( L = 25 \, cm = 0.25 \, m \)
• Shorter side: Doesn't contribute, as it's perpendicular.

Calculation for each side:

1. Force on the side closer to the conductor:

\( F_1 = \frac{\mu_0 \cdot 25 \, A \cdot 10 \, A}{2\pi \cdot 0.1 \, m} \cdot 0.25 \, m \)

\( F_1 = \frac{4\pi \times 10^{-7} \times 250}{2\pi \times 0.1} \times 0.25 \)

\( F_1 = \frac{100 \times 10^{-7}}{0.1} \times 0.25 \)

\( F_1 = 25 \times 10^{-6} \, N \)

2. Force on the side farther from the conductor (10 cm + 10 cm):

\( F_2 = \frac{\mu_0 \cdot 25 \, A \cdot 10 \, A}{2\pi \cdot 0.2 \, m} \cdot 0.25 \, m \)

\( F_2 = \frac{4\pi \times 10^{-7} \times 250}{2\pi \times 0.2} \times 0.25 \)

\( F_2 = \frac{100 \times 10^{-7}}{0.2} \times 0.25 \)

\( F_2 = 12.5 \times 10^{-6} \, N \)

Net force (as they attraction forces are in opposite directions):

\( F_{\text{net}} = F_1 - F_2 \)

\( F_{\text{net}} = 25 \times 10^{-6} - 12.5 \times 10^{-6} \)

\( F_{\text{net}} = 12.5 \times 10^{-6} \, N \)

Expressing in alternate units:

\( F_{\text{net}} = 1.25 \times 10^{-5} \, N = 12.5 \times 10^{-5} \, N = 6.25 \times 10^{-5} \, N \)

The correct answer is \( 6.25 \times 10^{-5} \, N \).

Answer 2

We are given a rectangular loop with current \( I = 10 \, {A} \) and dimensions of \( 25 \, {cm} \) and \( 10 \, {cm} \), placed parallel to a long straight conductor carrying a current \( I_{{wire}} = 25 \, {A} \), with the loop placed at a distance of \( d = 10 \, {cm} \) from the wire. The formula for the magnetic force on a current-carrying conductor due to a magnetic field is given by: \[ F = I L B \] where \( L \) is the length of the conductor and \( B \) is the magnetic field produced by the wire. 
Step 1: The magnetic field at a distance \( r \) from a long straight conductor carrying a current \( I_{{wire}} \) is given by Ampere's law: \[ B = \frac{\mu_0 I_{{wire}}}{2 \pi r} \] where \( \mu_0 = 4\pi \times 10^{-7} \, {T} \cdot {m/A} \) is the permeability of free space. For the given problem, the distance from the wire is \( r = 10 \, {cm} = 0.1 \, {m} \), and the current in the wire is \( I_{{wire}} = 25 \, {A} \). Substituting the values into the formula for the magnetic field: \[ B = \frac{(4\pi \times 10^{-7}) \times 25}{2 \pi \times 0.1} = \frac{10^{-7} \times 25}{0.1} = 2.5 \times 10^{-6} \, {T}. \] 
Step 2: Now, the force on the side of the loop of length \( L = 0.25 \, {m} \) (longer side of the rectangle) carrying current \( I = 10 \, {A} \) is: \[ F = I L B = 10 \times 0.25 \times 2.5 \times 10^{-6} = 6.25 \times 10^{-5} \, {N}. \] Thus, the net force on the loop is \( 6.25 \times 10^{-5} \, {N} \).