Question:
Two long and parallel straight wires A and B carrying current 5 A each in the same direction. If the force on a 5 cm section of wire B is \(5 \times 10^{-6}\, \text{N}\), then the separation between the two wires is (Given \( \mu_0 = 4\pi \times 10^{-7} \, \text{SI} \)).
Two long and parallel straight wires A and B carrying current 5 A each in the same direction. If the force on a 5 cm section of wire B is \(5 \times 10^{-6}\, \text{N}\), then the separation between the two wires is (Given \( \mu_0 = 4\pi \times 10^{-7} \, \text{SI} \)).
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For parallel wires, always use \(F = \frac{\mu_0 I_1 I_2}{2\pi d}L\) when length is given.
Updated On: Jun 20, 2026
- 2.5 cm
- 5 cm
- 7.5 cm
- 10 cm
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The Correct Option is B
Solution and Explanation
Step 1: Understand force between parallel current carrying wires.
Force per unit length between two long parallel currents is: \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]
Step 2: Write given values.
\[ I_1 = I_2 = 5\,A,\quad L = 5\,cm = 0.05\,m,\quad F = 5 \times 10^{-6}\,N \]
Step 3: Substitute in total force formula.
\[ F = \frac{\mu_0 I_1 I_2}{2\pi d} \cdot L \] \[ 5 \times 10^{-6} = \frac{4\pi \times 10^{-7} \cdot 25}{2\pi d} \cdot 0.05 \]
Step 4: Simplify expression.
Cancel \(\pi\): \[ 5 \times 10^{-6} = \frac{4 \times 10^{-7} \cdot 25}{2d} \cdot 0.05 \] \[ = \frac{100 \times 10^{-7}}{2d} \cdot 0.05 \] \[ = \frac{50 \times 10^{-7} \cdot 0.05}{d} \] \[ = \frac{2.5 \times 10^{-7}}{d} \]
Step 5: Solve for separation \(d\).
\[ d = \frac{2.5 \times 10^{-7}}{5 \times 10^{-6}} \] \[ d = 0.05\,m \]
Step 6: Convert to cm.
\[ d = 5\,cm \]
Step 7: Final conclusion.
\[ \boxed{5\,cm} \]
Force per unit length between two long parallel currents is: \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]
Step 2: Write given values.
\[ I_1 = I_2 = 5\,A,\quad L = 5\,cm = 0.05\,m,\quad F = 5 \times 10^{-6}\,N \]
Step 3: Substitute in total force formula.
\[ F = \frac{\mu_0 I_1 I_2}{2\pi d} \cdot L \] \[ 5 \times 10^{-6} = \frac{4\pi \times 10^{-7} \cdot 25}{2\pi d} \cdot 0.05 \]
Step 4: Simplify expression.
Cancel \(\pi\): \[ 5 \times 10^{-6} = \frac{4 \times 10^{-7} \cdot 25}{2d} \cdot 0.05 \] \[ = \frac{100 \times 10^{-7}}{2d} \cdot 0.05 \] \[ = \frac{50 \times 10^{-7} \cdot 0.05}{d} \] \[ = \frac{2.5 \times 10^{-7}}{d} \]
Step 5: Solve for separation \(d\).
\[ d = \frac{2.5 \times 10^{-7}}{5 \times 10^{-6}} \] \[ d = 0.05\,m \]
Step 6: Convert to cm.
\[ d = 5\,cm \]
Step 7: Final conclusion.
\[ \boxed{5\,cm} \]
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