Question

The currents through two thin long parallel wires, A and B carrying \(I\) and \(\frac{I}{3}\) respectively, in the same direction. The distance between two wires is '\(d\)'. The magnitude of force per unit length exerted by wire A on wire B is

• A. \(\frac{\mu _{0}I^{2}}{6\pi d}\)
• B. \(\frac{\mu _{0}I^{2}}{3\pi d}\)
• C. \(\frac{\mu _{0}I}{3\pi d}\)
• D. \(\frac{\mu _{0}I}{6\pi d}\)

Hint:

Use the force per unit length between parallel wires, \(\frac{\mu_{0}I_{1}I_{2}}{2\pi d}\), with currents \(I\) and \(\frac{I}{3}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The force per unit length between two long parallel wires is \(\frac{F}{l}=\frac{\mu_{0}I_{1}I_{2}}{2\pi d}\).

Step 2: Given Data
\(I_{1}=I\) and \(I_{2}=\frac{I}{3}\).

Step 3: Calculation
\(\frac{F}{l}=\frac{\mu_{0}\cdot I\cdot\frac{I}{3}}{2\pi d}=\frac{\mu_{0}I^{2}}{6\pi d}\).

Step 4: Compare the options
This is option (A).

Final Answer:
The force per unit length is \(\frac{\mu_{0}I^{2}}{6\pi d}\). \[ \boxed{\text{(A) }\dfrac{\mu_{0}I^{2}}{6\pi d}} \]