Question

Three very long parallel wires carrying current as shown. Find the force acting at 15 cm length of middle wire : 

• A. 1 \(\mu\)N
• B. 6 \(\mu\)N
• C. 7 \(\mu\)N
• D. 5 \(\mu\)N

Hint:

Direction is crucial. Always verify if the forces add up or cancel out based on current directions before calculating magnitudes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Parallel wires carrying currents in the same direction attract each other.
Parallel wires carrying currents in opposite directions repel each other.
Step 2: Key Formula or Approach:
The magnetic force per unit length between two wires is:
\[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r} \]
Force on length \(L\): \(F = \frac{\mu_0 I_1 I_2 L}{2\pi r}\).
Step 3: Detailed Explanation:
Let the wires be A (3A), B (1A), and C (2A). We need the force on wire B.
Distance \(AB = 3 \text{ cm} = 0.03 \text{ m}\).
Distance \(BC = 2 \text{ cm} = 0.02 \text{ m}\).
Length \(L = 15 \text{ cm} = 0.15 \text{ m}\).
1. Force on B due to A (\(F_{BA}\)): Currents are opposite (3A up, 1A down), so it is repulsive.
Wire B is pushed to the right by wire A.
\[ F_{BA} = \frac{2 \times 10^{-7} \times 3 \times 1 \times 0.15}{0.03} = \frac{0.9 \times 10^{-7}}{0.03} = 30 \times 10^{-7} = 3 \text{ \(\mu\)N (Right)} \]
2. Force on B due to C (\(F_{BC}\)): Currents are in the same direction (1A down, 2A down), so it is attractive.
Wire B is pulled to the right by wire C.
\[ F_{BC} = \frac{2 \times 10^{-7} \times 1 \times 2 \times 0.15}{0.02} = \frac{0.6 \times 10^{-7}}{0.02} = 30 \times 10^{-7} = 3 \text{ \(\mu\)N (Right)} \]
Since both forces act to the right, the net force is:
\[ F_{net} = 3 \text{ \(\mu\)N} + 3 \text{ \(\mu\)N} = 6 \text{ \(\mu\)N} \]
Step 4: Final Answer:
The net force acting on the 15 cm length of the middle wire is 6 \(\mu\)N.