Question

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by \(15\) cm length of wire \(Q\) is ________. (\( \mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1} \)) 

• A. \(6\times10^{-7}\,\text{N}\) towards \(P\)
• B. \(6\times10^{-6}\,\text{N}\) towards \(P\)
• C. \(6\times10^{-7}\,\text{N}\) towards \(R\)
• D. \(6\times10^{-6}\,\text{N}\) towards \(R\)

Hint:

Always determine attraction or repulsion first using current directions before applying the force formula.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Parallel wires with like currents attract; unlike currents repel. The total force on wire \(Q\) is the vector sum of forces from \(P\) and \(R\).
Step 2: Key Formula or Approach:
Force per unit length: \(F/L = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{2 \times 10^{-7} I_1 I_2}{d}\).
Step 3: Detailed Explanation:
Given: \(I_P = 3 \text{ A (down)}\), \(I_Q = 1 \text{ A (up)}\), \(I_R = 2 \text{ A (up)}\).
\(d_{PQ} = 0.03 \text{ m}\), \(d_{QR} = 0.02 \text{ m}\).
1. Force on Q due to P (\(F_{PQ}\)):
Opposite directions \(\Rightarrow\) Repulsion. Q is pushed to the Right (towards R).
\[ F_{PQ}/L = \frac{2 \times 10^{-7} \times 3 \times 1}{0.03} = 2 \times 10^{-5} \text{ N/m} \]
2. Force on Q due to R (\(F_{RQ}\)):
Same directions \(\Rightarrow\) Attraction. Q is pulled to the Right (towards R).
\[ F_{RQ}/L = \frac{2 \times 10^{-7} \times 2 \times 1}{0.02} = 2 \times 10^{-5} \text{ N/m} \]
3. Net Force Calculation:
Both forces act towards R.
\[ F_{net}/L = (2 + 2) \times 10^{-5} = 4 \times 10^{-5} \text{ N/m} \]
For \(L = 0.15 \text{ m}\):
\[ F = (4 \times 10^{-5}) \times 0.15 = 0.6 \times 10^{-5} = 6 \times 10^{-6} \text{ N} \]
Step 4: Final Answer:
The force is \(6 \times 10^{-6} \text{ N}\) towards \(R\).