Force Between Parallel Current Carrying Wires

 

❓ Question

Three long parallel wires carry currents:

• Left wire = $3\,A$
  
• Middle wire = $1\,A$
  
• Right wire = $2\,A$
  

Distances:

• Left ↔ Middle = $3\,cm$
  
• Middle ↔ Right = $2\,cm$
  

Find the net force on $15\,cm$ length of the middle wire.

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✍️ Short Explanation

Force between two parallel wires:

$$F=\frac{\mu_0 I_1 I_2 L}{2\pi d}$$

Since all currents are in the same direction (out of the plane), forces are attractive.

Thus:

• Left wire pulls middle wire to the left.
• Right wire pulls middle wire to the right.

Net force = difference of the two forces.

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🔷 Step 1 — Force due to Left Wire

$$I_1=3A,\quad I_2=1A$$
$$d=3\,cm=0.03\,m$$
$$L=15\,cm=0.15\,m$$
$$F_L= \frac{4\pi\times10^{-7}}{2\pi} \cdot \frac{(3)(1)(0.15)}{0.03}$$
$$= 2\times10^{-7}\times15$$
$$= 3\times10^{-6}\,N$$
$$F_L=3\,\mu N$$

(towards left)

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🔷 Step 2 — Force due to Right Wire

$$I_1=2A,\quad I_2=1A$$
$$d=2\,cm=0.02\,m$$
$$F_R= \frac{4\pi\times10^{-7}}{2\pi} \cdot \frac{(2)(1)(0.15)}{0.02}$$
$$= 2\times10^{-7}\times15$$
$$= 3\times10^{-6}\,N$$
$$F_R=3\,\mu N$$

(towards right)

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🔷 Step 3 — Net Force

$$F_{\text{net}} = |F_R-F_L|$$
$$= |3-3|\,\mu N$$
$$= 0$$

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✅ Final Answer

$$\boxed{0\ \mu N}$$

The forces on the middle wire due to the left and right wires are equal in magnitude and opposite in direction, so they cancel completely.

Note: The options shown in the image (1, 5, 6, 7 μN) do not include 0. Based on the given currents and distances, the correct physics calculation gives 0 μN, suggesting there may be a printing error in the question/options.

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