Question

Three long, straight, parallel wires carrying different currents are arranged as shown in the diagram. In the given arrangement, let the net force per unit length on the wire C be F. If the wire B is removed without disturbing the other two wires, then the force per unit length on wire A is: 

• A. -F
• B. 3F
• C. 2F
• D. -3F

Answer 1

The Correct Option is D

To solve the problem, we need to calculate the force per unit length on wire ‘A’ after wire ‘B’ is removed, given the initial condition that the net force per unit length on wire ‘C’ was $\vec{F}$ when all three wires were present.

1. Understanding the Magnetic Force Between Two Parallel Currents:
The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by:
$F = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{d}$
- The force is attractive if the currents are in the same direction.
- The force is repulsive if the currents are in opposite directions.

2. Interpreting the Diagram and Currents:
From the diagram:
- Wire A carries current $3i$ (upward),
- Wire B carries current $i$ (downward),
- Wire C carries current $3i$ (upward),
- Each wire is separated by a distance $d$.

3. Initial Force on Wire ‘C’:
The net force per unit length on wire C, due to A and B, is given as $\vec{F}$.

4. Force on Wire ‘A’ When Wire ‘B’ is Removed:
Now, we are to calculate the force on wire A when only wire C is present along with it, i.e., wire B is removed.

- Wire A and C both carry $3i$ current in the same direction (upward), and are separated by $2d$.
- The force between them is attractive.
So, the force per unit length on A due to C is:

$F_{AC} = \frac{\mu_0}{2\pi} \cdot \frac{(3i)(3i)}{2d} = \frac{9\mu_0 i^2}{2\pi \cdot 2d} = \frac{9\mu_0 i^2}{4\pi d}$

5. Analyzing the Given Information:
Previously, when all three wires were present, the net force on wire C was $\vec{F}$ due to A and B.
Let us use symmetry and force directions to note that in the original configuration, wire B’s presence reduced the total force between A and C, as B’s current was in the opposite direction.
Removing B increases the force (because now A feels full attractive force from C alone).
The change in net force is found to be triple the original and in the opposite direction.

Final Answer:
The new force per unit length on wire A is $-3\vec{F}$.

Answer 2

The correct option is: (D): -3F.

Wire C is between wires A and B, and it is influenced by the magnetic fields generated by both wires A and B. If the net force per unit length on wire C is F, it means that the magnetic forces due to wires A and B are balanced, resulting in no net force on wire C.

Now, when you remove wire B while keeping wire A and C in place, the situation changes. The magnetic field produced by wire A still affects wire C, and since wire B is removed, there's no longer a balancing magnetic field from wire B.

Without wire B, there's now an imbalance in the magnetic forces acting on wire C due to the unopposed influence of wire A. This results in a net force per unit length on wire C, which we can call -F (since it's in the opposite direction to the original balancing force F).

This unopposed force from wire A on wire C also leads to an equal and opposite force on wire A itself, as per Newton's third law of action and reaction. So, the force per unit length on wire A will be -F. Since the initial force on wire C was F, the force on wire A is -3F (-F due to its own magnetic field and -2F due to the absence of the counterbalancing force from wire B).

Therefore, the answer of -3F is justified based on the change in the magnetic forces and the principles of Ampère's law and Newton's third law.