Question

A, B and C are three parallel conductors of equal lengths carrying currents $I$, $I$ and $2I$ respectively. Distance between A and B is $x$ and that between B and C is also $x$. $F_1$ is the force exerted by conductor B on A. $F_2$ is the force exerted by conductor C on A. Current $I$ in A and $I$ in B are in the same direction and current $2I$ in C is in the opposite direction. Then

• A. $F_1 = F_2$
• B. $F_2 = 2F_1$
• C. $F_1 = 2F_2$
• D. $F_1 = -F_2$

Hint:

Notice that for wire C, both the current doubles ($2I$) and the distance doubles ($2x$). Since force is directly proportional to current but inversely proportional to distance ($F \propto \frac{I}{r}$), these two changes cancel out perfectly, meaning the magnitude of the force remains completely unchanged!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We have three parallel current-carrying wires arranged in a line: A, B, and C. The distance between adjacent wires is $x$, making the total distance between wire A and wire C equal to $2x$. We need to compare the vector forces $F_1$ (force on A due to B) and $F_2$ (force on A due to C).

Step 2: Key Formula or Approach:
The magnetic force per unit length between two parallel conductors carrying currents $I_a$ and $I_b$ separated by a distance $r$ is given by Ampere's Force Law: $$\frac{F}{L} = \frac{\mu_0 I_a I_b}{2\pi r} \implies F = \frac{\mu_0 I_a I_b L}{2\pi r}$$ Direction rules: 1. Parallel currents running in the

same direction attract each other. 2. Parallel currents running in

opposite directions repel each other.

Step 3: Detailed Explanation:
Let's analyze the direction of the forces acting on conductor A: Conductor A and B carry current $I$ in the same direction, so they attract. Therefore, the force $F_1$ exerted by B on A points to the right (towards B). Conductor A (current $I$) and C (current $2I$) carry currents in opposite directions, so they repel. Therefore, the force $F_2$ exerted by C on A points to the left (away from C). Since $F_1$ and $F_2$ point in opposite directions, they have opposite signs as vectors. Now let's compute and compare their magnitudes: For $F_1$ (force between A and B, distance $= x$): $$F_1 = \frac{\mu_0 \cdot I \cdot I \cdot L}{2\pi x} = \frac{\mu_0 I^2 L}{2\pi x}$$ For $F_2$ (force between A and C, distance $= 2x$): $$F_2 = \frac{\mu_0 \cdot I \cdot 2I \cdot L}{2\pi (2x)} = \frac{2\mu_0 I^2 L}{4\pi x} = \frac{\mu_0 I^2 L}{2\pi x}$$ The magnitudes are exactly equal ($|F_1| = |F_2|$). Combining this with their opposing directions yields the vector relationship: $$F_1 = -F_2$$

Step 4: Final Answer:
The relationship between the forces is $F_1 = -F_2$, corresponding to option (D).