Question

Two long conductors, separated by a distance $d$ carry currents $I_1$ and $I_2$ in the same directions. They exert a force $F$ on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to $3d$. The new value of the force between them is

• A. $-F$
• B. $\frac{F}{3}$
• C. $-\frac{2F}{3}$
• D. $-2F$

Hint:

Handle this problem instantly using quick ratios! Reversing the current switches the force polarity, yielding a negative sign. Doubling one current scales the numerator by 2, and tripling the distance scales the denominator by 3. Multiplying these factors gives $(-1) \times \frac{2}{3} = -\frac{2}{3}$ instantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem addresses the magnetic force per unit length operating between two infinitely long, parallel, current-carrying conductors.
Initially, they carry currents $I_1$ and $I_2$ in the same direction, separated by a distance $d$, yielding an attractive force magnitude $F$.
Then, one current is doubled ($2I_2$), its direction is reversed, and the separation is expanded to $3d$. We must find the modified force vector $F'$.

Step 2: Key Formula or Approach:
The magnetic force per unit length $F$ interacting between two parallel wires separated by distance $d$ is defined by Ampere's Law:
$$F = \frac{\mu_0 I_1 I_2}{2\pi d}$$ By convention, currents flowing in parallel directions create an attractive force ($+F$), while anti-parallel configurations create a repulsive force (represented by a negative sign, $-F$).

Step 3: Detailed Explanation:
Let's establish the initial state equation:
$$F = \frac{\mu_0 I_1 I_2}{2\pi d}$$ Now, let's construct the expression for the updated configuration based on the described parameter updates:
* New current 1: $I_1' = I_1$
* New current 2: $I_2' = 2I_2$ (with direction reversed)
* New separation: $d' = 3d$
Substitute these terms into the force relation:
$$F' = -\frac{\mu_0 I_1 (2I_2)}{2\pi (3d)}$$ The negative sign explicitly accounts for the transition from an attractive force to a repulsive force caused by reversing the current's direction.
Factoring out the numerical coefficients:
$$F' = -\frac{2}{3} \left( \frac{\mu_0 I_1 I_2}{2\pi d} \right)$$ Substitute the baseline force identity ($F$) back into the bracketed statement:
$$F' = -\frac{2}{3}F$$

Step 4: Final Answer:
The new value of the force between them is $-\frac{2F}{3}$, which corresponds to option (C).