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: State the force law.
The force per unit length between two long parallel wires is \[ F = \frac{\mu_0 I_1 I_2}{2\pi d}. \] Parallel currents attract (take this as positive), antiparallel currents repel (negative).
Step 2: Record the initial force.
With currents $I_1, I_2$ in the same direction at separation $d$, the force magnitude is $F$ and it is attractive.
Step 3: List the changes.
One current becomes $2I_2$, its direction is reversed, and the separation grows to $3d$.
Step 4: Build the new force.
\[ F' = -\frac{\mu_0 I_1 (2I_2)}{2\pi(3d)}, \] where the minus sign marks the switch from attraction to repulsion.
Step 5: Factor out the original force.
\[ F' = -\frac{2}{3}\cdot\frac{\mu_0 I_1 I_2}{2\pi d} = -\frac{2}{3}F. \]
Step 6: Conclude.
The new force is $-\dfrac{2F}{3}$, option (C); the negative sign tells us it is now repulsive. \[ \boxed{F' = -\frac{2F}{3}} \]