Two long conductors, separated by a distance d carry currents I1 and I2 in the same direction. They

Q: Why is the force attractive when currents are in the same direction?

You can determine this using a two-step process. First, use the right-hand grip rule to find the direction of the magnetic field created by wire 1 at the location of wire 2. It will point into the page. Then, use Fleming's left-hand rule (or the right-hand cross product rule for $\mathbf{F} = I\mathbf{l} \times \mathbf{B}$) on wire 2. The resulting force points towards wire 1, hence it's attractive.

Q: Why did we use a negative sign for the new current?

The negative sign is a mathematical tool to represent the reversal of direction. Since force is a vector, reversing one of the currents reverses the direction of the resulting force vector. If the initial force $F$ was attractive (positive in our convention), the new force will be repulsive (negative).

Q: Does it matter which wire's current is changed?

No, it doesn't. The force formula depends on the product of the two currents ($I_1 \times I_2$). Whether you change $I_1$ to $-2I_1$ or $I_2$ to $-2I_2$, the product becomes $-2I_1I_2$, leading to the exact same result. This is consistent with Newton's Third Law; the force is mutual.

Q: What is the length $l$ in the formula, and why didn't we need its value?

The formula gives the force acting on a specific segment of length $l$ of the wire. Since the problem asks for the ratio of the new force to the old force (by expressing $F'$ in terms of $F$), and we are considering the same length segment in both cases, the $l$ cancels out during the comparison.

Q: Why did the force decrease in magnitude even though one current doubled?

The force is directly proportional to the current but inversely proportional to the distance. While doubling the current tries to increase the force by a factor of 2, tripling the distance tries to decrease it by a factor of 3. The distance effect ($1/3$) outweighs the current effect ($2$), resulting in a net decrease in magnitude by a factor of $2/3$.

Question

Two long conductors, separated by a distance $d$ carry currents $I_1$ and $I_2$ in the same direction. 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

Options:

(A) $-2F$

(B) $\frac{F}{3}$

(C) $-\frac{2F}{3}$

(D) $-\frac{F}{3}$

Sigma Percentile · Accepted Answer

This problem is a beautiful demonstration of how physical laws scale with simple proportional changes. We are dealing with the magnetic force between two parallel current-carrying conductors, a classic scenario governed by Ampere's force law.

### The Initial Setup

Imagine two infinitely long, straight wires placed parallel to each other, separated by a distance $d$. They carry currents $I_1$ and $I_2$ in the same direction. According to the right-hand rule and the Lorentz force law, currents flowing in the same direction create an **attractive force** between the wires.

The magnitude of this force $F$ acting on a segment of length $l$ of either wire is given by the master equation:

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

This is our baseline. We will measure the new force against this initial value.

### The Transformation

Now, the universe of our problem undergoes a shift. Two things happen simultaneously:
1.  **Current Modification:** The current in one of the wires (let's say the first one) is doubled in magnitude and reversed in direction. Mathematically, $I_1$ transforms into $-2I_1$. The negative sign is crucial here; it mathematically encodes the reversal of direction.
2.  **Distance Modification:** The wires are pulled further apart, increasing their separation distance from $d$ to $3d$.

### The Final Calculation

Let's plug these transformed values into our force equation to find the new force, $F'$:

$$F' = \frac{\mu_0}{2\pi} \frac{(-2I_1)(I_2)}{3d} l$$

At first glance, it looks like a completely new expression. But the magic of physics lies in recognizing patterns. Let's rearrange the terms, pulling the numerical factors out to the front:

$$F' = -\frac{2}{3} \left( \frac{\mu_0}{2\pi} \frac{I_1 I_2}{d} l \right)$$

Look closely at the term inside the parentheses. It is exactly our initial force $F$! By substituting $F$ back into the equation, we arrive at our final, elegant result:

$$F' = -\frac{2}{3} F$$

The magnitude of the force has changed by a factor of $2/3$. More importantly, the **negative sign** tells a physical story: the force has flipped its nature. Because the currents are now flowing in opposite directions, the wires repel each other. The math perfectly mirrors the physical reality.

Select Answer:

Visualized Solution

Initial Setup

Initial Force

Final Setup

New Force Expression

Rearranging Terms

Final Answer

The Way Forward

The Sigma Insight: Magnetic Force on Current

The Initial Setup

The Transformation

The Final Calculation

Similar Questions

Two thin, long, parallel wires, separated by a distance $d$ carry a current of $I$ ampere in the same direction. They will

Two thin long parallel wires separated by a distance $b$ are carrying a current $i$ ampere each. The magnitude of the force per unit length exerted by one wire on the other is

Wires 1 and 2 carrying currents $I_{1}$ and $I_{2}$ respectively are inclined at an angle $θ$ to each other. What is the force on a small element $d l$ of wire 2 at a distance $r$ from wire 1 (as shown in figure) due to the magnetic field of wire 1?

A rigid square loop of side $a$ and carrying current $I_{2}$ is lying on a horizontal surface near a long current $I_{1}$ carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be

Two wires $A$ and $B$ are carrying currents $I_{1}$ and $I_{2}$ as shown in the figure. The separation between them is $d$ . A third wire $C$ carrying a current $I$ is to be kept parallel to them at a distance $x$ from $A$ such that the net force acting on it is zero. The possible values of $x$ are

Two coaxial solenoids of different radii carry current $I$ in the same direction. Let $F_{1}$ be the magnetic force on the inner solenoid due to the outer one and $F_{2}$ be the magnetic force on the outer solenoid due to the inner one. Then,

Due to the presence of the current $I_{1}$ at the origin in the figure,