A rigid square loop of side a and carrying current I2 is lying on a horizontal surface near a long c

Q: Why is the force on the top and bottom segments not zero?

Many textbooks mistakenly claim the force is zero because they assume the angle between the current and the magnetic field is zero. However, the magnetic field is directed *into* the page, while the current is in the plane of the page. The angle is actually $90^\circ$, so the force is non-zero. They just cancel each other out due to symmetry!

Q: How do we know the direction of the magnetic field?

By using the Right Hand Grip Rule. Point your right thumb in the direction of the current $I_1$ (upwards). Your fingers will curl into the page on the right side of the wire, which is where the loop is located.

Q: Why do opposite currents repel?

It comes down to the Lorentz force. The first wire creates a magnetic field. When the second wire's current flows in the opposite direction, the cross product $\vec{l} \times \vec{B}$ results in a force vector pointing away from the first wire.

Q: Why doesn't the loop rotate?

The forces on the top and bottom segments are equal and opposite, and they act along the same line of action in the plane of the loop. The forces on the left and right segments are also in the plane of the loop. Therefore, there is no net torque to cause rotation.

Q: What if the loop was placed on the left side of the wire?

The magnetic field would be pointing *out* of the page. However, the relative directions of the currents would still dictate the forces. The side closest to the wire would still have opposite current and repel, so the net force would still be repulsive!

Question

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

Options:

(A) repulsive and equal to $\frac{\mu_0 I_1 I_2}{2\pi}$

(B) attractive and equal to $\frac{\mu_0 I_1 I_2}{3\pi}$

(C) zero

(D) repulsive and equal to $\frac{\mu_0 I_1 I_2}{4\pi}$

Sigma Percentile · Accepted Answer

### The Beauty of Magnetic Interactions
Have you ever wondered how two completely separate wires can push or pull each other without touching? It is all thanks to the invisible, yet incredibly powerful, magnetic field.

In this problem, we are looking at a classic setup: a long straight wire and a square loop, both carrying current. Our mission is to find the net magnetic force acting on the loop.

To do this, we need to break the problem down into bite-sized pieces. We will analyze the force on each of the four segments of the square loop individually.

### Mapping the Magnetic Field
Before we can calculate any forces, we need to understand the environment the loop is sitting in. The long straight wire is the source of our magnetic field.

Using the **Right Hand Grip Rule**, if we point our thumb in the direction of the upward current $I_1$, our fingers curl into the page on the right side of the wire.

This means the entire square loop is bathed in a magnetic field that points directly into the screen. The strength of this field at a distance $r$ is given by:

$$B = \frac{\mu_0 I_1}{2\pi r}$$

### The Horizontal Segments: A Common Trap
Now, let's look at the top and bottom segments of the loop (AD and BC). Here is where many students, and even some textbooks, fall into a trap!

They often assume the force on these segments is zero because they mistakenly think the angle between the current and the magnetic field is zero. But remember, the magnetic field is pointing *into* the page, while the wires lie *on* the page.

The angle is actually $90^\circ$! This means there is a non-zero force acting on them.

However, because the magnetic field strength only depends on the horizontal distance from the straight wire, the force on the top segment is exactly equal in magnitude to the force on the bottom segment.

Using the right-hand rule for cross products, we find that these forces point in opposite vertical directions. They perfectly cancel each other out!

$$\vec{F}_{AD} + \vec{F}_{BC} = 0$$

### The Vertical Segments: The Real Players
With the horizontal segments out of the way, we can focus our attention on the vertical segments, AB and CD. These are parallel to the straight wire.

The force per unit length between two parallel current-carrying wires is a fundamental principle:

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

The golden rule here is simple: **currents in the same direction attract, while currents in opposite directions repel.**

Let's apply this to the left segment, AB. The current $I_2$ is flowing downwards, which is opposite to the upward current $I_1$.

This creates a **repulsive force**, pushing segment AB away from the straight wire. Since it is at a distance $a$ and has a length $a$, the force is:

$$F_{AB} = \frac{\mu_0 I_1 I_2}{2\pi a} a = \frac{\mu_0 I_1 I_2}{2\pi}$$

Now, let's look at the right segment, CD. Here, the current $I_2$ is flowing upwards, in the exact same direction as $I_1$.

This creates an **attractive force**, pulling segment CD towards the straight wire. This segment is further away, at a distance of $a + a = 2a$. The force is:

$$F_{CD} = \frac{\mu_0 I_1 I_2}{2\pi (2a)} a = \frac{\mu_0 I_1 I_2}{4\pi}$$

### Bringing It All Together
We now have two competing forces: a strong repulsive force pushing the loop away, and a weaker attractive force pulling it in.

To find the net force, we simply subtract the weaker force from the stronger one:

$$F_{net} = F_{AB} - F_{CD}$$

Substituting our values, we get:

$$F_{net} = \frac{\mu_0 I_1 I_2}{2\pi} - \frac{\mu_0 I_1 I_2}{4\pi}$$

Finding a common denominator and simplifying, we arrive at our final, elegant result:

$$F_{net} = \frac{\mu_0 I_1 I_2}{4\pi}$$

Because the repulsive force was stronger (since segment AB is closer to the source wire), the **net force is repulsive**. This perfectly matches option (d).

This problem is a beautiful demonstration of how breaking a complex system into simple, fundamental interactions leads us straight to the solution!

Animated Solution for Physics - Magnetic Effects of Current: 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

Select Answer:

Visualized Solution

\text{System Setup}

\text{Magnetic Field Direction}

\text{Forces on Horizontal Segments}

\text{Force Between Parallel Wires}

\text{Force on Segment AB}

\text{Force on Segment CD}

\text{Net Force Calculation}

\text{Final Answer}

The Sigma Insight: Magnetic Force on Current

The Beauty of Magnetic Interactions

Mapping the Magnetic Field Before we can calculate any forces, we need to understand the environment the loop is sitting in

The Horizontal Segments

The Vertical Segments

Bringing It All Together

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