An infinitely long current-carrying wire and a small current-carrying loop are in the plane of the p

Q: Why can we treat the loop as a magnetic dipole?

Because the distance $d$ is much greater than the radius $a$ ($d \gg a$), the magnetic field from the wire is nearly uniform across the small area of the loop, allowing us to use the dipole approximation for calculating the net force.

Q: What if the current in the loop was reversed?

The direction of the magnetic moment would reverse, which would flip the direction of the force (from attractive to repulsive, or vice versa), but the magnitude of the force and its proportionality to $a$ and $d$ would remain exactly the same.

Q: Why is the force not simply zero?

If the magnetic field were uniform, the net force on a current loop would be zero. However, the magnetic field from a straight wire decreases with distance, creating a gradient. This non-uniformity results in a net force on the dipole.

Q: Can we solve this by integrating the force on small segments of the loop?

Yes, absolutely! You can use $d\vec{F} = I (d\vec{l} \times \vec{B})$ and integrate around the loop. However, the dipole method is a powerful shortcut that saves a lot of complex calculus and is highly recommended for competitive exams.

Q: Does the loop exert the same force on the wire?

Yes! According to Newton's third law of motion, every action has an equal and opposite reaction. The force the loop exerts on the wire is equal in magnitude and opposite in direction to the force the wire exerts on the loop.

Question

An infinitely long current-carrying wire and a small current-carrying loop are in the plane of the paper as shown. The radius of the loop is $a$ and distance of its centre from the wire is $d$ ($d >> a$). If the loop applies a force $F$ on the wire, then

Options:

(A) $F \propto \left(\frac{a^2}{d^3}\right)$

(B) $F = 0$

(C) $F \propto \left(\frac{a}{d}\right)$

(D) $F \propto \left(\frac{a}{d}\right)^2$

Sigma Percentile · Accepted Answer

### The Setup: A Wire and a Loop

Imagine a long, straight wire carrying a steady current. This wire acts like a magnetic fountain, creating a magnetic field that swirls around it. Now, place a tiny circular loop of wire, also carrying a current, at a distance $d$ from the straight wire.

The magnetic field produced by the straight wire at a distance $x$ is given by the well-known formula:
$$B = \frac{\mu_0 I_1}{2\pi x}$$

Notice a crucial detail here: this magnetic field is **not uniform**. As you move further away from the wire (increasing $x$), the magnetic field gets weaker. This non-uniformity is the key to unlocking the entire problem.

### The Loop as a Magnetic Dipole

The problem states that the distance $d$ is much, much greater than the radius of the loop $a$ ($d \gg a$). This is a massive hint! Because the loop is so tiny compared to the distance, we don't need to worry about the slight variations of the magnetic field across the loop itself. Instead, we can treat the entire loop as a single, point-like **magnetic dipole**.

The magnetic moment $M$ of any current-carrying loop is simply the product of its current and its area:
$$M = I_2 \cdot A = I_2 \cdot \pi a^2$$

### The Force on a Dipole

Now, what happens when you place a magnetic dipole in a non-uniform magnetic field? It experiences a net force! The magnitude of this force is directly proportional to the dipole moment and the gradient (the rate of change) of the magnetic field. Mathematically, this is expressed as:
$$F = \left| M \frac{dB}{dx} ight|$$

To find this force, we first need to calculate the gradient of the magnetic field. We do this by differentiating our magnetic field equation with respect to $x$:
$$\frac{dB}{dx} = \frac{d}{dx} \left( \frac{\mu_0 I_1}{2\pi x} ight) = -\frac{\mu_0 I_1}{2\pi x^2}$$

The negative sign simply tells us that the magnetic field strength is decreasing as we move away from the wire. Since we are interested in the magnitude of the force, we will take the absolute value.

### The Final Calculation

Now, we bring it all together. We substitute our expressions for the magnetic moment $M$ and the magnetic field gradient $\frac{dB}{dx}$ into the force equation. We evaluate this at the location of the loop, which is at $x = d$:

$$F = (I_2 \pi a^2) 	imes \left( \frac{\mu_0 I_1}{2\pi d^2} ight)$$

Simplifying this expression, we get:
$$F = \frac{\mu_0 I_1 I_2 a^2}{2 d^2}$$

Looking closely at this final result, we can isolate the variables $a$ and $d$ to see their relationship with the force. The force is directly proportional to the square of the radius $a$ and inversely proportional to the square of the distance $d$.

Therefore, we can conclude:
$$F \propto \frac{a^2}{d^2} = \left(\frac{a}{d}ight)^2$$

And there we have it! By treating the small loop as a magnetic dipole and analyzing the gradient of the magnetic field, we've elegantly arrived at the correct proportionality.

Select Answer:

Visualized Solution

Visualizing the Setup

Magnetic Field of the Wire

Loop as a Magnetic Dipole

Force on a Dipole

Calculating the Gradient

Evaluating the Force

Final Proportionality

The Sigma Insight: Magnetic Moment of Current Loop

The Setup

The Loop as a Magnetic Dipole

The Force on a Dipole

The Final Calculation

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