A square loop of side 2a and carrying current I is kept in xy-plane with its centre at origin. A lon

Q: Why did we assume the loop is in the xz-plane when the question says xy-plane?

If the loop is in the xy-plane, the magnetic field from the wire (which is parallel to the Z-axis) lies entirely in the xy-plane. The resulting forces on the loop would be in the Z-direction, producing zero torque about the Z-axis. This was a typographical error in the official JEE paper, which was corrected in the very next shift!

Q: Why do only two sides of the loop contribute to the torque?

The sides of the loop parallel to the X-axis experience forces that are either zero or act along lines passing through the Z-axis, meaning their lever arm is zero. Only the sides parallel to the Z-axis experience forces that create a twisting motion around the Z-axis.

Q: How did we determine the direction of the forces $F_1$ and $F_2$?

We used the principle that parallel currents attract and anti-parallel currents repel. Since the loop is a closed circuit, the current in one side is parallel to the long wire, while the current in the opposite side is anti-parallel.

Q: Why can we approximate $r^2 \approx b^2$?

The problem explicitly states that $b \gg a$ (b is much greater than a). In the expression $r^2 = a^2 + b^2$, the $a^2$ term becomes negligibly small compared to $b^2$, allowing us to simplify the math without losing significant accuracy.

Q: Is there a faster way to solve this during the exam?

Absolutely! Because $b \gg a$, the magnetic field is nearly uniform across the small loop. You can treat the loop as a magnetic dipole with moment $M = I(2a)^2$. The torque is simply $\vec{\tau} = \vec{M} \times \vec{B}$, which gives the answer in just two lines of calculation!

Question

A square loop of side $2a$ and carrying current $I$ is kept in $xy$-plane with its centre at origin. A long wire carrying the same current $I$ is placed parallel to the $Z$-axis and passing through the point $(0, b, 0)$, $(b >> a)$. The magnitude of the torque on the loop about $Z$-axis is given by

Options:

(A) $\frac{2\mu_0 I^2 a^3}{\pi b^2}$

(B) $\frac{2\mu_0 I^2 a^2}{\pi b}$

(C) $\frac{\mu_0 I^2 a^3}{2\pi b^2}$

(D) $\frac{\mu_0 I^2 a^2}{2\pi b}$

Sigma Percentile · Accepted Answer

### The Famous JEE Typo

Imagine sitting in the examination hall, reading this question, and realizing something is physically impossible. The question states the square loop is in the $xy$-plane, and the long wire is parallel to the $Z$-axis.

If you apply the right-hand rule, the magnetic field from the wire lies entirely within the $xy$-plane. Consequently, the magnetic force on the loop would point strictly in the $Z$-direction. A force in the $Z$-direction cannot produce a torque about the $Z$-axis!

This was a famous typographical error by the NTA. For the given options to be mathematically derivable, the loop must actually be in the **$xz$-plane**. Interestingly, they corrected this exact typo in the very next shift's paper! Let's proceed by placing the loop in the $xz$-plane.

### Analyzing the Setup

Let's look at the cross-section of our setup in the $xy$-plane. The long wire passes through $(0, b, 0)$. The square loop, now in the $xz$-plane, appears as a line segment extending from $x = -a$ to $x = a$.

The sides of the loop that are parallel to the $Z$-axis (located at $x = a$ and $x = -a$) will interact with the long wire. The distance $r$ from the wire to each of these sides forms a right-angled triangle, where $r = \sqrt{a^2 + b^2}$.

### The Magnetic Forces

The magnetic field $B$ produced by the long wire at the location of these sides is given by Ampere's Law:

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

The force $F$ on each side of length $2a$ is simply the product of current, length, and magnetic field:

$F = I (2a) B = \frac{\mu_0 I^2 (2a)}{2\pi r}$

Because the current flows in a closed loop, the currents in these two sides are anti-parallel. One side experiences an **attractive force** towards the wire, while the other experiences a **repulsive force** away from it.

### Calculating the Torque

Even though one force is attractive and the other is repulsive, they both create a torque that tries to rotate the loop in the **same direction** about the origin!

To find the torque, we need the component of the force that is perpendicular to the position vector $a$. From our geometric setup, this perpendicular component is $F \cos	heta$.

The torque from one side is $	au_1 = F a \cos	heta$. Since both sides contribute equally, the total net torque is:

$	au_{net} = 2 F a \cos	heta$

From our right-angled triangle, we can see that $\cos	heta = \frac{b}{r}$.

### The Final Substitution

Let's substitute our expressions for $F$ and $\cos	heta$ into the torque equation:

$	au_{net} = 2 \left( \frac{\mu_0 I^2 (2a)}{2\pi r} ight) a \left( \frac{b}{r} ight)$

Simplifying this, we get:

$	au_{net} = \frac{2\mu_0 I^2 a^2 b}{\pi r^2}$

We know that $r^2 = a^2 + b^2$. However, the problem gives us a powerful approximation: $b \gg a$. This means the wire is very far away compared to the size of the loop, allowing us to approximate $r^2 \approx b^2$.

Substituting this approximation yields our final answer:

$	au_{net} = \frac{2\mu_0 I^2 a^2}{\pi b}$

### The Pro-Tip Shortcut

Whenever you see a condition like $b \gg a$, it's a massive hint that you can use dipole approximations! Because the wire is far away, the magnetic field is nearly uniform across the loop.

We can treat the square loop as a magnetic dipole with a magnetic moment $M = I A = I(2a)^2 = 4Ia^2$. 
The magnetic field at the origin is $B = \frac{\mu_0 I}{2\pi b}$.

The torque is simply the cross product $	au = M 	imes B$. 
Multiplying the magnitudes gives:

$	au = (4Ia^2) \left( \frac{\mu_0 I}{2\pi b} ight) = \frac{2\mu_0 I^2 a^2}{\pi b}$

Two lines of math, and you arrive at the exact same elegant result!

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Visualized Solution

$Visualizing the Setup$

$Distance to Loop Sides$

$Magnetic Force Magnitude$

$Direction of Forces$

$Torque from One Side$

$Total Torque$

$Substituting Values$

$Applying Approximation$

$The Way Forward$

The Sigma Insight: Magnetic Force on Current

The Famous JEE Typo

Analyzing the Setup

The Magnetic Forces

Calculating the Torque

The Final Substitution

The Pro-Tip Shortcut

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance to Loop Sides

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnetic Force Magnitude

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Direction of Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque from One Side

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Torque

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Approximation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward