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

Q: Why is the torque from the horizontal segments (parallel to X-axis) zero?

The magnetic field from the long wire lies in the XY plane. For segments parallel to the X-axis, the cross product $d\mathbf{l} \times \mathbf{B}$ results in a force directed purely along the Z-axis. Since torque is $\mathbf{r} \times \mathbf{F}$, a force parallel to the axis of rotation produces zero torque about that axis.

Q: How do we determine the direction of the forces on the vertical segments?

Use the fundamental rule: parallel currents attract, and anti-parallel currents repel. The segment at $x=a$ has current in the same direction as the long wire, so it's attracted towards it. The segment at $x=-a$ has current in the opposite direction, so it's repelled.

Q: Why do we only take the Y-component of the force to calculate the torque?

The position vector $\mathbf{r}$ for the vertical segments lies along the X-axis ($\mathbf{r} = \pm a\hat{i}$). Torque is $\mathbf{r} \times \mathbf{F}$. The cross product of $\hat{i}$ with the X-component of force is zero. Only the cross product with the Y-component ($\hat{j}$) yields a torque along the Z-axis ($\hat{k}$).

Q: How does the approximation $b \gg a$ simplify the result?

When $b$ is much greater than $a$, the term $a^2$ becomes negligibly small compared to $b^2$. Therefore, in the denominator, $(a^2 + b^2)$ can be approximated simply as $b^2$. This allows one $b$ to cancel out, simplifying the final expression.

Q: What would happen if the current in the long wire was reversed?

If the current in the long wire is reversed, the attractive and repulsive forces on the vertical segments would swap. However, because the current in the loop also circulates, the net rotational effect (the direction of the torque) would reverse, but its magnitude would remain exactly the same.

Question

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

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

Imagine you are standing in a 3D coordinate system, looking at a fascinating interplay of magnetic forces. We have a square loop of side $2a$ placed perfectly in the $xz$-plane, centered right at the origin. Now, place a long straight wire parallel to the $z$-axis, passing through the $y$-axis at a distance $b$. Both the loop and the long wire carry the same current $I$. Our mission is to find the magnitude of the torque exerted on this loop about the $z$-axis.

### Analyzing the Setup

To find the torque about the $z$-axis, we must first analyze the magnetic forces acting on each side of the square loop. The magnetic field produced by the long wire circles around it in the $xy$-plane.

For the segments of the loop that are parallel to the $x$-axis, the current element $d\mathbf{l}$ is along the $x$-direction. The cross product of $d\mathbf{l}$ and the magnetic field $\mathbf{B}$ results in a force directed purely along the $z$-axis. Since these forces are parallel to the axis of rotation (the $z$-axis), they produce absolutely zero torque. Therefore, we can completely ignore these horizontal segments and focus our attention solely on the vertical segments parallel to the $z$-axis.

### The Master Equation

Let's look closely at the two vertical segments located at $x = a$ and $x = -a$. The perpendicular distance from the long wire (at $y=b, x=0$) to either of these segments forms the hypotenuse of a right-angled triangle with sides $a$ and $b$. Thus, the distance $d$ is $\sqrt{a^2 + b^2}$.

The magnitude of the force on each segment of length $2a$ is given by the standard formula for the force between parallel currents:

$$F = \frac{\mu_0 I^2}{2\pi d} (2a) = \frac{\mu_0 I^2 a}{\pi \sqrt{a^2 + b^2}}$$

Now, let's determine the directions. The segment at $x = a$ carries current parallel to the long wire, resulting in an attractive force $\mathbf{F}_1$ directed towards the wire. Conversely, the segment at $x = -a$ carries anti-parallel current, resulting in a repulsive force $\mathbf{F}_2$ directed away from the wire.

### Final Calculation

To calculate the torque about the $z$-axis, we only need the $y$-component of these forces. By simple geometry, the $y$-component is $F_y = F \sin	heta = F \left( \frac{b}{\sqrt{a^2 + b^2}} ight)$.

Both forces create a torque in the same rotational sense. The torque from each segment is $	au_1 = a F_y$ and $	au_2 = a F_y$. Adding them up, the total torque is:

$$	au = 2a F_y = 2a \left( \frac{\mu_0 I^2 a}{\pi \sqrt{a^2 + b^2}} ight) \left( \frac{b}{\sqrt{a^2 + b^2}} ight) = \frac{2\mu_0 I^2 a^2 b}{\pi (a^2 + b^2)}$$

Finally, the problem provides a crucial approximation: $b \gg a$. This means $a^2$ is negligible compared to $b^2$ in the denominator. Applying this, we get:

$$	au \approx \frac{2\mu_0 I^2 a^2 b}{\pi b^2} = \mathbf{\frac{2\mu_0 I^2 a^2}{\pi b}}$$

And there we have it! A beautiful result derived purely from visualizing the 3D geometry and applying the fundamental laws of electromagnetism.

Select Answer:

Visualized Solution

3D Geometry of the Setup

Identifying Torque-Producing Forces

Distance and Force Magnitude

Direction of Forces

Calculating Torque Components

Total Torque Expression

Applying the Approximation

The Way Forward

The Sigma Insight: Magnetic Force on Current

Analyzing the Setup

The Master Equation

Final Calculation

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