Animated Solution for Physics - Magnetic Effects of Current: Comprehension Passage

This is a brilliant two-part comprehension problem from JEE Advanced 2014 that tests your spatial reasoning, mastery of the Biot-Savart Law, and ability to calculate magnetic torque. Let's break it down step-by-step.

Imagine a 3D coordinate system. We have a circular loop of radius $a$ lying flat in the $xy$-plane, carrying a counter-clockwise current $I$. By the right-hand rule, this loop produces a magnetic field along its central axis (the $z$-axis) that points straight up in the $+\hat{k}$ direction.

Flanking the loop are two long straight wires, parallel to the $y$-axis, located at $x=-d$ and $x=d$. We are tasked with finding a point on the $z$-axis at height $h$ where the net magnetic field is exactly zero. For this to happen, the two straight wires must conspire to produce a combined magnetic field that points straight down in the $-\hat{k}$ direction to perfectly cancel the loop's upward field.

Using the right-hand grip rule, if Wire 1 (at $x=-d$) carries current in the $+y$ direction (from $P$ to $Q$), its magnetic field at $(0,0,h)$ will have a downward $z$-component. Similarly, if Wire 2 (at $x=d$) carries current in the $-y$ direction (from $S$ to $R$), its magnetic field will also contribute a downward $z$-component. Thus, the currents must flow in the directions $PQ$ and $SR$.

Let's calculate the exact magnitude of this downward field. The distance from either wire to the point $(0,0,h)$ is $r=\sqrt{d^2+h^2}$. The magnetic field from a single wire is $B=\frac{{\mu }_{0}I}{2\pi r}$.

A Classic Textbook Trap: Many solutions manuals incorrectly use the sine component instead of the cosine component when resolving the magnetic field of the wires. They mistakenly write the $z$-component as $B\frac{h}{r}$ instead of $B\frac{d}{r}$. If you follow that flawed logic, you end up with $h\approx 1.07a$, which doesn't match the correct option! The true geometry dictates that the magnetic field vector is perpendicular to the position vector $\mathbf{r}$, making the $z$-component proportional to $d$.

The total downward field from both wires is:

Equating this to the upward field of the loop:

The problem states that $d\approx a$. Substituting this approximation, the equation simplifies beautifully:

Canceling terms yields:

Squaring both sides:

Using the standard approximation ${\pi }^2\approx 9.87$, we get $\frac{{\pi }^2}{4}\approx 2.467$. Therefore:

This perfectly matches option (c)!

Torque on the Rotated Loop

Now, we shift gears. The distance $d$ is now much greater than $a$ ($d\gg a$), and the currents in the wires are opposite. We need to find the torque on the loop when it is rotated by $30^{\circ }$ about its diameter.

First, let's find the uniform magnetic field at the center of the loop due to the wires. Since the currents are opposite, their magnetic fields at the origin reinforce each other, both pointing in the $+\hat{k}$ direction:

The magnetic moment of the loop is $M=I\pi a^2$. When the loop is rotated by $30^{\circ }$, its area vector (and thus its magnetic moment $\mathbf{M}$) tilts by $30^{\circ }$ relative to the $z$-axis. The torque is given by the cross product:

Substituting our values:

This elegant result leads us straight to option (b).