Animated Solution for Physics - Magnetic Effects of Current: Two long current carrying thin wires, both with current $I$, are held by insulating threads of length $L$ and are in equilibrium as shown in the figure, with threads making an angle $\theta$ with the vertical. If wires have mass $\lambda$ per unit length then, the value of $I$ is ($g=$ gravitational acceleration)

Magnetic Repulsion vs. Gravity

Imagine two long wires hanging from the ceiling like swings. They are carrying current, and because of the magnetic force between them, they push each other apart and settle at an angle $\theta$ in equilibrium. This is a classic physics problem that beautifully marries mechanics with electromagnetism. Let's break down the forces at play to find the exact current $I$ flowing through these wires.

To understand the equilibrium, we must isolate one of the wires and draw its free body diagram. Let's focus on the right wire. What forces are acting on it?

First, gravity pulls it down with a force $mg$. Since we are given the mass per unit length $\lambda$, the mass of a segment of length $l$ is $m=\lambda l$. Thus, the downward force is $\lambda lg$.

Second, the string pulls it up and to the left with a tension $T$.

Finally, the magnetic force pushes it to the right with a force $F_B$. Because the wires are repelling each other, we know the currents must be flowing in opposite directions. Since the wire is in equilibrium, these three forces must perfectly balance each other.

To balance the forces mathematically, we resolve the tension $T$ into its horizontal and vertical components.

The upward vertical component is $T\cos \theta$, which must balance the downward weight:

The horizontal component is $T\sin \theta$, which must balance the rightward magnetic repulsion:

But what exactly is the magnetic force $F_B$ between two parallel wires? According to Ampere's force law, the force on a length $l$ of wire is given by:

Here, $d$ is the distance between the two wires. By looking at the geometry of our setup, we can see that the distance from the center line to one wire is $L\sin \theta$. Therefore, the total distance $d$ between the two wires is simply $2L\sin \theta$.

Now, we have a system of equations. Let's divide the horizontal force equation by the vertical force equation. This is a standard trick in mechanics that elegantly eliminates the unknown tension $T$:

This simplifies to:

Notice how the length of the wire, $l$, beautifully cancels out! This tells us that the equilibrium angle is independent of how long the wire segment we consider is.

Finally, we just need to rearrange the terms to solve for the current $I$. We can write $\tan \theta$ as $\frac{\sin \theta }{\cos \theta }$ to make the algebra cleaner:

Taking the square root of both sides, we arrive at our final, elegant expression:

This perfectly matches option (b). Always remember, if the currents were flowing in the same direction, the wires would attract each other instead of repelling, and they would swing inwards. The interplay of forces here is a fantastic reminder of how interconnected different branches of physics truly are.