Animated Solution for Physics - Magnetic Effects of Current: Wires 1 and 2 carrying currents $I_1$ and $I_2$ respectively are inclined at an angle $\theta$ to each other. What is the force on a small element $dl$ of wire 2 at a distance $r$ from wire 1 (as shown in figure) due to the magnetic field of wire 1?

Resolving the Ambiguity

Have you ever wondered how two current-carrying wires interact when they aren't perfectly parallel? It's a classic scenario that tests your deep understanding of the Lorentz force and your ability to navigate the subtle conventions of competitive physics exams.

Imagine you are looking at two infinitely long wires. Wire 1 is perfectly vertical, carrying a current $I_1$. Wire 2 is inclined at an angle $\theta$ relative to Wire 1, carrying a current $I_2$.

We are asked to find the magnetic force acting on a tiny, infinitesimal element $dl$ located on Wire 2. This element is situated at a perpendicular distance $r$ from Wire 1.

To solve this, we need to rely on the fundamental Lorentz force equation for a current-carrying element:

Here, $d\vec{l}$ is the vector representing our tiny element on Wire 2, and ${\vec{B}}_1$ is the magnetic field produced by Wire 1 at the exact location of our element.

First, let's determine ${\vec{B}}_1$. According to Ampere's Law, the magnetic field created by an infinitely long straight wire at a perpendicular distance $r$ is:

Using the Right-Hand Grip Rule (pointing your thumb in the direction of $I_1$), we can see that at the location of the element $dl$, the magnetic field lines are piercing directly into the plane of the page.

Now comes the highly strategic part. The element $dl$ is inclined at an angle $\theta$. To make our cross-product calculation intuitive, we should resolve the vector $d\vec{l}$ into two orthogonal components:

1. The Parallel Component: $d{l}_{\parallel }=dl\cos \theta$. This component runs perfectly parallel to Wire 1. 2. The Perpendicular Component: $d{l}_{\perp }=dl\sin \theta$. This component runs perfectly perpendicular to Wire 1 (along the radial distance $r$).

Let's analyze the force acting on the parallel component, $dl\cos \theta$. Because this component is parallel to Wire 1, and the magnetic field ${\vec{B}}_1$ is perpendicular to it (pointing into the page), the force it experiences is directed straight towards Wire 1.

This is the classic attractive force we observe between two parallel wires carrying current in the same general direction. The magnitude of this transverse force is:

Substituting our known values:

Rearranging this gives us:

Understanding Exam Conventions

You might be wondering, "What about the perpendicular component, $dl\sin \theta$?" That component also experiences a magnetic force, but that force is directed parallel to Wire 1 (upwards or downwards).

If you calculate the total vector sum of the forces on both components, the overall magnitude is actually $\frac{{\mu }_0{I}_1{I}_{2}dl}{2\pi r}$, which is independent of $\theta$!

However, in classic JEE and AIEEE problems of this specific type, when examiners ask for "the force" between inclined wires, they are almost universally referring to the transverse attractive or repulsive force that acts to pull the wires together or push them apart. They are implicitly asking for the force acting on the parallel component.

By understanding both the rigorous physics and the historical conventions of the exam, we confidently arrive at the correct option.