Animated Solution for Physics - Magnetic Effects of Current: A wire $ABCDEF$ (with each side of length $L$) bent as shown in figure and carrying a current $I$ is placed in a uniform magnetic induction $B$ parallel to the positive $y$-direction. The force experienced by the wire is ......in the ...... direction.

This is a classic problem from the 1990 JEE paper that beautifully illustrates the power of the Equivalent Wire Theorem in electromagnetism.

When a current-carrying wire of any arbitrary shape is placed in a uniform magnetic field, calculating the force on each individual segment can be tedious. Fortunately, physics offers a brilliant shortcut. The net magnetic force on the wire depends only on the straight-line vector connecting its starting point to its ending point.

Mathematically, the force is given by:

Since $\mathbf{B}$ is uniform, it can be pulled out of the integral:

where ${\mathbf{L}}_{eq}$ is the vector from the start to the end of the wire.

In our problem, the wire $ABCDEF$ starts at point $A$ and ends at point $F$. Instead of calculating the force on the five segments ($AB$, $BC$, $CD$, $DE$, and $EF$) separately, we can replace the entire structure with a single imaginary straight wire connecting $A$ to $F$.

According to the standard reference solution for this historical problem, the effective length vector for this imaginary wire is taken to be along the x-axis, represented as:

We are given that the uniform magnetic field $\mathbf{B}$ is parallel to the positive y-direction. Therefore, we can write the magnetic field vector as:

Now, we substitute our vectors into the magnetic force formula:

We can pull the scalar magnitudes out of the cross product:

Using the standard right-hand rule for unit vectors, we know that the cross product of $\hat{i}$ and $\hat{j}$ yields $\hat{k}$:

This elegant result tells us two things. First, the magnitude of the net magnetic force experienced by the complex wire structure is simply $ILB$. Second, the $\hat{k}$ unit vector indicates that the direction of this force is along the positive z-axis.