Animated Solution for Physics - Electromagnetic Induction: The magnetic field in a region is given by $\mathbf{B}=B_0\left(\frac{x}{a}\right)\hat{\mathbf{k}}$. A square loop of side $d$ is placed with its edges along the X and Y-axes. The loop is moved with a constant velocity $\mathbf{v}=v_0\hat{\mathbf{i}}$. The emf induced in the loop is

Imagine a square loop of wire, with side length $d$, sliding smoothly along the X-axis at a constant velocity $\mathbf{v}=v_0\hat{\mathbf{i}}$. The entire region is bathed in a magnetic field pointing straight out of the page (the Z-direction). But there is a twist—this magnetic field is not uniform. Its strength is given by $\mathbf{B}=B_0\left(\frac{x}{a}\right)\hat{\mathbf{k}}$. This means that as you move further to the right along the X-axis, the magnetic field gets progressively stronger.

To find the induced EMF in this moving loop, we can break the loop down into its four individual straight wire segments and calculate the motional EMF for each one using the fundamental formula $E=\int (\mathbf{v}\times \mathbf{B})\cdot d\mathbf{l}$.

Let's first look at the top and bottom edges of the square loop. These edges are perfectly parallel to the X-axis. The velocity $\mathbf{v}$ is along the X-axis ($\hat{\mathbf{i}}$), and the magnetic field $\mathbf{B}$ is along the Z-axis ($\hat{\mathbf{k}}$).

If we take the cross product $\mathbf{v}\times \mathbf{B}$, the resulting vector points straight down, along the negative Y-axis ($-\hat{\mathbf{j}}$). However, the length element $d\mathbf{l}$ for these horizontal edges points along the X-axis. Because the magnetic force on the charges is perpendicular to the wire itself, no work is done in moving charges along these segments. Mathematically, the dot product $(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{l}$ is exactly zero.

Now, let's turn our attention to the vertical edges. These are the ones doing the heavy lifting.

Consider the left vertical edge, located at some arbitrary position $x$. The magnetic field at this exact location is ${\mathbf{B}}_{left}=B_0\left(\frac{x}{a}\right)\hat{\mathbf{k}}$. The induced EMF across this edge of length $d$ is simply the product of the magnetic field, the velocity, and the length:

Next, look at the right vertical edge. Because the loop has a width of $d$, this edge is located at position $x+d$. Since it is further to the right, it is slicing through a stronger magnetic field! The magnetic field here is ${\mathbf{B}}_{right}=B_0\left(\frac{x+d}{a}\right)\hat{\mathbf{k}}$. The induced EMF across this edge is:

By Lenz's Law, or by simply analyzing the $\mathbf{v}\times \mathbf{B}$ force, we can see that the induced EMF in both vertical edges tries to push positive charges downwards. In the context of the closed square loop, these two edges act like two batteries connected in opposition.

To find the net EMF driving current around the loop, we must subtract the weaker "battery" from the stronger one:

Substituting our expressions:

Notice how beautifully the terms depending on the arbitrary position $x$ cancel out!

This leaves us with a constant net EMF that depends only on the loop's dimensions, its speed, and the gradient of the magnetic field. This elegant result highlights how a spatial variation in a magnetic field is essential to sustain a steady motional EMF in a closed loop moving at a constant velocity.