Animated Solution for Physics - Electromagnetic Induction: A thin semicircular conducting ring of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $B$. At the position $MNQ$ the speed of the ring is $v$ and the potential difference developed across the ring is

The phenomenon of motional EMF is one of the most elegant consequences of electromagnetic induction. When a conductor moves through a magnetic field, the charges within it experience a magnetic Lorentz force, leading to a separation of charges and the creation of a potential difference. But what happens when the conductor isn't a simple straight wire, but a curved shape like a semicircle? Let's dive into this fascinating problem!

Imagine a thin, semicircular conducting ring $MNQ$ of radius $R$. It is falling vertically downwards with a speed $v$.

The region it is falling through contains a uniform horizontal magnetic field $B$, which points directly into the plane of the screen (or paper).

Our goal is to find the potential difference developed across the ends of this ring, and to determine which end ($M$ or $Q$) is at a higher potential.

If we were to calculate the induced EMF rigorously, we would need to integrate the motional EMF contribution $dϵ=(\vec{v}\times \vec{B})\cdot d\vec{l}$ along the entire curved path of the semicircle.

However, because the magnetic field $\vec{B}$ is uniform and the velocity $\vec{v}$ is constant for all parts of the ring, the term $(\vec{v}\times \vec{B})$ is a constant vector.

This gives us a powerful shortcut: we can replace the curved conductor with an imaginary straight wire connecting its endpoints. This straight-line distance is called the effective length (${\vec{l}}_{eff}$).

For our semicircular ring, the endpoints are $M$ and $Q$. The straight line connecting them is simply the diameter of the semicircle.

Therefore, the effective length is:

Now that we have the effective length, calculating the magnitude of the induced EMF is straightforward. We use the standard formula for motional EMF in a straight conductor:

Substituting our effective length into the equation, we get:

This is the magnitude of the potential difference developed across the ring.

We have the magnitude, but we still need to figure out which end is at a higher potential. To do this, we must determine the direction in which positive charges are pushed by the magnetic force.

The Lorentz force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by:

Let's apply the right-hand rule: 1. The velocity vector $\vec{v}$ is pointing downwards ($-\hat{j}$). 2. The magnetic field vector $\vec{B}$ is pointing inwards ($-\hat{k}$).

Taking the cross product $\vec{v}\times \vec{B}$:

The resulting force vector points to the right ($+\hat{i}$). This means that free positive charges inside the conducting ring will be pushed towards the right end, which is $Q$.

Consequently, positive charge accumulates at $Q$, leaving a net negative charge at $M$.

Combining our findings, the potential difference developed across the ring is $2RBv$, and the end $Q$ is at a higher potential.

This perfectly matches option (d).

Always remember the power of the effective length concept—it turns complex integrations into simple geometry!