Animated Solution for Physics - Electromagnetic Induction: A rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R\Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1,I_2$ and $I$ are

A Tale of Motional EMF and Parallel Circuits

Imagine you are looking down at a perfectly rectangular conducting track. Right in the middle of this track lies a sliding connector, a rod named $PQ$. This rod isn't just sitting there; it's being pulled to the right with a constant velocity $v$. To make things interesting, this entire setup is bathed in a uniform magnetic field $B$ that plunges straight into the screen.

Our mission is to find the currents flowing through the three distinct branches of this circuit: the left loop ($I_1$), the right loop ($I_2$), and the moving rod itself ($I$).

Motional EMF

When a conductor moves through a magnetic field, it slices through the invisible magnetic field lines. This action forces the free electrons inside the conductor to move, generating what we call a Motional EMF.

The magnitude of this induced EMF is given by the elegant equation:

But which end of the rod becomes positive? We can determine this using the Lorentz force law or Fleming's Right-Hand Rule. The force on a positive charge $q$ is given by $\vec{F}=q(\vec{v}\times \vec{B})$. Since velocity $\vec{v}$ is to the right and magnetic field $\vec{B}$ is into the page, the cross product points upwards, from $Q$ to $P$. Therefore, positive charges are pushed towards $P$, making $P$ the positive terminal of our newly formed "battery."

Now, let's translate this physical setup into a standard electrical circuit. The moving rod $PQ$ acts exactly like a battery with an EMF of $e=Bvl$. However, the rod itself has a resistance $R$, which acts as the internal resistance of our battery.

This battery is connected to two external paths: the left side of the rectangular loop and the right side. Both of these paths have a resistance of $R$. In our equivalent circuit, these two paths are connected in parallel across the terminals of our moving rod.

To find the currents, we must first determine the total resistance of the circuit. Let's start with the external parallel branches. The equivalent resistance $R_p$ of two identical resistors $R$ in parallel is:

Next, we add the internal resistance of the rod to find the total resistance $R_{total}$ of the entire circuit:

With the total EMF and total resistance in hand, Ohm's law gives us the total current $I$ flowing through the main branch (the rod $PQ$):

This total current $I$ flows up the rod to point $P$, where it faces a junction. It must split into $I_1$ (going left) and $I_2$ (going right). Because the left and right branches offer the exact same resistance $R$, the current divides perfectly in half.

And there we have it! By breaking down the physical motion into an equivalent electrical circuit, we've elegantly solved for all three currents.