Animated Solution for Physics - Electromagnetic Induction: A long solenoid of radius $a$ and number of turns per unit length $n$ is enclosed by cylindrical shell of radius $R$, thickness $d$ ($d<<R$) and length $L$. A variable current $i=i_0\sin \omega t$ flows through the solenoid. If the resistivity of the material of cylindrical shell is $\rho$, find the induced current in the shell.

Imagine you are looking at a fascinating electromagnetic system. We have a thick cylindrical shell with a radius $R$, a thickness $d$, and a length $L$. Right down the central axis of this shell, a long solenoid of radius $a$ is perfectly positioned.

An alternating current, given by $i=i_0\sin \omega t$, is pumped through the solenoid. This isn't just a static setup; it's a dynamic, breathing system where the current is constantly changing, and as we know from the principles of electromagnetism, a changing current is the spark that ignites induction.

Let's dive into the core of the solenoid. The magnetic field produced by a long, ideal solenoid is beautifully confined entirely within its interior. Outside the solenoid, the magnetic field is practically zero. The magnitude of this internal magnetic field is given by the classic formula:

Now, we need to find the magnetic flux $\varphi$ passing through the cross-section of the outer cylindrical shell. Here is where many students fall into a trap! You might be tempted to use the entire cross-sectional area of the shell ($\pi R^2$). However, remember that the magnetic field only exists inside the solenoid. Therefore, the flux only passes through the area of the solenoid, which is $\pi a^2$.

With our flux expression in hand, we can invoke Faraday's Law of Electromagnetic Induction. This law tells us that the induced EMF is the negative rate of change of magnetic flux. We are primarily interested in the magnitude of this EMF, so we differentiate our flux expression with respect to time:

Since the derivative of $\sin \omega t$ is $\omega \cos \omega t$, the magnitude of the induced EMF becomes:

Now comes the most crucial and often misunderstood part of the problem: calculating the resistance of the cylindrical shell. The changing magnetic flux induces an electric field that drives current in circular loops around the circumference of the shell.

To find the resistance $R_{\text{shell}}=\frac{\rho l}{A}$, we must carefully define the path length $l$ and the cross-sectional area $A$ for this specific current flow.

1. Path Length ($l$): The current travels in a full circle around the cylinder, so the length of its path is the circumference, $l=2\pi R$. 2. Cross-Sectional Area ($A$): The current spreads out over the entire length $L$ of the shell and flows through its thickness $d$. Therefore, the area it flows through is $A=L\cdot d$.

Plugging these into our resistance formula gives:

We are finally ready to bring it all together. By Ohm's Law, the induced current $i_{\text{ind}}$ is simply the induced EMF divided by the resistance of the shell:

Substituting our expressions for $|e|$ and $R_{\text{shell}}$:

Simplifying this fraction by canceling the $\pi$ terms and rearranging, we arrive at our elegant final answer:

This result beautifully ties together the geometry of the system, the material properties of the shell, and the dynamic nature of the driving current.