Animated Solution for Physics - Magnetic Effects of Current: If in a circular coil $A$ of radius $R$, current $I$ is flowing and in another coil $B$ of radius $2R$ a current $2I$ is flowing, then the ratio of the magnetic fields, $B_A$ and $B_B$ produced by them will be

The beauty of electromagnetism often lies in its elegant symmetries and proportionalities. In this problem, we are presented with a classic scenario involving two circular coils, each carrying a current and producing a magnetic field at its center. This is a fantastic opportunity to see how scaling different physical parameters—like radius and current—affects the final outcome.

Let's dive into the physics and unravel the relationship between these two coils.

Imagine you are looking at two distinct circular loops of wire, which we will call Coil $A$ and Coil $B$.

For Coil $A$, the physical parameters are straightforward: - The radius of the coil is $R$. - The current flowing through the wire is $I$.

For Coil $B$, everything is scaled up by a factor of two: - The radius of the coil is $2R$. - The current flowing through the wire is $2I$.

Our objective is to determine the ratio of the magnetic fields produced at the exact center of each coil, denoted as $B_A$ and $B_B$.

To solve this, we need to recall the fundamental formula derived from the Biot-Savart Law for the magnetic field at the center of a current-carrying circular loop. The magnitude of this magnetic field $B$ is given by:

Here, ${\mu }_0$ is the permeability of free space, $N$ is the number of turns in the coil, $I$ is the current, and $r$ is the radius of the coil. Since the problem does not specify the number of turns for either coil, it is standard practice to assume they both consist of a single turn, meaning $N=1$.

Let's apply our master equation to Coil $A$. Substituting its specific parameters into the formula, we get:

This is our baseline magnetic field.

Now, let's turn our attention to Coil $B$. We must be careful to substitute its scaled parameters correctly. The current is $2I$ and the radius is $2R$. Plugging these into the formula yields:

At first glance, the expression for $B_B$ looks a bit bulkier. However, let's simplify it. Notice that we have a factor of $2$ in the numerator (from the current) and a factor of $2$ in the denominator (from the radius).

These factors of $2$ perfectly cancel each other out:

Wow! Look at that result. Despite Coil $B$ being twice as large and carrying twice as much current, the magnetic field it produces at its center is exactly the same as that of Coil $A$.

This happens because the magnetic field at the center of a loop is directly proportional to the current ($B\propto I$) and inversely proportional to the radius ($B\propto \frac{1}{R}$). By doubling both the current and the radius, the effects perfectly neutralize each other.

We are asked to find the ratio of the two magnetic fields, $\frac{B_A}{B_B}$. Since we have established that $B_A=B_B$, the calculation is trivial:

This problem serves as a brilliant reminder that in physics, understanding proportionalities can often lead you to the answer much faster than brute-force calculation. Always keep an eye out for how scaling one variable might be offset by scaling another!