Animated Solution for Physics - Electromagnetic Induction: A circular loop of radius $0.3\text{ cm}$ lies parallel to a much bigger circular loop of radius $20\text{ cm}$. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is $15\text{ cm}$. If a current of $2.0\text{ A}$ flows through the smaller loop, then the flux linked with bigger loop is

The problem of finding the magnetic flux linked with a large loop due to a tiny current-carrying loop might initially seem like an insurmountable mathematical challenge. If you try to calculate the magnetic field produced by the small loop and integrate it over the vast area of the large loop, you will quickly find yourself drowning in complex elliptic integrals. But physics is beautiful, and it often provides elegant shortcuts.

Imagine the physical scenario: we have two circular loops facing each other, sharing the same central axis.

The smaller loop has a minuscule radius of $r=0.3\text{ cm}$ and carries a current of $I=2.0\text{ A}$.

The larger loop has a much bigger radius of $R=20\text{ cm}$.

They are separated by a distance of $x=15\text{ cm}$. Our objective is to find the total magnetic flux ${\Phi }_{big}$ linked with the larger loop.

Reciprocity Theorem

This is where we invoke one of the most powerful symmetry principles in electromagnetism: The Reciprocity Theorem of Mutual Inductance.

The theorem states that the mutual inductance between two coils is perfectly symmetric. Mathematically, $M_{12}=M_{21}=M$.

What does this mean for us? It implies that the magnetic flux linked with the big loop due to a $2.0\text{ A}$ current in the small loop is exactly equal to the magnetic flux linked with the small loop if that same $2.0\text{ A}$ current were flowing in the big loop instead!

By pretending the current is in the big loop, we transform an impossible surface integral into a trivial multiplication.

Let's assume the $2.0\text{ A}$ current is flowing through the big loop. We need to find the magnetic field it produces at its axis, exactly where the small loop is located.

The standard formula for the magnetic field on the axis of a circular loop is:

Because the small loop is so tiny ($r\ll R$), the magnetic field $B$ produced by the big loop is practically uniform over the entire area of the small loop.

Thus, the flux linked with the small loop is simply the uniform magnetic field multiplied by its area:

Let's break down the calculation step-by-step to avoid any silly mistakes. First, we evaluate the denominator term $(x^2+R^2{)}^{3/2}$.

Converting our distances to meters, we have $x=0.15\text{ m}$ and $R=0.20\text{ m}$.

Notice that $0.0625$ is exactly the square of $0.25$. Therefore:

Now, let's calculate the magnetic field $B$. Substituting ${\mu }_0=4\pi \times 10^{-7}\text{ T}\cdot \text{m/A}$ and $I=2\text{ A}$:

Next, we calculate the area of the small loop. With $r=0.3\text{ cm}=3\times 10^{-3}\text{ m}$:

Finally, we multiply the magnetic field and the area to find the total flux:

In competitive exams, we often approximate ${\pi }^2\approx 10$ to quickly reach the final answer.

Rounding to the nearest given option, we get $9.2\times 10^{-11}\text{ Wb}$.

By leveraging the symmetry of mutual inductance, we elegantly bypassed a brutal integration. Always keep an eye out for symmetry—it is the physicist's ultimate weapon!