Animated Solution for Physics - Electromagnetic Induction: A transmitting station releases waves of wavelength $960\text{ m}$. A capacitor of $2.56\mu \text{F}$ is used in the resonant circuit. The self-inductance of coil necessary for resonance is ......... $\times 10^{-8}\text{ H}$.

Imagine you are standing next to a massive radio transmitting station. It is broadcasting signals into the air using a towering antenna. But what exactly is driving that antenna? Deep inside the station, there is a resonant $LC$ circuit—a beautiful dance of energy between an inductor and a capacitor. In this problem, we are going to tune that circuit to broadcast a very specific wave.

The station releases electromagnetic waves with a wavelength of $\lambda =960\text{ m}$. To understand how the circuit generates this wave, we first need to find the wave's frequency. Remember the fundamental wave equation? The speed of a wave is the product of its frequency and its wavelength:

Since these are electromagnetic waves, they travel at the cosmic speed limit—the speed of light, $c=3\times 10^8\text{ m/s}$. By rearranging our equation, we can express the frequency as:

For the station to transmit efficiently at this frequency, the $LC$ circuit must be in a state of resonance. This happens when the inductive reactance perfectly cancels out the capacitive reactance. The resonant frequency of an $LC$ circuit is given by the elegant formula:

Now, let's bridge our two worlds—the physical wave traveling through space and the electrical circuit generating it. We equate the two expressions for frequency:

Our goal is to find the self-inductance, $L$. Let's square both sides to eliminate that pesky square root:

Rearranging the terms to isolate $L$, we get our master equation for this problem:

Now comes the execution phase. We carefully substitute our known values into the equation. We have $\lambda =960\text{ m}$, $v=3\times 10^8\text{ m/s}$, and $C=2.56\mu \text{F}$. Crucially, we must convert the capacitance into standard SI units (Farads), so $C=2.56\times 10^{-6}\text{ F}$.

Let's break down the arithmetic. Squaring the numerator gives $921600$. In the denominator, squaring the speed of light gives $9\times 10^{16}$.

Combining the constants in the denominator ($4\times 9\times 2.56=92.16$) and the powers of ten ($10^{16}\times 10^{-6}=10^{10}$), we get:

Notice how beautifully the numbers align! $921600$ divided by $92.16$ is exactly $10000$, or $10^4$.

To find the final numerical value, we use a classic physicist's approximation: ${\pi }^2\approx 10$.

The question asks for the answer in the format of $...\times 10^{-8}\text{ H}$. We can rewrite $10^{-7}$ as $10\times 10^{-8}$. Therefore, the required value is 10.