Animated Solution for Physics - Electromagnetic Induction: In a uniform magnetic field of induction $B$, a wire in the form of semi-circle of radius $r$ rotates about the diameter of the circle with angular frequency $\omega$. If the total resistance of the circuit is $R$, the mean power generated per period of rotation is

Imagine a semi-circular wire loop placed in a uniform magnetic field. Now, we start spinning it around its straight edge, which acts as the diameter, with a constant angular frequency $\omega$. This rotation is going to continuously change the magnetic flux passing through the loop. This is the fundamental principle behind electric generators!

To find the induced EMF, we first need to write down the expression for the magnetic flux. The flux $\varphi$ is simply the dot product of the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$. This gives us:

where $\theta$ is the angle between the magnetic field and the normal to the area.

Now, let's plug in what we know. The area of our semi-circle is half the area of a full circle, so $A=\frac{1}{2}\pi r^2$. And since it's rotating with a constant angular velocity $\omega$, the angle $\theta$ at any time $t$ is simply $\omega t$. Substituting these into our flux equation, we get:

According to Faraday's Law of Electromagnetic Induction, the induced EMF is the negative rate of change of magnetic flux. So, we differentiate our flux expression with respect to time. The derivative of $\cos (\omega t)$ is $-\omega \sin (\omega t)$. The negative signs cancel out, leaving us with this beautiful expression for the induced EMF:

Next, we need to find the power generated. The electrical power dissipated in a circuit with resistance $R$ is given by $P=\frac{e^2}{R}$. Let's carefully square our EMF expression. The $\frac{1}{2}$ becomes $\frac{1}{4}$, and we get a ${\sin }^2(\omega t)$ term. This is the instantaneous power at any given moment:

The question asks for the mean power over a full period of rotation. To find this, we take the average of our instantaneous power. The only part that changes with time is the ${\sin }^2(\omega t)$ term. A very famous and useful result in AC circuits is that the average value of ${\sin }^2(\omega t)$ over a full cycle is exactly $\frac{1}{2}$.

Substituting $⟨{\sin }^2(\omega t)⟩=\frac{1}{2}$, we get:

We can then group the terms neatly into a single perfect square in the numerator to match the given options:

And there we have it! The mean power generated perfectly matches option (b). Always pay close attention to the geometry given in the problem—if it were a full circle, the area would double, and the mean power would become four times larger!