A charge Q is uniformly distributed over the surface of non-conducting disc of radius R. The disc ro

Q: Why do we consider an elemental ring instead of the whole disc at once?

Because the magnetic field formula $B = \frac{\mu_0 I}{2R}$ only applies to a thin circular loop where all charges are at the exact same distance from the center. In a disc, charges are spread at different radii, so we must integrate their individual contributions.

Q: How did we get the time period $T = \frac{2\pi}{\omega}$?

Angular velocity $\omega$ is the angle covered per unit time. A full circle is $2\pi$ radians. So, the time taken to complete one full circle (the time period $T$) is distance over speed, which is $\frac{2\pi}{\omega}$.

Q: Why does the radius $r$ cancel out in the $dB$ equation?

It's a beautiful mathematical coincidence! The charge $dq$ (and thus current $dI$) is proportional to the area of the ring, which grows linearly with $r$. But the magnetic field of a loop is inversely proportional to $r$. These two effects perfectly cancel each other out.

Q: What if the disc was a conducting material?

If it's a solid conductor, the charges would redistribute themselves to the edges due to mutual repulsion, making the surface charge density non-uniform. The problem specifies a non-conducting disc so the charge remains uniformly distributed.

Q: Why is the graph a rectangular hyperbola?

Our final expression is $B = \frac{\mu_0 \omega Q}{2\pi R}$. Since all terms except $R$ are constant, $B$ is inversely proportional to $R$ ($B \propto \frac{1}{R}$). In coordinate geometry, the equation $y = \frac{k}{x}$ represents a rectangular hyperbola.

Question

A charge $Q$ is uniformly distributed over the surface of non-conducting disc of radius $R$. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity $\omega$. As a result of this rotation, a magnetic field of induction $B$ is obtained at the centre of the disc.
If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc, then the variation of the magnetic induction at the centre of the disc will be represented by the figure

Sigma Percentile · Accepted Answer

### The Spinning Disc: A Symphony of Moving Charges

Imagine a non-conducting disc uniformly charged with a total charge $Q$, spinning like a record with an angular velocity $\omega$. Our goal is to find the magnetic field right at its center. To tackle this, we can't just use a single formula because the charges are spread out at different distances from the center. We need to slice the disc into infinitely many thin concentric rings. Let's pick one such elemental ring of radius $r$ and thickness $dr$.

### Slicing the Disc: The Power of Calculus

First, how much charge does this tiny ring hold? Since the charge is uniformly distributed, the surface charge density $\sigma$ is the total charge divided by the total area, which is $\frac{Q}{\pi R^2}$. The area of our elemental ring is its circumference multiplied by its thickness, $dA = 2\pi r dr$. So, its charge $dq$ is the density times this area:

$$dq = \sigma dA = \left(\frac{Q}{\pi R^2}\right) 2\pi r dr = \frac{2Q}{R^2} r dr$$

### From Charge to Current: The Equivalent Loop

Now, remember that a charge moving in a circle is basically an electric current! The equivalent current $dI$ is the charge $dq$ divided by the time period of one revolution, $T$. And we know $T$ is $\frac{2\pi}{\omega}$. Let's plug in the values. $dI$ becomes $dq$ times $\omega$ over $2\pi$. Substituting our expression for $dq$, the $2\pi$ cancels out nicely, leaving us with:

$$dI = \frac{\omega}{2\pi} \left( \frac{2Q}{R^2} r dr \right) = \frac{\omega Q}{\pi R^2} r dr$$

### The Magnetic Field of a Tiny Ring

We know the magnetic field at the center of a circular current loop is $\mu_0$ times the current, divided by twice its radius. So for our elemental ring, the tiny magnetic field $dB$ is $\frac{\mu_0 dI}{2r}$. Let's substitute our $dI$ into this equation. Notice how the radius $r$ in the numerator of $dI$ perfectly cancels the $r$ in the denominator!

$$dB = \frac{\mu_0}{2r} \left( \frac{\omega Q}{\pi R^2} r dr \right) = \frac{\mu_0 \omega Q}{2\pi R^2} dr$$

### Integrating to the Final Answer

To find the total magnetic field $B$, we just need to add up the contributions from all such rings. We integrate $dB$ from the center, where $r = 0$, to the edge, where $r = R$. The integral of $dr$ is simply $R$. One $R$ cancels out, and we get our final expression:

$$B = \int_0^R \frac{\mu_0 \omega Q}{2\pi R^2} dr = \frac{\mu_0 \omega Q}{2\pi R^2} [R] = \frac{\mu_0 \omega Q}{2\pi R}$$

Since $Q$ and $\omega$ are constant, $B$ is inversely proportional to $R$ ($B \propto \frac{1}{R}$). This means the graph representing the variation of $B$ with $R$ is a rectangular hyperbola, which perfectly matches option (a).

Select Answer:

Visualized Solution

The Rotating Charged Disc

Elemental Ring

Charge on Elemental Ring

Equivalent Current

Calculating Equivalent Current

Magnetic Field of a Loop

Magnetic Field of Elemental Ring

Total Magnetic Field

Final Expression and Graph

The Way Forward

The Sigma Insight: Biot-Savart Law

The Spinning Disc

Slicing the Disc

From Charge to Current

The Magnetic Field of a Tiny Ring

Integrating to the Final Answer

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