Problem 21 - Master Resource Book in JEE Main Physics

Question

A thin disc having radius \(r\) and charge \(q\) distributed uniformly over the disc is rotated \(n\) rotations per second about its axis. The magnetic field at the centre of the disc is (a) \(\frac{\mu_{0} q n}{2 r}\) (b) \(\frac{\mu_{0} q n}{r}\) (c) \(\frac{\mu_{0} q n}{4 r}\) (d) \(\frac{3 \mu_{0} q n}{4 r}\)

Step-by-Step Solution

Answer

The correct answer is (a) \( \frac{\mu_0 q n}{2 r} \).

Step 1: Understand the Concept of a Rotating Charged Disc

A rotating charged disc acts as a circular current loop, and we need to find the magnetic field at the center due to this current loop. When the disc rotates, it creates a magnetic moment due to charges moving along circular paths.

Step 2: Determine Surface Charge Density

Since the charge is uniformly distributed over the disc, the surface charge density, \( \sigma \), is given by \( \sigma = \frac{q}{\pi r^2} \), where \( q \) is the total charge and \( r \) is the radius of the disc.

Step 3: Find Current Due to Rotating Disc

The current, \( I \), produced by a rotating charged disc can be calculated using the formula \( I = \sigma \cdot \text{area swept per second} \). Since the disc completes \( n \) rotations per second, the area swept is the whole disc area, \( \pi r^2 \), per rotation. Hence, \( I = q n \).

Step 4: Apply Magnetic Field Formula for a Loop

The magnetic field at the center of a circular loop (here formed by the rotating disc) is given by the formula \( B = \frac{\mu_0 I}{2r} \). Substitute \( I = q n \) to find \( B = \frac{\mu_0 q n}{2r} \).

Step 5: Choose the Correct Answer

From the options given, we can see that the correct expression for the magnetic field at the center of the disc is \( B = \frac{\mu_0 q n}{2r} \). Therefore, the correct choice is option (a).

Key Concepts

Surface Charge Density

When considering a uniformly charged disc, we first need to understand the concept of **surface charge density**. Surface charge density, denoted by \( \sigma \), is a measure of how much charge \( q \) is distributed over a given area \( A \). For our disc, where the charge is spread uniformly, the surface charge density is given by the formula:

• \( \sigma = \frac{q}{\pi r^2} \)

In this equation, \( q \) represents the total charge present on the disc, and \( r \) is the radius of the disc. Breaking it down, this formula calculates the amount of charge per unit area by dividing the total charge by the total area of the disc, which is \( \pi r^2 \). This uniform distribution is a key aspect in analyzing the magnetic effects of the rotating disc.

Circular Current Loop

The next key concept to understand is how a rotating charged disc can be equated to a **circular current loop**. When the charged disc rotates at a speed of \( n \) rotations per second, it effectively creates a loop of moving charge, which acts like a current. This is because current, by definition, is the flow of charge over time.
To determine the current \( I \) associated with this rotating disc, consider the entire disc area \( \pi r^2 \) being swept out once per rotation:

• The formula \( I = \sigma \cdot \text{area swept per second} \)

Substituting in the surface charge density \( \sigma \) and the total charge area, the current becomes:

• \( I = q n \)

Since the charge completes \( n \) rotations each second, every point on the disc contributes to this loop current. This current helps us find the magnetic field at the center of the disc.

Magnetic Moment

Finally, the **magnetic moment** is an essential concept when studying the magnetic effects created by moving charges, such as those in a rotating disc. A magnetic moment is a vector that represents the magnitude and direction of a magnetic source.
The rotating current loop sets up a magnetic field whose strength at the center can be determined by a simple formula:

• \( B = \frac{\mu_0 I}{2r} \)

Here, \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space, \( I \) is the current we already calculated for the rotating disc, and \( r \) is the radius. With \( I = q n \), substituting this into the magnetic field equation gives us:

• \( B = \frac{\mu_0 q n}{2r} \)

Thus, the rotating charged disc results in a magnetic field at its center, forming a complete picture of how charges in motion can create magnetism. This equation tells us the influence of the current generated by the spinning charge and is crucial for understanding magnetic effects in physics.