Magnetic field due to a ring having n turns at a distance `x` on its axis is proportional to (if `r =` radius of ring)

Options

A. (a) `(r)/((x^(2)+r^(2)))`
B. (b) `(r)/((x^(2)+r^(2))^(3//2))`
C. (c) `(nr^(2))/((x^(2)+r^(2))^(3//2))`
D. (d) `(n^(2)r^(2))/((x^(2)+r^(2))^(3//2))`

Answer

To find the magnetic field due to a ring with \( n \) turns at a distance \( x \) on its axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a circular ring of radius \( r \) carrying a current \( I \).
- We want to find the magnetic field at a point \( P \) located at a distance \( x \) along the axis of the ring.

2. **Use Biot-Savart Law**:
- According to the Biot-Savart Law, the magnetic field \( dB \) due to a small current element \( dL \) is given by:
\[ dB = \frac{\mu_0}{4\pi} \frac{I \, dL \sin \theta}{r^2} \]
- Here, \( \theta \) is the angle between the current element and the line connecting the element to the point \( P \).

3. **Determine \( \sin \theta \)**:
- In the right triangle formed by the radius \( r \), the distance \( x \), and the distance from the current element to point \( P \):
\[ \sin \theta = \frac{r}{\sqrt{x^2 + r^2}} \]

4. **Substituting \( \sin \theta \) into \( dB \)**:
- Substitute \( \sin \theta \) into the expression for \( dB \):
\[ dB = \frac{\mu_0}{4\pi} \frac{I \, dL \cdot \frac{r}{\sqrt{x^2 + r^2}}}{(x^2 + r^2)} \]

5. **Integrate Over the Ring**:
- The total magnetic field \( B \) is obtained by integrating \( dB \) around the entire ring:
\[ B = \int dB = \int \frac{\mu_0 I r}{4\pi} \frac{dL}{(x^2 + r^2)^{3/2}} \]
- The integral of \( dL \) around the ring is \( 2\pi r \):
\[ B = \frac{\mu_0 I r}{4\pi} \cdot \frac{2\pi r}{(x^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2 (x^2 + r^2)^{3/2}} \]

6. **Considering \( n \) Turns**:
- For \( n \) turns, the magnetic field becomes:
\[ B = \frac{\mu_0 n I r^2}{2 (x^2 + r^2)^{3/2}} \]

7. **Identify Proportionality**:
- From the final expression, we can see that the magnetic field \( B \) is proportional to:
\[ B \propto \frac{r^2}{(x^2 + r^2)^{3/2}} \]

### Final Answer:
The magnetic field due to a ring having \( n \) turns at a distance \( x \) on its axis is proportional to:
\[ \frac{r^2}{(x^2 + r^2)^{3/2}} \]