A point P lies on the axis of a flat coil carryinga current. The mangetc moment of the coil is `mu`. What will be the magnetic field at point P ? It is given that the distance of P from the centre of coil is d, which is large compared to the radius of the coil.

Options

A. `(mu_(0))/( 2pi ) ((mu)/(d^(3)))`
B. `(mu_(0))/( 4pi ) ((mu)/(d^(3)))`
C. `(mu_(0))/( 6pi ) ((mu)/(d^(2)))`
D. `(mu_(0))/( 8pi ) ((mu)/(d^(2)))`

Answer

To find the magnetic field at point P, which lies on the axis of a flat coil carrying a current, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a flat circular coil of radius \( r \) carrying a current \( I \).
- The magnetic moment \( \mu \) of the coil is given.
- The point P is located at a distance \( d \) from the center of the coil along its axis, where \( d \) is much larger than \( r \) (i.e., \( d \gg r \)).

2. **Magnetic Field Due to a Small Element**:
- Consider a small current element \( dl \) on the coil. The magnetic field \( dB \) at point P due to this element can be expressed using the Biot-Savart Law:
\[ dB = \frac{\mu_0 I \, dl}{4 \pi r^2} \]
- Here, \( r \) is the distance from the current element to point P, which can be approximated as:
\[ r = \sqrt{d^2 + r^2} \]
- However, since \( d \gg r \), we can simplify this to:
\[ r \approx d \]

3. **Components of the Magnetic Field**:
- The magnetic field has both vertical and horizontal components. The vertical components from opposite sides of the coil will cancel each other out, leaving only the horizontal component.
- The horizontal component of the magnetic field \( dB_h \) can be expressed as:
\[ dB_h = dB \cdot \cos(\theta) \]
- Using the small angle approximation, where \( \theta \) is the angle between the radius of the coil and the line connecting the current element to point P, we can express \( \cos(\theta) \) in terms of \( r/d \).

4. **Integrating Over the Coil**:
- To find the total magnetic field \( B \) at point P, we need to integrate \( dB_h \) over the entire coil:
\[ B = \int dB_h = \int \frac{\mu_0 I \, dl \cdot \cos(\theta)}{4 \pi d^2} \]
- The total length of the coil is \( 2\pi r \), and we can substitute \( dl \) with \( r \, d\theta \) to perform the integration.

5. **Substituting Magnetic Moment**:
- The magnetic moment \( \mu \) of the coil is defined as:
\[ \mu = I \cdot A = I \cdot \pi r^2 \]
- We can express \( I \) in terms of \( \mu \):
\[ I = \frac{\mu}{\pi r^2} \]

6. **Final Expression for Magnetic Field**:
- Substituting \( I \) back into the expression for \( B \) and simplifying gives:
\[ B = \frac{\mu_0 \mu}{2 \pi d^3} \]
- This is the magnetic field at point P, where \( d \) is much larger than the radius of the coil.

### Final Answer:
The magnetic field at point P is given by:
\[ B = \frac{\mu_0 \mu}{2 \pi d^3} \]