Find the magnetic field induction at a point on the axis of a circular coil carrying current and hence find the magnetic field at the centre of circular coil carrying current.

Options

A. `mu_(0)/2`
B. `mu_(0)i/10sqrt(5)a`
C. `mu_(0)i/4a`
D. `mu_(0)i`

Answer

To find the magnetic field induction at a point on the axis of a circular coil carrying current and then determine the magnetic field at the center of the coil, we can follow these steps:

### Step 1: Understand the Setup
Consider a circular coil of radius \( a \) carrying a current \( I \). We want to find the magnetic field \( B \) at a point \( P \) located at a distance \( x \) along the axis of the coil.

### Step 2: Use Biot-Savart Law
According to the Biot-Savart Law, the magnetic field \( dB \) due to a small current element \( dL \) at a point in space is given by:
\[ dB = \frac{\mu_0 I}{4\pi} \frac{dL \times \hat{R}}{R^2} \]
where \( \hat{R} \) is the unit vector pointing from the current element to the point \( P \), and \( R \) is the distance from the current element to point \( P \).

### Step 3: Define the Geometry
For a circular coil:
- The distance \( R \) from the current element \( dL \) to point \( P \) can be expressed as:
\[ R = \sqrt{a^2 + x^2} \]
- The angle \( \theta \) between the current element and the line connecting the element to point \( P \) can be defined such that:
\[ \sin \theta = \frac{a}{R} \]

### Step 4: Resolve the Magnetic Field Components
The magnetic field component \( dB \) at point \( P \) due to the current element can be resolved into components. The vertical components (along the axis) will add up, while the horizontal components will cancel out due to symmetry.

Thus, the total magnetic field \( B \) at point \( P \) is given by:
\[ B = \int dB \sin \theta \]

### Step 5: Substitute Values
Substituting the expressions for \( dB \) and \( \sin \theta \):
\[ B = \int \frac{\mu_0 I}{4\pi} \frac{dL \cdot a}{(a^2 + x^2)^{3/2}} \]
The integral \( \int dL \) over the entire circular coil gives the total length of the coil, which is \( 2\pi a \).

### Step 6: Final Expression for Magnetic Field
Thus, we can write:
\[ B = \frac{\mu_0 I a}{4\pi (a^2 + x^2)^{3/2}} \cdot 2\pi a = \frac{\mu_0 I a}{2 (a^2 + x^2)^{3/2}} \]

### Step 7: Find the Magnetic Field at the Center
To find the magnetic field at the center of the coil, we set \( x = 0 \):
\[ B_{\text{center}} = \frac{\mu_0 I a}{2 (a^2)^{3/2}} = \frac{\mu_0 I}{2a} \]

### Conclusion
The magnetic field induction at the center of the circular coil carrying current is:
\[ B = \frac{\mu_0 I}{2a} \]