A circular coil of n turns and radius r carries a current I. The magnetic field at the centre is

Options

A. `(mu_(0)i)/(2R)`
B. `(i)/(2c^(2)epsilon_(0)R)`
C. `(mu_(0)i)/(2piR)`
D. `(ic^(2))/(2epsilon_(0)R)`

Answer

To find the magnetic field at the center of a circular coil of n turns and radius r carrying a current I, we can follow these steps:

### Step 1: Understand the Magnetic Field Contribution
The magnetic field at the center of a circular coil is derived from the contributions of each infinitesimal segment of the coil. For a small segment of the coil carrying current, the magnetic field can be expressed using the Biot-Savart Law.

### Step 2: Apply the Biot-Savart Law
The Biot-Savart Law states that the magnetic field \( dB \) due to a small current element \( dl \) at a distance \( r \) is given by:
\[ dB = \frac{\mu_0}{4\pi} \frac{I \, dl \times r}{r^3} \]
In this case, since we are considering the magnetic field at the center of the coil, the distance \( r \) is constant for all elements of the coil.

### Step 3: Simplify the Expression
Since the angle between the current element \( dl \) and the radius vector \( r \) is 90 degrees, we can simplify the expression:
\[ dB = \frac{\mu_0 I}{4\pi r^2} \, dl \]
Here, \( r \) is the radius of the coil.

### Step 4: Integrate Over the Entire Coil
To find the total magnetic field \( B \) at the center, we need to integrate \( dB \) over the entire coil. The total length of the coil is \( 2\pi r \) (the circumference of the circle):
\[ B = \int dB = \frac{\mu_0 I}{4\pi r^2} \int dl \]
The integral of \( dl \) over one complete turn of the coil is \( 2\pi r \):
\[ B = \frac{\mu_0 I}{4\pi r^2} \cdot 2\pi r \]

### Step 5: Factor in the Number of Turns
Since the coil has \( n \) turns, we multiply the result by \( n \):
\[ B = n \cdot \frac{\mu_0 I}{4\pi r^2} \cdot 2\pi r \]
This simplifies to:
\[ B = \frac{n \mu_0 I}{2 r} \]

### Final Result
Thus, the magnetic field at the center of the circular coil is:
\[ B = \frac{n \mu_0 I}{2 r} \]