A coil having `N` turns is would tightly in the form of a spiral with inner and outer radii a and `b` respectively. When a current `I` passes through the coil, the magnetic field at the centre is.

Options

A. `(mu_(0)NI)/(b)`
B. `(mu_(0)NI)/(a)`
C. `(mu_(0)NI)/(2(b-a))ln((b)/(a))`
D. `(mu_(0)NI)/((b-a))ln((b)/(a))`

Answer

To find the magnetic field at the center of a coil wound in the form of a spiral with inner radius \( a \) and outer radius \( b \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Coil Structure**:
- The coil has \( N \) turns and is tightly wound in a spiral shape with inner radius \( a \) and outer radius \( b \).

2. **Consider an Element of the Coil**:
- We can consider a thin ring element of thickness \( dr \) at a distance \( r \) from the center. The number of turns in this thin ring can be expressed as \( dn \).

3. **Finding the Number of Turns in the Element**:
- The total number of turns \( N \) is distributed over the radial distance from \( a \) to \( b \). Therefore, the number of turns per unit length can be given by:
\[ \frac{N}{b - a} \]
- Thus, the number of turns \( dn \) in the thickness \( dr \) is:
\[ dn = \frac{N}{b - a} \cdot dr \]

4. **Magnetic Field Contribution from the Element**:
- The magnetic field \( dB \) at the center due to this thin ring is given by:
\[ dB = \frac{\mu_0 \cdot I \cdot dn}{2 \pi r} \]
- Substituting \( dn \):
\[ dB = \frac{\mu_0 \cdot I}{2 \pi r} \cdot \frac{N}{b - a} \cdot dr \]

5. **Integrating to Find Total Magnetic Field**:
- To find the total magnetic field \( B \) at the center, we integrate \( dB \) from \( r = a \) to \( r = b \):
\[ B = \int_{a}^{b} dB = \int_{a}^{b} \frac{\mu_0 \cdot I \cdot N}{2 \pi (b - a)} \cdot \frac{dr}{r} \]
- This integral simplifies to:
\[ B = \frac{\mu_0 \cdot I \cdot N}{2 \pi (b - a)} \cdot \left[ \ln(r) \right]_{a}^{b} \]
- Evaluating the integral gives:
\[ B = \frac{\mu_0 \cdot I \cdot N}{2 \pi (b - a)} \cdot (\ln(b) - \ln(a)) = \frac{\mu_0 \cdot I \cdot N}{2 \pi (b - a)} \cdot \ln\left(\frac{b}{a}\right) \]

6. **Final Expression for the Magnetic Field**:
- Thus, the magnetic field at the center of the coil is:
\[ B = \frac{\mu_0 \cdot N \cdot I}{2 \pi (b - a)} \cdot \ln\left(\frac{b}{a}\right) \]