A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil:

Options

A. `(mu_(0)I)/(4(a-b))[(1)/(a)-(1)/(b)]`
B. `(mu_(0)I)/(8)[(a+b)/(a-b)]`
C. `(mu_(0)IN)/(2(b-a)) log_(e )(b)/(a)`
D. `(mu_(0)I)/(8)(a-b)/(a+b)`

Answer

To find the magnetic field at the center of a tightly wound spiral coil with inner radius \( a \), outer radius \( b \), and \( N \) turns carrying a current \( I \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Coil Structure**:
- The coil is wound in a spiral form with inner radius \( a \) and outer radius \( b \). The total number of turns is \( N \).

2. **Determine the Width of a Small Section**:
- Consider a small radial section of the coil with a width \( dR \). The radius of this section will be \( R \), where \( R \) varies from \( a \) to \( b \).

3. **Calculate the Number of Turns in the Small Section**:
- The total length of the coil at radius \( R \) can be approximated as \( 2\pi R \). The number of turns \( dN \) in the small section \( dR \) can be calculated as:
\[ dN = \frac{N}{b - a} \cdot dR \]
- This is because the total number of turns \( N \) is uniformly distributed across the width \( b - a \).

4. **Magnetic Field Contribution from the Small Section**:
- The magnetic field \( dB \) at the center due to the small section can be expressed using the formula for the magnetic field due to a circular loop:
\[ dB = \frac{\mu_0 I}{2R} dN \]
- Substituting \( dN \):
\[ dB = \frac{\mu_0 I}{2R} \cdot \frac{N}{b - a} dR \]

5. **Integrate to Find Total Magnetic Field**:
- To find the total magnetic field \( B \) at the center, integrate \( dB \) from \( R = a \) to \( R = b \):
\[ B = \int_a^b dB = \int_a^b \frac{\mu_0 I N}{2(b - a) R} dR \]
- This integral simplifies to:
\[ B = \frac{\mu_0 I N}{2(b - a)} \int_a^b \frac{1}{R} dR \]

6. **Evaluate the Integral**:
- The integral \( \int \frac{1}{R} dR \) is \( \ln R \):
\[ B = \frac{\mu_0 I N}{2(b - a)} \left[ \ln R \right]_a^b = \frac{\mu_0 I N}{2(b - a)} (\ln b - \ln a) \]
- Using the property of logarithms, \( \ln b - \ln a = \ln \frac{b}{a} \):
\[ B = \frac{\mu_0 I N}{2(b - a)} \ln \frac{b}{a} \]

### Final Answer:
The magnetic field at the center of the spiral coil is given by:
\[ B = \frac{\mu_0 I N}{2(b - a)} \ln \frac{b}{a} \]