Question

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 :

• A. \(\frac{\mu_0 IN}{2(b - a)} \log_e \left(\frac{b}{a}\right)\)
• B. \(\frac{\mu_0 I}{8} \left(\frac{a - b}{a + b}\right)\)
• C. \(\frac{\mu_0 I}{4(a - b)} \left[\frac{1}{a} - \frac{1}{b}\right]\)
• D. \(\frac{\mu_0 I}{8} \left[\frac{a + b}{a - b}\right]\)

Hint:

For continuous current distributions in circular geometries, always use the formula for a single loop and integrate over the density of turns.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The magnetic field at the center of a single circular loop of radius \( r \) carrying current \( I \) is given by \( B = \frac{\mu_0 I}{2r} \).
For a spiral coil, we treat it as a collection of infinitesimal concentric loops and integrate their contributions.
Step 2: Key Formula or Approach:
Number of turns per unit radial thickness is \( n = \frac{N}{b-a} \).
For a small radial element \( dr \) at distance \( r \) from the center, the number of turns is \( dN = n \, dr = \frac{N}{b-a} dr \).
Step 3: Detailed Explanation:
The infinitesimal magnetic field \( dB \) at the center due to the turns in width \( dr \) is:
\[ dB = \frac{\mu_0 (dN) I}{2r} = \frac{\mu_0 I}{2r} \left( \frac{N}{b-a} dr \right) \]
Total magnetic field \( B \) is obtained by integrating from \( r = a \) to \( r = b \):
\[ B = \int_a^b \frac{\mu_0 NI}{2(b-a)} \frac{1}{r} dr \]
\[ B = \frac{\mu_0 NI}{2(b-a)} \int_a^b \frac{1}{r} dr \]
\[ B = \frac{\mu_0 NI}{2(b-a)} \left[ \ln r \right]_a^b \]
\[ B = \frac{\mu_0 NI}{2(b-a)} (\ln b - \ln a) = \frac{\mu_0 NI}{2(b-a)} \ln \left( \frac{b}{a} \right) \]
Step 4: Final Answer:
The magnetic field at the center is \(\frac{\mu_0 IN}{2(b - a)} \log_e \left(\frac{b}{a}\right)\).