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

The problem involves calculating the magnetic field at the center of a spiral coil with N turns, when a current I is passing through it. The coil is tightly wound with inner radius a and outer radius b. Let's solve this step by step:

1. The spiral coil can be thought of as a series of tight concentric loops between the radii a and b. Each small loop at a radius r contributes to the magnetic field at the center.
2. For a single loop of radius r, the magnetic field at the center is given by the expression: \[\frac{\mu_0 I}{2r}\], where \mu_0 is the permeability of free space.
3. Considering a differential element dr at radius r, the number of turns per unit length is \frac{N}{b-a} because the total number of turns is N across a radial distance from a to b.
4. Thus, the infinitesimal contribution to the magnetic field by an elemental ring is: dB = \frac{\mu_0 I}{2r} \cdot \frac{N}{b-a} \, dr
5. To find the total magnetic field at the center, integrate dB from the inner to the outer radius: \[ B = \int_a^b \frac{\mu_0 I}{2r} \cdot \frac{N}{b-a} \, dr = \frac{\mu_0 IN}{2(b-a)} \int_a^b \frac{1}{r} \, dr \]
6. The integral of \frac{1}{r} is: \[\int_a^b \frac{1}{r} \, dr = \log_e(b) - \log_e(a) = \log_e\frac{b}{a}\]
7. Substitute back the integral value: B = \frac{\mu_0 IN}{2(b-a)} \log_e \left(\frac{b}{a}\right)

Thus, the magnetic field at the center of the spiral coil is \frac{\mu_0 IN}{2(b-a)} \log_e \left(\frac{b}{a}\right), which is the correct answer.