Question:
A coil of n number of turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current of strength I is passed through the coil, the magnetic field at its centre is
A coil of n number of turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current of strength I is passed through the coil, the magnetic field at its centre is
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For a spiral, integrate the circular coil formula $B = \mu_0 I / 2r$ over the range of radii.
Updated On: Apr 10, 2026
- $\frac{\mu_{0}nI}{(b-a)}log_{e}\frac{a}{b}$
- $\frac{\mu_{0}nI}{2(b-a)}$
- $\frac{2\mu_{o}nI}{b}$
- $\frac{\mu_{0}nI}{2(b-a)}log_{e}\frac{b}{a}$
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The Correct Option is D
Solution and Explanation
Step 1: Setup
Consider a small element $dr$ at distance $r$. Turns per unit length $= \frac{n}{b-a}$.
Step 2: Field Element
Field due to element $dr$: $dB = \frac{\mu_0 I dn}{2r} = \frac{\mu_0 I}{2r} (\frac{n dr}{b-a})$.
Step 3: Integration
$B = \int_{a}^{b} \frac{\mu_0 n I}{2(b-a)} \frac{dr}{r} = \frac{\mu_0 n I}{2(b-a)} [log_e r]_a^b = \frac{\mu_0 n I}{2(b-a)} log_e \frac{b}{a}$.
Final Answer: (d)
Consider a small element $dr$ at distance $r$. Turns per unit length $= \frac{n}{b-a}$.
Step 2: Field Element
Field due to element $dr$: $dB = \frac{\mu_0 I dn}{2r} = \frac{\mu_0 I}{2r} (\frac{n dr}{b-a})$.
Step 3: Integration
$B = \int_{a}^{b} \frac{\mu_0 n I}{2(b-a)} \frac{dr}{r} = \frac{\mu_0 n I}{2(b-a)} [log_e r]_a^b = \frac{\mu_0 n I}{2(b-a)} log_e \frac{b}{a}$.
Final Answer: (d)
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