Question:
An insulated wire is wound so that it forms a flat coil with \(N = 200\) turns. The radius of the innermost turn is \(r_1 = 3\) cm, and of the outermost turn \(r_2 = 6\) cm. If 20 mA current flows in it then the magnetic moment will be \(\alpha \times 10^{-2}\) A.m². The value of \(\alpha\) is ______.
An insulated wire is wound so that it forms a flat coil with \(N = 200\) turns. The radius of the innermost turn is \(r_1 = 3\) cm, and of the outermost turn \(r_2 = 6\) cm. If 20 mA current flows in it then the magnetic moment will be \(\alpha \times 10^{-2}\) A.m². The value of \(\alpha\) is ______.
Updated On: Apr 10, 2026
- 4.4
- 2.64
- 3.25
- 1.2
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
For a flat spiral coil, the radius of the turns varies linearly from the inner radius $r_1$ to the outer radius $r_2$. The magnetic moment of a single loop is $I \times A$. For a spiral, we must integrate the moments of all infinitesimal turns across the radius.
Step 2: Key Formula or Approach:
1. Number of turns per unit radial length $n = \frac{N}{r_2 - r_1}$. 2. Magnetic moment of a small element $dr$: $dm = (n \, dr) I (\pi r^2)$. 3. Total magnetic moment $M = \int_{r_1}^{r_2} n I \pi r^2 \, dr$.
Step 3: Detailed Explanation:
1. Substitute the constants: \[ M = \frac{N I \pi}{r_2 - r_1} \int_{r_1}^{r_2} r^2 \, dr = \frac{N I \pi}{r_2 - r_1} \left[ \frac{r^3}{3} \right]_{r_1}^{r_2} \] 2. Simplify the expression using $r_2^3 - r_1^3 = (r_2 - r_1)(r_2^2 + r_1^2 + r_1 r_2)$: \[ M = \frac{N I \pi}{3} (r_2^2 + r_1^2 + r_1 r_2) \] 3. Plug in the values ($N=200, I=0.02, r_1=0.03, r_2=0.06$): \[ M = \frac{200 \times 0.02 \times 3.14}{3} (0.0036 + 0.0009 + 0.0018) \] \[ M = \frac{12.56}{3} (0.0063) = 12.56 \times 0.0021 = 0.026376 \text{ A.m}^2 \] 4. Converting to $\alpha \times 10^{-2}$: $M \approx 2.64 \times 10^{-2}$. So, $\alpha = 2.64$.
Step 4: Final Answer:
The value of $\alpha$ is 2.64.
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