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 ______.

• A. 4.4
• B. 2.64
• C. 3.25
• D. 1.2

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A flat coil can be considered as a collection of infinitesimally thin circular current loops. The total magnetic moment is the integral of the magnetic moments of all these infinitesimal loops from the inner radius to the outer radius.
Step 2: Key Formula or Approach:
The magnetic moment of a single loop of radius \(r\) carrying current \(I\) is \(M = I \cdot A = I(\pi r^2)\).
Since the \(N\) turns are uniformly distributed between \(r_1\) and \(r_2\), the number of turns per unit radial width is \(n = \frac{N}{r_2 - r_1}\).
The number of turns in an elemental ring of thickness \(dr\) is \(dN = n \, dr = \frac{N}{r_2 - r_1} dr\).
The elemental magnetic moment is \(dM = (dN) \cdot I \cdot (\pi r^2)\).
Integrating this from \(r_1\) to \(r_2\) gives the total magnetic moment.
Step 3: Detailed Explanation:
Set up the integral:
\[ M = \int_{r_1}^{r_2} \left( \frac{N}{r_2 - r_1} dr \right) I (\pi r^2) = \frac{N I \pi}{r_2 - r_1} \int_{r_1}^{r_2} r^2 dr \] \[ M = \frac{N I \pi}{r_2 - r_1} \left[ \frac{r^3}{3} \right]_{r_1}^{r_2} = \frac{N I \pi}{3(r_2 - r_1)} (r_2^3 - r_1^3) \] Using the algebraic identity \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\):
\[ M = \frac{N I \pi}{3(r_2 - r_1)} (r_2 - r_1)(r_2^2 + r_1 r_2 + r_1^2) = \frac{N I \pi}{3} (r_2^2 + r_1 r_2 + r_1^2) \] Given values:
\(N = 200\)
\(I = 20 \text{ mA} = 20 \times 10^{-3} \text{ A}\)
\(r_1 = 3 \text{ cm} = 0.03 \text{ m}\)
\(r_2 = 6 \text{ cm} = 0.06 \text{ m}\)
Calculate the term \((r_2^2 + r_1 r_2 + r_1^2)\) in \(\text{m}^2\):
\(r_2^2 + r_1 r_2 + r_1^2 = (0.06)^2 + (0.03)(0.06) + (0.03)^2 = 0.0036 + 0.0018 + 0.0009 = 0.0063 \text{ m}^2\).
Now, substitute all values into the magnetic moment formula:
\[ M = \frac{200 \times 20 \times 10^{-3} \times 3.14}{3} \times 0.0063 \] \[ M = \frac{4 \times 3.14}{3} \times 0.0063 = 4 \times 3.14 \times 0.0021 = 12.56 \times 0.0021 = 0.026376 \text{ A.m}^2 \] Converting this to the given format \(\alpha \times 10^{-2}\):
\(0.026376 = 2.6376 \times 10^{-2} \approx 2.64 \times 10^{-2} \text{ A.m}^2\).
Step 4: Final Answer:
Comparing this with \(\alpha \times 10^{-2}\), we get \(\alpha = 2.64\).