Question

There is a spiral which has \(r_i=3\,\text{cm\), \(r_{ext}=6\,\text{cm}\), \(I=20\,\text{mA}\) and \(N=200\), where \(r_i\) : internal radius \(r_{ext}\) : external radius \(N\) : number of turns \(I\) : current. Find the magnetic moment of the spiral.}

• A. \(2.64\times10^{-2}\,\text{A m}^2\)
• B. \(4.87\times10^{-2}\,\text{A m}^2\)
• C. \(3.65\times10^{-2}\,\text{A m}^2\)
• D. \(6.67\times10^{-2}\,\text{A m}^2\)

Answer 1

The Correct Option is A

Concept:
Magnetic moment of a current loop is \[ M = I \times A \] For a spiral coil with continuously varying radius, the total magnetic moment is obtained by integrating the contribution of small circular loops.

Step 1: Magnetic moment of a small ring. For a small ring of radius \(r\), \[ dM = I\,dA \] \[ dM = I(\pi r^2) \] If the number of turns changes continuously, \[ dM = I(dN)\pi r^2 \] Step 2: Relate \(dN\) with radius. Since \(N\) turns are distributed between \(r_i\) and \(r_{ext}\), \[ dN=\frac{N}{r_{ext}-r_i}\,dr \] Substitute: \[ dM = I\pi r^2 \frac{N}{r_{ext}-r_i}dr \] Step 3: Integrate between the limits. \[ M = \int_{r_i}^{r_{ext}} I\pi r^2 \frac{N}{r_{ext}-r_i}dr \] \[ M = \frac{I\pi N}{r_{ext}-r_i}\int_{r_i}^{r_{ext}} r^2 dr \] \[ M = \frac{I\pi N}{r_{ext}-r_i}\left[\frac{r^3}{3}\right]_{r_i}^{r_{ext}} \] \[ M = \frac{I\pi N}{3(r_{ext}-r_i)}(r_{ext}^3-r_i^3) \] Step 4: Substitute numerical values. \[ I = 20\times10^{-3}\,A \] \[ N = 200 \] \[ r_{ext}=6\times10^{-2}\,m \] \[ r_i=3\times10^{-2}\,m \] \[ M= \frac{20\times10^{-3}\times3.14\times200}{3(6-3)\times10^{-2}} (6^3-3^3)\times10^{-6} \] \[ M = 2.64\times10^{-2}\,\text{A m}^2 \]