Question

The magnetic field at a point, 10 cm away from a magnetic dipole, is found to be 2.0 × 10⁻⁴ T. Find the magnetic moment of the dipole if the point is (a) in end-on position of the dipole and (b) in broadside-on position of the dipole.

Answer 1

Given:
Magnetic field strength, B = 2 × 10⁻⁴ T
Distance of the point from the dipole, d = 10 cm = 0.1 m

(a) If the point is at the end-on position:
The magnetic field $\left(B\right)$ on the axial point of the dipole is given by
$$\begin{gathered} B=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2M}{d^3} \\ \mathrm{Here},M\mathrm{is}\mathrm{the}\mathrm{magnetic}\mathrm{moment}\mathrm{of}\mathrm{the}\mathrm{dipole}\mathrm{that}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{find}\mathrm{out}. \\ \therefore 2\times {10}^{-4}=\frac{{10}^{-7}\times 2M}{{\left({10}^{-1}\right)}^3} \\ \Rightarrow M=\frac{2\times {10}^{-4}\times {10}^{-3}}{{10}^{-7}\times 2} \\ \Rightarrow M=1\mathrm{A}-{\mathrm{m}}^2 \end{gathered}$$

(b) If the point is at broadside-on position (equatorial position):
The magnetic field $\left(B\right)$ is given by
$$\begin{gathered} B=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathit{M}}{{\mathit{d}}^{\mathit{3}}} \\ \Rightarrow 2\times {10}^{-4}=\frac{{10}^{-7}\times M}{{\left({10}^{-1}\right)}^3} \\ \Rightarrow M=2\mathrm{A}-{\mathrm{m}}^2 \end{gathered}$$