Question

Using the electrostatic analogue, obtain the magnetic field at a distance x on the perpendicular bisector of a magnetic dipole \[\vec{m}\]. For \[x\gg R\], verify that `vec B = mu_0/(4 pi) vec m/x^3`.

Answer 1

Consider a magnetic dipole consisting of two poles \[+m_p\] and \[-m_p\], separated by a distance 2R. Let P be a point on its perpendicular bisector at a distance x from the centre.

The distance of P from each pole is

r = \[\sqrt{x^2+R^2}\]

The magnetic field due to each pole is

\[B_1=B_2=\frac{\mu_0}{4\pi}\frac{m_p}{r^2}\]

The components perpendicular to the dipole axis cancel, while the components along the dipole axis add.

Hence,

B = \[2B_1\cos\theta\]

where

cos⁡ θ = \[Rr\cos\theta=\frac{R}{r}\]

Therefore,

B = \[2\left(\frac{\mu_0}{4\pi}\frac{m_p}{r^2}\right)\frac{R}{r}\]

 B = \[μ04π2mpRr3B=\frac{\mu_0}{4\pi}\frac{2m_pR}{r^3}\]

But the magnetic dipole moment is

m = \[2m_pR\]

Hence,

B = \[\frac{\mu_0}{4\pi} \frac{m}{(x^2+R^2)^{3/2}}\]

The field on the perpendicular bisector is directed opposite to \[\vec m\]. Therefore, in vector form,

\[\vec B= -\frac{\mu_0}{4\pi} \frac{\vec m}{(x^2+R^2)^{3/2}}\]

For x ≫ R,

\[x^2+R^2\approx x^2\]

Therefore,

\[\vec B= -\frac{\mu_0}{4\pi}\frac{\vec m}{x^3}\]

Thus, the magnitude of the magnetic field is

\[B=\frac{\mu_0}{4\pi}\frac{m}{x^3}\]

The minus sign in the vector form indicates that the field on the perpendicular bisector is opposite to the magnetic dipole moment.