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प्रश्न
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`.
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उत्तर
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.
