i. Consider a bar magnet of magnetic moment \(\overset\rightarrow{m}\) with centre at O as shown in figure and let P be any point in its magnetic field.
ii. Magnetic moment \(\overset\rightarrow{m}\) is resolved into components along \(\overset\rightarrow{r}\) and perpendicular to \(\overset\rightarrow{r}\).
iii. For the component m cos θ along \(\overset\rightarrow{r}\), the point P is an axial point.
iv. For the component m sin θ perpendicular to \(\overset\rightarrow{r}\), the point P is an equatorial point at the same distance \(\overset\rightarrow{r}\).
v. For a point on the axis, Ba = \(\frac{\mu_0}{4\pi}\,\frac{2m}{r^3}\)
Here,
directed along m cosθ.
vi. For point on equator,
directed opposite to m sin θ
vii. Thus, the magnitude of the resultant magnetic field B, at point P is given by
viii. Let a be the angle made by the direction of \(\overset\rightarrow{B}\) with \(\overset\rightarrow{r}.\) Then, by using equation (1) and equation (2),
tan α = \(\frac{B_{eq}}{B_a}\) = \(\frac{1}{2}\) (tan θ)
The angle between directions of \(\overset\rightarrow{B}\) and \(\overset\rightarrow{m}\) is then (θ + a).
