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Assume the dipole model for earth’s magnetic field B which is given by BV = vertical component of magnetic field

$=\frac{\mu _{0}}{ 4 \pi}\frac{2m \cos \theta }{r^{3}}$

BH = Horizontal component of magnetic field

$=\frac{\mu _{0}}{ 4 \pi}\frac{ \sin \theta m }{r^{3}}$
$\theta = 90^{\circ}$ – latitude as measured from magnetic equator.
Find loci of points for which (i) B is minimum; (ii) dip angle is zero; and (iii) dip angle is $\pm 45^{\circ}$

Answers (1)

$(a)B_v=\mu _{0}2m\frac{\cos \theta}{ 4\pi r^3 }$

$B_H=\frac{\mu _{0}}{4 \pi}\frac{m\sin \theta}{ r^{3} }$

These are the components of B to a net magnetic field will be

$B=\sqrt{B_v^2+ B_H^2}=\frac{\mu_{0}m}{4\pi r^3}[3\cos 2 \theta +1]^{\frac{1}{}2}$

From the above equation, the value of B is minimum, if

$\cos \theta =\frac{\pi}{}2$

$\theta =\frac{\pi}{}2$

Thus, B is minimum at the magnetic equator.

(b) Angle of dip

$\tan \delta =\frac{B_v}{B_H}=\frac{\mu _{0}}{4 \pi }\frac{.2m \frac{\cos \theta }{r^3}}{\frac{\mu _{0}}{4\pi }\sin \theta .\frac{m}{r^{3}}} =2\cot \theta .........i$

$\tan \delta =2\cot \theta$

For dip angle is zero $\tan \delta =0$

$\cot \theta =0 , \theta =\frac{\pi}{}2$

For this value of θ angle of dip is vanished. It means that locus is again magnetic equator.

(c)

$tan \delta =\frac{B_v}{B_H}$

Angle of dip

$\delta =\pm 45^{\circ}$

$\frac{ B_v}{B_H}=tan \pm 45^{\circ}$

$\frac{ B_v}{B_H}=1$

$2 \cot \theta =1$

$\tan \theta =\frac{1}{}2$

$\theta =\tan ^{-1}({\frac{1}{}2})$

Thus, $\theta =\tan ^{-1}({\frac{1}{}2})$ is the locus.

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