Assume the dipole model for earth's magnetic field B which is given by B_V = vertical component of magnetic field = _0 4 2m r^3 B_H = Horizontal component of magnetic field = _0 4 r^3 θ = 90º - lattitude as measured from magnetic equator. Find loci of points for which, 1. |B| is minimum, 2. Dip angle is zero, 3. Dip angle is ± 45º.

1. B_V= _02m 4 B_H= _0 4 m r^3 These are the components of B to net magnetic field will be B= B_v^2+B_H^2 = _0m 4 ^3 [3 ^ 2 +1 ]^1 2 From above equation, the value of B is minimum, if cos = 2 . = 2 . Thus, B is minimum at the magnetic equator. 2. Angle of dip, = B_V B_H = _0 4 . 2m r^3 _0 4 . .m r^3 =2 .....(i) =2 For dip angle is zero i.e., = 0 =0 = 2 For this value of angle fo dip is vanished. It means that locus is again magnetic equator. 3. = B_V B_H Angle of dip i.e., = 45^ B_V B_H = ( 45^ ) B_V B_H =1 2 =1 [From Eq. (i) ] =1 2 =1 2 = ^ -1 (2) Thus, = ^ -1 (2) is the locus.

Assume the dipole model for earth's magnetic field B which is giv…

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Question

Assume the dipole model for earth's magnetic field B which is given by $B_V =$ vertical component of magnetic field $=\frac{\mu_0}{4\pi}\frac{2\text{m}\cos\theta}{\text{r}^3}$
$B_H =$ Horizontal component of magnetic field $=\frac{\mu_0}{4\pi}\frac{\sin\theta\text{m}}{\text{r}^3}$
θ = 90º - lattitude as measured from magnetic equator. Find loci of points for which,

|B| is minimum,
Dip angle is zero,
Dip angle is ± 45º.

✓

Answer

$\text{B}_\text{V}=\frac{\mu_02\text{m}\cos\theta}{4\pi}$

$\text{B}_\text{H}=\frac{\mu_0}{4\pi}\frac{\text{m}\sin\theta}{\text{r}^3}$
These are the components of B to net magnetic field will be
$\text{B}=\sqrt{\text{B}_\text{v}^2+\text{B}_\text{H}^2}=\frac{\mu_0\text{m}}{4\pi\text{r}^3}\big[3\cos^{2}\theta+1\big]^\frac{1}{2}$
From 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.

Angle of dip,

$\tan\delta=\frac{\text{B}_\text{V}}{\text{B}_\text{H}}=\frac{\frac{\mu_0}{4\pi}.\frac{2\text{m}\cos\theta}{\text{r}^3}}{\frac{\mu_0}{4\pi}.\frac{\sin\theta.\text{m}}{\text{r}^3}}=2\cot\theta\ .....(\text{i})$
$\tan\delta=2\cot\theta$
For dip angle is zero i.e., $\delta = 0$
$\cot\theta=0$
$\theta=\frac{\pi}{2}$
For this value of $\theta$ angle fo dip is vanished. It means that locus is again magnetic equator.

$\tan\delta=\frac{\text{B}_\text{V}}{\text{B}_\text{H}}$

Angle of dip i.e., $\delta=\pm45^\circ$
$\frac{\text{B}_\text{V}}{\text{B}_\text{H}}=\tan(\pm45^\circ)$
$\frac{\text{B}_\text{V}}{\text{B}_\text{H}}=1$
$2\cot\theta=1\ \big[\text{From Eq. (i)}\big]$
$\cot\theta=\frac{1}{2}$
$\tan\theta=\frac{1}{2}$
$\Rightarrow\ \theta=\tan^{-1}(2)$
Thus, $\theta=\tan^{-1}(2)$ is the locus.

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