Question

Assume the dipole model for earth's magnetic field B which is given by the vertical component of magnetic field, $B_v=\frac{{\mu }_0}{4\pi }\frac{2mcos\theta }{r^3}$ and the horizontal component of magnetic field is $B_H=\frac{{\mu }_0}{4\pi }\frac{msin\theta }{r^3}$
where $\theta ={90}^0$ - latitude as measured from magnetic equator, then the loci of point for which dip angle is ${45}^0$.

• A. $ta{n}^{-1}\left(3\right)$
• B. $ta{n}^{-1}\left(2\right)$
• C. $ta{n}^{-1}\left(0.5\right)$
• D. $ta{n}^{-1}\left(1\right)$

Answer 1

The correct option is B $ta{n}^{-1}\left(2\right)$

Given that,

Vertical component of Earth's magnetic field, $B_V=\frac{{\mu }_0}{4\pi }\frac{2m\cos \theta }{r^3}$

Horizontal component of Earth's magnetic field, $B_H=\frac{{\mu }_0}{4\pi }\frac{msin\theta }{r^3}$

Angle of dip, $\delta ={45}^o$

$\because tan\delta =\frac{B_V}{B_H}$

$⟹tan{45}^o=\frac{2cos\theta }{sin\theta }$

$⟹1=\frac{2}{tan\theta }$

$⟹tan\theta =2$

or $\theta =ta{n}^{-1}\left(2\right)$ is the required loci of points.

So, $\left(B\right)$ is correct option.