Question

The earth's magnetic field at the geometric poles is $\sqrt{10} \times 10^{-5} T$. The magnitude of the field at a point on the earth's surface where the radius makes an angle $\theta$ with the axis of earth's assumed magnetic dipole is $5 \times 10^{-5} T$. The magnitude of $\theta$ in degree is :

• A. $30^{\circ}$
• B. $60^{\circ}$
• C. $45^{\circ}$
• D. $75^{\circ}$

Answer 1

Answer: (C)

According to the question,
$B_{v}=\sqrt{10} \times 10^{-5} T$

In $\Delta \, OBC$,
$\because \, cos \theta=\frac{B_{V}}{R_{R}}$
$=\frac{\sqrt{10} \times 10^{-5}}{5 \times 10^{-5}}$
$=\frac{\sqrt{2}}{\sqrt{5}}$
or $\theta=50.76^{\circ} \approx 45^{\circ}$
Hence, the magnitude of $\theta$ is $45^{\circ}$.