Question

At a certain place, the angle of dip is $ 60^{\circ} $ and the horizontal component of the earth’s magnetic field $ (B_{H}) $ is $ 0.8 \times 10^{-4} \, T $ . The earth’s overall magnetic field is

• A. $ 1.5 \times 10^{-4} \, T $
• B. $ 1.6 \times 10^{-3} \, T $
• C. $ 1.5 \times 10^{-3} \, T $
• D. $ 1.6 \times 10^{-4} \, T $

Answer 1

Answer: (D)

Given,
$B_{H} =0.8 \times 10^{-4} T$
$\theta =60^{\circ} $
$B_{e} =?$

We know that,
$ B_{H} =B_{e} \cos \theta $
$0.8 \times 10^{-4} =B_{e} \cos 60^{\circ} $
$ B_{e} =\frac{0.8 \times 10^{-4}}{\frac{1}{2}} $
$=1.6 \times 10^{-4} T $