Question

At a certain place, the angle of dip is $60^\circ $ and the horizontal component of the earth's magnetic field $\left(B_{H}\right)$ 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 $
$0.8\times 10^{- 4}=B_{e}\cos 60^{\circ}$
$B_{e}=\frac{0.8 \times 10^{- 4}}{\frac{1}{2}}$
$=1.6\times 10^{- 4} \, T$