Q 12-05-015NEETJEE MainEasy
At a place, the horizontal component of the Earth's field is $3 \times 10^{-5}$ T and the angle of dip is $60°$. The total field there is
Answer: (B) $6 \times 10^{-5}$ T
Concept. The horizontal component is $B_H = B\cos\delta$, where $\delta$ is the angle of dip.
Step 1. Solve for $B$.
$$B = \frac{B_H}{\cos\delta} = \frac{3 \times 10^{-5}}{\cos 60°} = \frac{3 \times 10^{-5}}{0.5} = 6 \times 10^{-5}\ \text{T}$$
Answer. $6 \times 10^{-5}$ T (Option B).
Solution by Sreeraj P, M.Sc Physics