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
$B_H = B\cos\delta \Rightarrow B = \dfrac{3 \times 10^{-5}}{\cos 60°} = 6 \times 10^{-5}$ T.
Solution by Sreeraj P, M.Sc Physics