At a place, vertical component of earth's magnetic field is \( \sqrt{3} \) times its horizontal component. If total intensity of earth's magnetic field at the place is 0.4 G, then the horizontal component of earth's magnetic field?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.2 G
The horizontal and vertical components of the earth’s field are perpendicular, so the total field is their resultant: \( B_E = \sqrt{B_H^{2} + B_V^{2}} \).
\( B_V = \sqrt{3}\,B_H \)
\( B_E = \sqrt{B_H^{2} + 3B_H^{2}} = 2B_H \)
\( \displaystyle B_H = \frac{B_E}{2} = \frac{0.4\,\mathrm{G}}{2} = 0.2\,\mathrm{G} \)
Equivalently, \( \tan I = B_V/B_H = \sqrt{3} \) gives a dip of \( 60^\circ \), and \( B_H = B_E\cos60^\circ = 0.2\,\mathrm{G} \).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Magnetic Declination and Dip – “… Thus, dip is the angle that the total magnetic field BE of the earth makes with the surface of the earth. … The total magnetic field at P can be resolved into a horizontal component HE and a vertical component ZE. … Representing the verticle component by ZE, we have
\( \displaystyle Z_E = B_E \sin I \qquad [5.10(a)] \)
\( \displaystyle H_E = B_E \cos I \qquad [5.10(b)] \)”