In the magnetic meridian at a certain place the horizontal component of earths magnetic field is 0.32 G and the angle of dip is \( 60^\circ \). The total magnetic field of the earth at this location, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.64 G
The angle of dip \( \delta \) is the angle between the earth's total field \( B_E \) and the horizontal, in the magnetic meridian. So the horizontal component is
\( \displaystyle H = B_E\cos\delta \;\Rightarrow\; B_E = \frac{H}{\cos\delta} \)
\( H = 0.32\,\mathrm{G} \)
\( \delta = 60^\circ,\ \cos 60^\circ = 0.5 \)
\( \displaystyle B_E = \frac{0.32\,\mathrm{G}}{0.5} = 0.64\,\mathrm{G} \)
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
\( \displaystyle Z_E = B_E \sin I \qquad [5.10(a)] \)
\( \displaystyle H_E = B_E \cos I \qquad [5.10(b)] \)”