At a place, the horizontal component of earth's magnetic field is \( 4\times10^{-4}\,\mathrm{T} \) and angle of dip is 60°. The value of horizontal component of earth's magnetic field at the equator is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 8\times10^{-4}\,\mathrm{T} \)
The angle of dip \( \delta \) is the angle that the earth's total magnetic field \( B_E \) makes with the horizontal, so the horizontal component is \( H = B_E\cos\delta \).
\( \displaystyle B_{E} = \frac{H}{\cos\delta} = \frac{4\times10^{-4}\,\mathrm{T}}{\cos 60^\circ} = \frac{4\times10^{-4}\,\mathrm{T}}{1/2} = 8\times10^{-4}\,\mathrm{T} \)
At the magnetic equator the dip is zero, so the whole field is horizontal. Taking the total field to have the same value there, as the question assumes,
\( \displaystyle H_{\mathrm{eq}} = B_{E}\cos 0^\circ = 8\times10^{-4}\,\mathrm{T} \)
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. … Representing the verticle component by ZE, we have
\( \displaystyle Z_E = B_E \sin I \qquad [5.10(a)] \)
\( \displaystyle H_E = B_E \cos I \qquad [5.10(b)] \)”