A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at \( 30^\circ \) with the horizontal. The horizontal component of the earth's magnetic field at the place is B. The strength of the earth's magnetic field at the place is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{2B}{\sqrt{3}} \)
A needle free to turn in the vertical plane of the magnetic meridian sets itself along the earth's total field. The angle it makes with the horizontal is the angle of dip \( \delta \), here \( 30^\circ \).
The horizontal component is the projection of the total field \( B_E \):
\( \displaystyle B_H = B_E\cos\delta \;\Rightarrow\; B_E = \frac{B_H}{\cos\delta} \)
\( \displaystyle B_E = \frac{B}{\cos 30^\circ} = \frac{B}{\sqrt{3}/2} = \frac{2B}{\sqrt{3}} \)
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
\( \displaystyle Z_E = B_E \sin I \qquad [5.10(a)] \)
\( \displaystyle H_E = B_E \cos I \qquad [5.10(b)] \)”