CBSE Class 12 · 2011 · Magnetic Properties of MatterShort Answer

A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at

6 0^{\circ}

with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be

0.4 G

. Determine the magnitude of the earth's magnetic field at the place.

Answer: $0.8 G$ .

Solution

Since the needle rotates in the magnetic meridian and dips by

6 0^{\circ}

, the angle of dip is

δ = 6 0^{\circ}

.
The horizontal component relates to the total field by

B_{H} = B_{E} cos δ .

Therefore

B_{E} = \frac{B _{H}}{cos δ} = \frac{0.4}{cos 6 0 ^{\circ}} = \frac{0.4}{1/2} = 0.8 G .

B_{E} = 0.8 G

CBSE Class 12 Board · 2011 · Subjective