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Q. At certain place, the horizontal component of earth’s magnetic field is $3.0\, G$ and the angle dip at that place is $30^{\circ}$. The magnetic field of earth at that location

KCETKCET 2016Magnetism and Matter

Solution:

Given, $H=3.0$ gauss
Angle of dip, $ \delta=30^{\circ} $
$\frac{H}{B} =\cos\, \delta $
$H =B \,\cos \,\delta $
$3.0 =B \times \cos\, 30^{\circ} $
$B =\frac{3.0}{\cos \,30^{\circ}}=\frac{3}{\sqrt{3} / 2}=\frac{2 \times 3}{\sqrt{3}} $
$=\frac{6}{1.73}=3.5 \,G$