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(B) The angle of dip (or inclination) $	heta$ is given by the relation $	an 	heta = \frac{B_V}{B_H}$,where $B_V$ is the vertical component and $B_H

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$60^{\circ}$ below horizontal

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(B) The angle of dip (or inclination) $	heta$ is given by the relation $	an 	heta = \frac{B_V}{B_H}$,where $B_V$ is the vertical component and $B_H$ is the horizontal component of the Earth's magnetic field.Given that $B_V = \sqrt{3} B_H$.Substituting this into the formula,we get $	an 	heta = \frac{\sqrt{3} B_H}{B_H} = \sqrt{3}$.Since $	an 60^{\circ} = \sqrt{3}$,the angle of dip is $	heta = 60^{\circ}$.Therefore,the magnetic needle will incline $60^{\circ}$ below the horizontal.

At a certain place,the vertical component of the Earth's magnetic field is $\sqrt{3}$ times the horizontal component of the Earth's magnetic field. If a magnetic needle is suspended freely in the air,then it will incline:

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