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(D) The horizontal component of the earth's magnetic field is given by $B_H = 0.35 	imes 10^{-4} \,T$.The angle of dip is $\delta = 60^{\circ}$.The r

Vedclass · Accepted Answer

$0.61$

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(D) The horizontal component of the earth's magnetic field is given by $B_H = 0.35 	imes 10^{-4} \,T$.The angle of dip is $\delta = 60^{\circ}$.The relationship between the vertical component $(B_V)$,horizontal component $(B_H)$,and the angle of dip $(\delta)$ is given by the formula: $B_V = B_H 	an \delta$.Substituting the given values into the formula:$B_V = (0.35 	imes 10^{-4}) 	imes 	an(60^{\circ})$.Since $	an(60^{\circ}) = \sqrt{3} \approx 1.732$,we have:$B_V = 0.35 	imes 1.732 	imes 10^{-4} \,T$.$B_V \approx 0.6062 	imes 10^{-4} \,T$.Rounding to two decimal places,we get $B_V \approx 0.61 	imes 10^{-4} \,T$.Therefore,the correct option is $D$.

The value of horizontal component of earth's magnetic field at a place is $0.35 \times 10^{-4} \,T$. If the angle of dip is $60^{\circ}$,the value of vertical component of earth's magnetic field is nearly ............. $\times 10^{-4} \,T$.

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