The horizontal component of the earth's magne | vedclass

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Question

(D) The horizontal component of the earth's magnetic field is given by $B_H = 0.36 	imes 10^{-4} \; Wb/m^2$.The angle of dip is $\delta = 60^{\circ}$

Vedclass · Accepted Answer

$0.622$

Vedclass · Answer

(D) The horizontal component of the earth's magnetic field is given by $B_H = 0.36 	imes 10^{-4} \; Wb/m^2$.The angle of dip is $\delta = 60^{\circ}$.The vertical component $B_V$ is related to the horizontal component $B_H$ by the formula:$B_V = B_H 	an \delta$Substituting the given values:$B_V = (0.36 	imes 10^{-4}) 	imes 	an 60^{\circ}$Since $	an 60^{\circ} = \sqrt{3} \approx 1.732$:$B_V = 0.36 	imes 10^{-4} 	imes 1.732$$B_V \approx 0.6235 	imes 10^{-4} \; Wb/m^2$.Rounding to the provided option,the value is $0.622 	imes 10^{-4} \; Wb/m^2$.

The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} \; Wb/m^2$. If the angle of dip at that place is $60^{\circ}$,then the value of the vertical component of the earth's magnetic field will be ........ $\times 10^{-4} \; Wb/m^2$.

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