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Question

(D) The horizontal component of the earth's magnetic field is given by $B_H = B \cos \phi$,and the vertical component is given by $B_V = B \sin \phi$,

Vedclass · Accepted Answer

$0.62 	imes 10^{-4}$

Vedclass · Answer

(D) The horizontal component of the earth's magnetic field is given by $B_H = B \cos \phi$,and the vertical component is given by $B_V = B \sin \phi$,where $\phi$ is the angle of dip.Dividing the two equations,we get $\frac{B_V}{B_H} = 	an \phi$.Therefore,the vertical component is $B_V = B_H 	an \phi$.Given $B_H = 0.36 	imes 10^{-4} \ Wb/m^2$ and $\phi = 60^o$.Substituting the values: $B_V = (0.36 	imes 10^{-4}) 	imes 	an(60^o)$.Since $	an(60^o) = \sqrt{3} \approx 1.732$,$B_V = 0.36 	imes 1.732 	imes 10^{-4} \approx 0.623 	imes 10^{-4} \ Wb/m^2$.Thus,the correct option is $D$.

At a place,the earth's horizontal component of magnetic field is $0.36 \times 10^{-4} \ Wb/m^2$. If the angle of dip at that place is $60^o$,then the vertical component of the earth's magnetic field at that place in $Wb/m^2$ will be approximately:

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