The angle of dip at a certain place on earth | vedclass

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(C) The horizontal component of the earth's magnetic field is given by $H_{E} = B_{E} \cos \delta$,where $B_{E}$ is the total magnetic field of the ea

Vedclass · Accepted Answer

$0.52$

Vedclass · Answer

(C) The horizontal component of the earth's magnetic field is given by $H_{E} = B_{E} \cos \delta$,where $B_{E}$ is the total magnetic field of the earth and $\delta$ is the angle of dip.Given,$H_{E} = 0.26 \, G$ and $\delta = 60^{\circ}$.Substituting the values into the formula:$B_{E} = \frac{H_{E}}{\cos \delta}$$B_{E} = \frac{0.26 \, G}{\cos 60^{\circ}}$Since $\cos 60^{\circ} = 0.5$,we have:$B_{E} = \frac{0.26 \, G}{0.5} = 0.52 \, G$.Therefore,the magnetic field at the place on earth is $0.52 \, G$.

The angle of dip at a certain place on earth is $60^{\circ}$ and the magnitude of earth's horizontal component of magnetic field is $0.26 \, G$. The magnetic field at the place on earth is.....$G$

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