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(A) The horizontal component of the earth's magnetic field is given as $B_{H} = 0.35 \; G$.The angle of dip $\delta$ is the angle that the total magne

Vedclass · Accepted Answer

$0.38$

Vedclass · Answer

(A) The horizontal component of the earth's magnetic field is given as $B_{H} = 0.35 \; G$.The angle of dip $\delta$ is the angle that the total magnetic field makes with the horizontal,which is given as $\delta = 22^{\circ}$.The relationship between the total magnetic field $B$ and its horizontal component $B_{H}$ is given by $B_{H} = B \cos \delta$.Rearranging the formula to solve for $B$,we get $B = \frac{B_{H}}{\cos \delta}$.Substituting the given values: $B = \frac{0.35}{\cos 22^{\circ}}$.Using $\cos 22^{\circ} \approx 0.9272$,we get $B = \frac{0.35}{0.9272} \approx 0.377 \; G$.Rounding to two decimal places,the magnitude of the earth's magnetic field is approximately $0.38 \; G$.

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