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Question

Physics Question on the earth's magnetic field

At a certain place, the angle of dip is 60∘60^{\circ} and the horizontal component of the earth?? magnetic field (BH)(B_{H}) is 0.8×10−4 T0.8 \times 10^{-4} \, T . The earth?? overall magnetic field is

A

1.5×10−4 T1.5 \times 10^{-4} \, T

B

1.6×10−3 T1.6 \times 10^{-3} \, T

C

1.5×10−3 T1.5 \times 10^{-3} \, T

D

1.6×10−4 T1.6 \times 10^{-4} \, T

Answer

1.6×10−4 T1.6 \times 10^{-4} \, T

Explanation

Solution

Given, BH=0.8×10−4TB_{H} =0.8 \times 10^{-4} T θ=60∘\theta =60^{\circ} Be=?B_{e} =? We know that, BH=Becos⁡θB_{H} =B_{e} \cos \theta 0.8×10−4=Becos⁡60∘0.8 \times 10^{-4} =B_{e} \cos 60^{\circ} Be=0.8×10−412B_{e} =\frac{0.8 \times 10^{-4}}{\frac{1}{2}} =1.6×10−4T=1.6 \times 10^{-4} T