The horizontal component of Earth's magnetic field at a place is $3 \times 10^{-5}$ T. If the dip at the place is $45^{\circ}$, what is the resultant magnetic field at that location?

Question

$\frac{3}{2} \sqrt{3} \times 10^{-5}$ T

$3 \sqrt{2} \times 10^{-5}$ T

$3 \times 10^{-5}$ T

$\frac{3}{\sqrt{2}} \times 10^{-5}$ T

Solution

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