Question

The angle of dip at a certain place on earth is 60 and the magnitude of earths horizontal component of magnetic field is 0.26 g. The net magnetic field at the place on earth is:

Answer 1

Answer:

The net magnetic field at the place on earth is 5.2×10⁻⁵T.

Explanation:

The relation between horizontal and net magnetic field is given as,

$B_H=Bcos\delta$     (1)

Where,

$B_H$=magnetic field in the horizontal direction

B=net magnetic field

δ=angle of dip

From the question we have,

The angle of dip(δ)=60°

The horizontal component of magnetic field($B_H$)=0.26G=0.26×10⁻⁴T

(1 Gauss(G)=10⁻⁴(T))

By inserting the entities in equation (1) we get;

$0.26\times 10^-^4=B\times cos60\textdegree$

$0.26\times 10^-^4=B\times \frac{1}{2}$

$B=0.26\times 10^-^4\times 2$

$B=0.52\times 10^-^4$

$B=5.2\times 10^-^5\ T$

Hence, the net magnetic field at the place on earth is 5.2×10⁻⁵T.

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