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In the magnetic meridian of a certain place the horizontal component of earth’s magnetic field is 0.25G and the dip angle is $60^\circ $. The magnetic field of the earth at this location is:
A) 0.50G.
B) 0.52G.
C) 0.54G.
D) 0.56G.

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint:The magnetic field is defined as the vector field influence on the moving charge particle or electric currents. The force exerted by the magnetic field will be always perpendicular to the moving charge particle.

Complete step by step solution:
It is given the problem that in the magnetic meridian of a certain place the horizontal component of earth’s magnetic field is 0.25G and dip angle is $60^\circ $ we need to find the magnetic field of the earth at this location.
Let the angle between the horizontal component of the magnetic field and the earth’s magnetic field be $\theta $.
Then the horizontal component of earth’s component will equal to,

$ \Rightarrow {H_x} = H\cos \theta $

Where the horizontal component of the magnetic field is ${H_x}$, the magnetic field is H and the angle between them is $\theta $.
It is given that the earth’s magnetic field is 0.25G and dip angle is $60^\circ $ therefore we get,
$ \Rightarrow {H_x} = H\cos \theta $.
$ \Rightarrow H = {H_x} \times \sec \theta $.
$ \Rightarrow H = 0 \cdot 25G \times \sec 60^\circ $.
$ \Rightarrow H = 0 \cdot 25G \times 2$.
$ \Rightarrow H = 0 \cdot 5G$.

The earth’s magnetic field is equal to $H = 0 \cdot 5G$.

The correct answer for this problem is option A.

Additional information:The earth’s magnetic field arises due to the molten metal which is spinning in the earth’s core the North and the South Pole is present vertically opposite to each other and the angle of tilt of the axis from the vertical axis is $23 \cdot 5^\circ $.

Note:The earth’s magnetic field is at some angle to the earth’s surface and the trigonometry can be used to calculate the magnetic field on the earth surface. The magnetic field of the earth is also responsible for holding the atmosphere.