At a place, the horizontal component of earth's magnetic field is B and angle of dip is 60°. What is the value of horizontal component of the earth's magnetic field at equator?

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#### Solution

The horizontal component of the electric field is given by

BH=Becos(θ)

where θ is the angle of dip at the given place

Be is the net magnetic field

At θ=60°

BH=Becos(60)=Be×1/2

Be=2BH=2B (given BH=B)

Now horizontal component at equator

BH=B_{e}cos(θ)

angle of dip at equator is 0°

BH=B_{e}cos(0)=Be=2B

BH=2B

Concept: Magnetic Field on the Axis of a Circular Current Loop

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