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=Becos(θ)
angle of dip at equator is 0°
BH=Becos(0)=Be=2B
BH=2B
Concept: Magnetic Field on the Axis of a Circular Current Loop
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