Answer:
The correct magnitude of the coil's magnetic field= =50μT
Explanation :
The magnetic field takes place as a result of movement of charge I e current, can also occurs from magnetised material, magnetic field as a result of charge movement can be deducted using right hand grip rule.
At the equator magnetic field lines are parallel towards the earth's surface and the angle of inclination of the magnetic lines of force at the horizontal position is referred to as the angle of dip at the point.
As the current is produced then the varying magnetic field is opposed ,then there is induced current when the coil is positioned at varying magnetic field.
Given from the question,
angle below horizontal θ=60-degree
The Earth's magnetic field B=50μT
The horizontal magnetic field can be expressed in terms of the formula below;
BH=Bcosθ
B(H) = earth's horizontal component of magnetic field
Ø is the angle between coil's field
B=the magnetic field in Tesla
Then,
BH=50μT×cos60∘
=50μT× 0.5
As the current is passed through the coil to produce a field , when combined with the earth's field, which creates a net field with the same strength and dip angle (60 degrees below horizontal) as the earth's field.
It can be deducted that B has the same magnitude and angle which makes the those vertical component to cancel each other since they are the same.
For the magnetic field to be pointed out at North direction, we can calculate the corrected magnetic field using the formula below
Bc=2BH
Bc=2×50μT× 0.5=50μT
The correct magnitude of the coil's magnetic field= =50μT
