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Question
The magnetic field due to the earth has a horizontal component of 26 μT at a place where the dip is 60°. Find the vertical component and the magnitude of the field.
Solution
Given :
Horizontal component of Earth's magnetic field, B = 26 μT
Angle of dip, δ = 60°
The horizontal component of Earth's magnetic field `(B_H)` is given by
`B_H = Bcosδ`
Here,
B = Total magnetic field of Earth
On substituting the respective values, we get
`26 xx 10^-6 = B xx 1/2`
⇒ B = `52 xx 10^-6 = 52 "uT"`
The vertical component of Earth's magnetic field `(B_y)` is given by
`B_y = Bsinδ`
⇒ `B_y = 52 xx 10^-6 xx (sqrt(3))/2`
⇒ `B_y = 44.98 "uT" = 45 "uT"`
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