NEET-XII-Physics
36: Permanent Magnets
- #13The 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.Ans : Given:
Horizontal component of Earth's magnetic field, B = 26 μT
Angle of dip, δ = 60°
The horizontal component of Earth's magnetic field `` \left({B}_{H}\right)`` is given by
BH = Bcosδ
Here,
B = Total magnetic field of Earth
On substituting the respective values, we get
`` 26\times {10}^{-6}=B\times \frac{1}{2}``
`` \Rightarrow B=52\times {10}^{-6}=52\,\mathrm{\,\mu T\,}``
`` ``
The vertical component of Earth's magnetic field `` \left({B}_{y}\right)`` is given by
`` {B}_{y}\mathit{=}B\,\mathrm{\,sin\,}\delta ``
`` \Rightarrow {B}_{y}=52\times {10}^{-6}\times \frac{\sqrt{3}}{2}``
`` \Rightarrow {B}_{y}=44.98\,\mathrm{\,\mu T\,}=45\,\mathrm{\,\mu T\,}``
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