> Q. The horizontal component of the earth’s magnetic field is $3.6 times 10^{-5}$ tesla where the dip angle is $60^{circ}$. The magnitude of the earth’s magnetic field is – LIVE ANSWER TODAY

Q. The horizontal component of the earth’s magnetic field is $3.6 times 10^{-5}$ tesla where the dip angle is $60^{circ}$. The magnitude of the earth’s magnetic field is

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The horizontal component of the earth’s magnetic field is $3.6 times 10^{-5}$ tesla where the dip angle is $60^{circ}$. The magnitude of the earth’s magnetic field is

Solution:

Horizontal component of earth’s field,

$H = Bcos,theta,$ since, $theta = 60^{circ}$

$3.6times10^{-5} = Btimesfrac{1}{2} Rightarrow ,B = 7.2times10^{-5} ,Tesla$

Solution Image

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