In This post you will get answer of 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
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:
Horizontal component of earth’s field,
$H = Bcos,theta,$ since, $theta = 60^{circ}$
$3.6times10^{-5} = Btimesfrac{1}{2} Rightarrow ,B = 7.2times10^{-5} ,Tesla$