A magnetic needle free to rotate …
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A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. The horizontal components of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place.
Posted by Rahul Singh Bhandari 6 years, 2 months ago
- 1 answers
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Sia ? 6 years, 2 months ago
Given that,
Angle of dip, {tex}\delta = 60^\circ = \frac{\pi }{3}{/tex}
and horizontal component of the earth's magnetic field is,
BH = 0.4 G (G=10-4 T)
BH = 0.4 {tex}\times{/tex} 10-4 T
Magnetic field of earth (B) =?
{tex}\because {/tex} Horizontal component of the earth's magnetic field,
{tex}B_H = B cos \delta{/tex}
{tex}B = \frac { B_H } { \cos \delta } \\ B = \frac { 0.4 \ \times\ 10^{-4} \ T } { \cos 60 ^ { \circ } } \\ B = \frac { 0.4 \ \times\ 10^{-4} \ T } { 1 / 2 } \\ \therefore B = 0.8 \ \times\ 10^{-4} \ T \\ \therefore B = 8 \ \times\ 10^{-5} \ T{/tex}
0Thank You