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Sia ? 6 years, 2 months ago
At magnetic equator, vertical component of earth's magnetic field will be zero. As vertical component of field is Bv = Bsinδ and at equator δ = 0o
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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}
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Sia ? 6 years, 2 months ago
Here, l= 80 cm = 0.8 m, N = 5 {tex}\times{/tex} 400 = 2000
I = 8.0 A, D = 1.8 cm, n = no. of turns per unit length
Magnitude of magnetic field inside a solenoid near its center
{tex}n = \frac{{Total\;turn}}{{length}}{/tex}
{tex}n = \frac{{2000}}{{0.80}}{/tex}
{tex}B = {\mu _0}nI = \frac{{4\pi \times {{10}^{ - 7}} \times 2000 \times 8.0}}{{0.80}}{/tex}
{tex} = 8\pi \times {10^{ - 3}}T = 2.5 \times {10^{ - 2}}T{/tex}
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Řøhăň Řąjpůť ✌️✊ 6 years, 2 months ago
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