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3 years ago

Who thinks there is a 9th planet (not counting Pluto) and tell me why

Physics

1 answer:

LenKa [72]3 years ago

4 0

Because there is a gravitational force affecting the orbit of the far outer planets. (Neptune, Uranus)

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Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add

jok3333 [9.3K]

Answer:

When x = 2.8 cm, $B_{x1} = 0.0265 T$

When x = 5.5 cm, $B_{x2} = 0.0209 T$

when x = 7.3 cm, $B_{x3} = 0.0169 T$

When x = 11.0 cm, $B_{x4} = 0.0103 T$

Explanation:

According to Biot-Savart law,

$B_{x} = \frac{N \mu_{o}IR^{2} }{2(x^{2} +R^{2} )^{3/2} }\\$ .......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

$\mu_{o} = 4\pi * 10^{-7} N/A^{2}$

When x₁ = 2.8 cm = 0.028 m

$B_{x1} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.028^{2} +0.11^{2} )^{3/2} }\\B_{x1} = 0.0265 T$

When x₂ = 5.5cm = 0.055 m

$B_{x2} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.055^{2} +0.11^{2} )^{3/2} }\\B_{x2} = 0.0209 T$

When x₃ = 7.3 cm = 0.073 m

$B_{x3} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.073^{2} +0.11^{2} )^{3/2} }\\B_{x3} = 0.0169 T$

When X₄ = 11.0 cm = 0.11 m

$B_{x4} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.11^{2} +0.11^{2} )^{3/2} }\\B_{x4} = 0.0103 T$

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3 years ago

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bixtya [17]

Answer:

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