I am confused on this question. I am plagued by this. Please help. I need it. t)

Question

I am confused on this question. I am plagued by this. Please help. I need it.

t) The earth's radius is typically given us 638 x 10,n. The field strength of the earth is 5×10-ST at de surface. The spice shuttles used to have a height of 14.12m. They orbited the earth at a speed of 7860 /s. They typically orbited at an altitude cf 322 km. If we assarme that a single current loop at half the earth's radius is generating the magnctic field. then what is that current? Using that current determine the niagnetic field at the shuitles arbital altitude. What is the electric potential from the top to bottom of lhe Khuttle"

Explanation / Answer

3) radius of earthR = 6.38*10^6 m

B = 5*10^-5 T

h = 14.12 m

v = 7860 m/s

H = 322,000 m

assuming current I ar radius R/2

then

B = k*2*pi*(R/2)^2*I/(R^2 + (R/2)^2)^3/2

B = k*2*pi*R^2*I/4(5R^2/4)^3/2

5*10^-5 = 4pi*10^-7*I/6.38*10^6*(5)^3/2

I = 2838153141.87599 A

magnetic field at shuttle's orbit = B'

B' = k*2*pi*(R/2)^2*I/((R + H)^2 + (R/2)^2)^3/2

B' = 10^-7*pi*(6.38*10^6)^2*2838153141.8759/2((6.38*10^6 + 322,000)^2 + (6.38/2 *10^6)^2)^3/2

B' = 4.4376*10^-5 T

EMF induced = B'*h*v = 4.925 V

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