Physics: Asatellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod. the antenna is oriented perpendicular to the earth s surface. at the equator the earth s magnetic field is essentially horizontal and has a value of 8.0×10−5t; ignore any changes in b with altitude. assuming the orbit is circular, determine the induced emf between the tips of the antenna.

For him being such a fat bear, it would have taken alot of workExplanation:...

Asatellite, orbiting the earth at the equator at

ebzloera

10.01.2022

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

For the orbital velocity of the satellite at an altitude, H:

$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

where

G = Gravitational constant

$m_{e} = 5.972\times 10^{24}\ kg$ = mass of earth

$R_{E} = 6371\ km$ = radius of earth

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s$

Using this value value in eqn (1):

$e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V$

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