number SEVEN please! I\'m trying to understand the steps. thanks! What velocity

Question

number SEVEN please!

I'm trying to understand the steps.

thanks!

What velocity would a proton need to circle Earth 1 000 km above the magnetic equator, where Earth's magnetic field is directed horizontally north and has a magnitude of 4.00 times 10^-8 T? An electron is accelerated through 2 400 V from rest and then enters a region where there is a uniform 1.70-T magnetic field. What are (a) the maximum and (b) the minimum magnitudes of the magnetic force acting on this electron? A proton moving at 4.00 times 10^6 m/s through a magnetic field of magnitude 1.70 T experiences a magnetic

Explanation / Answer

7.

Magnetic equator is an img circle parallel to geograph equator.

so the radius of orbit for proton (under B)
R = 6400 km (earth radius+ 1000 km =7400x10^3 m
for describing circular path under cross mag field (B in z diection and v, F and x-y plane parallel to equator
m v^2 / R = q v B
v = q R B /m
or, v = 1.6*x0^-19 x7400*10^3 x4.00 10^-8 / (1838x 9.1x10^-31 )

or, v = 2.831 x10^7 m/s.................................................................ans

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