I don\'t even know how to go about starting this problem. Please help, and thank

Question

I don't even know how to go about starting this problem. Please help, and thank you in advance!

Assume the Earth’s magnetic dipole moment is aligned with the Earth’s rotational axis, and the Earth’s magnetic field is cylindrically symmetric (like an ideal bar magnetic).
What speed would a proton need to orbit in an exact circle around the Earth at a height of 1440 km, where the Earth’s magnetic field has an intensity of 4.65 × 10^-8 T? The mass of a proton is 1.67262 × 10^-27 kg and the radius of Earth is 6.37 × 10^6 m.
Answer in units of m/s.

Explanation / Answer

The centripetal force due to circular motion on proton is F = mv^2/r The magnetic force on the proton is F = Bqv From above two equations Bqv = mv^2/r The speed of the proton is v= (Bqr)/m Here Earth's magnetic field B = 4.65 × 10^-8 T mass of the proton is m = 1.67262 × 10^-27 kg charge of the proton q = 1.60 x 10^-19 C radius r =( 6.37 × 10^6 m) + (1440 x 10^3m) On substitution v = 3.47397 x10^+7 m/s

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