3 From a point on the equator where the earth magnetic field is 50 microT , poin

Question

3 From a point on the equator where the earth magnetic field is 50 microT , pointing due north parallel to the surface of the earth, a proton is launched straight up, with a speed of 3 x10^6 / m/ s

a. Calculate the magnitude of the magnetic force on the proton.

b. What is the magnitude of the acceleration of the proton?

c. The proton will follow a curved path until it will hit the surface of the earth again. What is the maximum height that the proton will reach? What is its speed at that point?

d. Relative to the launch point, where will the proton hit the surface, and what is its velocity (speed and direction) at that point?

Explanation / Answer

  In this situation, F = q*v*B*sin(theta), where is q is the charge of a proton (1.60 x 10^-19 C), v is the velocity, B is the magnetic field, and theta is the angle.

(a) F = (1.60 x 10^-19 C) x (3.00 x 10^6 m/s) x (50 x 10^-3 T) x sin(90.0°)

= 2.4 x 10^-14 N

b)

The acceleration is given by F / m, where F is the magnetic field (already calculated) and m is the mass of a proton (1.67 x 10^-27 kg).

a = (2.4 x 10^-14 N) / (1.67 x 10^-27 kg) = 1.5x10^13 m/s2
C)

P.E = K.E

mgh = 1/2 mv2

h = 0.5mv2/mg

h = 0.5(3x10^6 m/s)2 / 9.8 m/s

h = 4.5x1011m.

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