1. An alpha particle (the nucleus of a helium atom formed by 2 protons and 2 neu

Question

1. An alpha particle (the nucleus of a helium atom formed by 2 protons and 2 neutrons with charge q = +2(1.6 x 10-19 C ) is moving perpendicular to a magnetic field of 0.710 tesla at a velocity of 3.60E+6 m/s ; calculate the magnitude of the magnetic force exerted on it.

2.Calculate the acceleration of the alpha particle

3.What will be the radius of curvature of the path followed by the alpha particle in the magnetic field?

4.What would be the force on the alpha particle if it were traveling parallel to the magnetic field lines?

Explanation / Answer

1) With B perpendicular to v, F = qvB = 3.2E-19*3.6E6*0.71 = 8.1792E-13 N
2) m(alpha) = 6.6446E-27 kg
a = F/m =8.1792E-13/6.6446E-27 = 1.23E+14 m/s^2
3) mv^2/r = F ==> r = mv^2/F = 0.1053 m
4) F = 0, since F = qvB*sin(theta) where theta is angle between v and B. When v||B, theta = 0.

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