Question

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 0.680 tesla at a velocity of 2.50E+6 m/s ; calculate the magnitude of the magnetic force exerted on it.

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Calculate the acceleration of the alpha particle.

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

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

Answer 1

Part -A -

With magnetic field perpendicular to the velocity, F = qvB = 2 x 1.6 x 10^-19 x 2.50 x 10^6 x 0.68

= 5.44 x 10^-13 N

Part -B -  
Mass of alpha particle, m = 4 x 1.675 x 10^-27 = 6.70 x 10^-27 kg

So, acceleration of the particle, a = F/m = (5.44 x 10^-13) / (6.70 x 10^-27) = 8.12 x 10^13 m/s^2

Part -C-

Write the expression -

mv^2/r = F

=> r = mv^2/F = [(6.70 x 10^-27) x (2.50 x 10^6)^2] / (5.44 x 10^-13) = 7.70 x 10^-2 = 0.077 m

Part -D -

In this condition, force on the alpha particle, F = 0

Because, F = qvB*sin(theta) where theta is angle between v and B. When v||B, theta = 0.