An alpha particle (the nucleus of a helium atom formed by 2 protons and 2 neutro

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.

Calculate the acceleration of the alpha particle.

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

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

Explanation / Answer

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.

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