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 0.600 *tesla* at a
velocity of 3.20E+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

Given Charge Q = + 2 * 1.6 * 10^-19 C

mass of alpha particle is m = 6.64424 * 10^-27 kg

magnetic field B = 0.6 T

velocity v = 3.2 * 10^6 m/s

a)

magnetic force F = q * v * B * sin(theta)

here Theta = 90

F = (+ 2 * 1.6 * 10^-19) * (3.2 * 10^6) * 0.6 * sin(90)

**F = 6.144 * 10^-13 N**

b)

acceleration a = F / m

a = (6.144 * 10^-13) / (6.64424 * 10^-27)

**a = 9.247 * 10^13 m/s^2**

c)

This acceleration is the centripetal acceleration so

a = v^2 / r

9.247 * 10^13 = (3.2 * 10^6)^2 / r

radius r = 0.1107 m

**r = 11.07 cm**

d)

If 'v' and 'B' are parallel then

sin(0) = 0 therefore **F = 0**

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