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 times 10-19 is moving perpendicular to a magnetic field of 0.680 tesla at a velocity of 2.90E+6 m/s calculate the magnitude of the magnetic force exerted on it. 6.31 times 10^-13 N You are correct Your receipt no. is 152-7655 2) Previous Tries Calculate the acceleration of the alpha particle 835 10^14/s^2 You can assume proton and neutron have the same mass of 1.67e-27 kg Submit Answer Incorrect. Tries 3/10 Previous Tries What will be the radius of curvature of the path followed by the alpha particle in the magnetic field? Submit Answer Tries 0/10 What would be the force on the alpha particle if it were traveling parallel to the magnetic field lines? Submit Answer Tries 0/10

Explanation / Answer

here you have calculated the force which is correct

in part b.) a = F/m = 6.31 x 10-13 / 1.67 x 10-27   = 3.778 x 10 14 m/s2

c.) F = mv2/ r

=> r = mv2/F = 1.67 x 10-27 x (2.9 x 106)2/ 6.31 x 10-13

= 0.022 m = 22 mm

d.) if the alpha particle were moving parallel to the field, force = 0

in calculating force, we take the cross product of velocty & field . two parallel vectors have 0 cross product so F=0

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