6. An alpha particle (mass malpha = 6.64 x 10^27 kg, charge = 2e) traveling hori

Question

6. An alpha particle (mass malpha = 6.64 x 10^27 kg, charge = 2e) traveling horizontally with the speed of 28.0 km/s enters an area with the uniform magnetic field B=0.650 T directed into the plane of the drawing (see diagram below). (a) Calculate the force on the particle in the magnetic field and indicate the direction of the initial trajectory deflection as a result of the magnetic force. (b) What is the acceleration of the particle? (c) What is the radius of curvature of the particle?s trajectory in the field?

Explanation / Answer

a) F = q v B = 2*1.6E-19*28.0E3*0.65= 5.82E-15 N

v into B is up so force is up

b) F = m a

a = F/m = 5.82E-15/6.64E-27= 8.77E11 m/s^2

c)

a = v^2/r

r = v^2/a = 28.0E3^2/8.77E11= 8.94E-4 m

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