A particle with charge 5.3 Coulombs and velocity v = 63 m/s enters a region in w

Question

A particle with charge 5.3 Coulombs and velocity v = 63 m/s enters a region in which the magnetic field of magnitude 4.3 T is uniform and directed into the paper as shown.

1) What is the magnitude of the force on the charge immediately after it enters this region? If the particle exits this region through the hole at the top and then has velocity v1 which is perpendicular to the original velocity v, what is the mass of the particle? If the particle exits this region through the hole at the top and then has velocity v1 which is perpendicular to the original velocity v, what is the mass of the particle?

Explanation / Answer

1) magnetic force on the moving charged particle,

F = q*v*B*sin(90)

= 5.3*10^-6*63*4.3

= 1.44*10^-3 N <<<<<<<<<<-------------Answer

Apply, F = q*v*B

m*a = q*v*B

m*v^2/r = q*v*B

m = B*q*r/v

= 4.3*5.3*10^-6*0.7/63

= 2.53*10^-7 kg <<<<<<<<<<-------------Answer

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