A particle with charge 4.3 Coulombs and velocity v = 53 m/s enters a region in w

Question

A particle with charge 4.3 Coulombs and velocity v = 53 m/s enters a region in which the magnetic field of magnitude 3.3 T is uniform and directed into the paper as shown.

a)What is the magnitude of the force on the charge immediately after it enters this region?

b)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?

c)If in addition to the magnetic field a constant electric field is now created within the region where the magnetic field is located so that the charge proceeds in a straight line and exits with velocity v2 = v, what must be the magnitude of this electric field?

Explanation / Answer

a)

q = charge = 4.3 C

V = speed = 53 m/s

B = magnetic field = 3.3 T

magnetic force is given as

F = q V B Sin90 = 4.3 x 53 x 3.3 = 752.1 N

b)

r = radius of the path = 0.7 m

r = mv/qB

0.7 = m (53) / (4.3 x 3.3)

m = 0.187 kg

c)

electric force balances the magnetic force

hence Fe = Fb

qE = qVB

E = VB = 53 x 3.3 = 175 N/C

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