A negative charge of q = -2.1 times 10^-17 C and m = 8.1 times 10^-26 kg enters

Question

A negative charge of q = -2.1 times 10^-17 C and m = 8.1 times 10^-26 kg enters a magnetic field B = 0.25 T with initial velocity v = 25 m/s, as shown in the figure. The magnetic field points into the screen. q = -2.1 times 10^-17 C m = 8.1 times 10^-26 kg B= 0.25 T v = 25 m/s Part (a) Which direction will the magnetic force be on the charge? Part (b) Express the magnitude of the magnetic force, F. in terms of q, v, and B. Part (c) Calculate the magnitude of the force F, in newtons. Part (d) Under such a magnetic force, which kind of motion will the charge undergo? Part (e) Express the centripetal acceleration of the particle in terms of the force F and the mass m. Part (f) Calculate the magnitude of a. in meters per square second. Part (g) Express the radius, R, of the circular motion in terms of the centripetal acceleration a and the speed v. Part (h) Calculate the numerical value of the radius R, in meters.

Explanation / Answer

a)
Using Lorentz left hand rule, direction of force is downward.
Note that here the direction of current is to the left since conventionally, the positive direction of current is taken as the opposite direction of the movement of negative particles.

b)
F = qvB

c)
F = (2.1 x 10-17) x 25 x 0.25
= 13.125 x 10-17 N

d)
The charge will undergo a circular motion. There will be a force directed perpendicular to the motion of the particle as long as it moves.

e)
F = ma
a = F/m

f)
a = F/m
= (13.125 x 10-17) / (8.1 x 10-26)
= 1.62 x 109 m/s2.

g)
a = v2/r
r = v2/a

h)
r = v2/a
= (25)2 / (1.62 x 109)
= 3.86 x 10-7 m

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