A charged particle of mass m = 5.7X10 -8 kg, moving with constant velocity in th

Question

A charged particle of mass m = 5.7X10-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 1.2T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.95 m, 0) and leaves the region at (x,y) = 0, 0.95 m a time t = 710 s after it entered the region.

a.) With what speed v did the particle enter the region containing the magnetic field?
b.) What is Fx, the x-component of the force on the particle at a time t1 = 236.7 s after it entered the region containing the magnetic field.
c.) What is Fy, the y-component of the force on the particle at a time t1 = 236.7 s after it entered the region containing the magnetic field.
d.) What is q, the charge of the particle? Be sure to include the correct sign.

Explanation / Answer

a)

radius of circular path r = 0.95 m

distance travelled = length of of arc

s = r*theta = 0.95*pi/2

speed of particle v = s/t = 0.95*pi/(2*710*10^-6)

speed of particle v = 2.1*10^3 m/s

=====================

b)

at time t1 = distance travelled s = v*t1 = 2.1*10^3*236.7*10^-6 = 0.497 m

angle subtanted theta = s/r = 0.523 radians = 0.523*360/(2pi) = 30 degrees

the -component of the force on the particle at a time t1

Fx = -m*v^2*cos30/r = -5.7*10^-8*(2.1*10^3)^2*cos30/0.95 = -0.23 N

(c)

the y-component of the force on the particle at a time t1

Fy = -m*v^2*sin30/r = -5.7*10^-8*(2.1*10^3)^2*sin30/0.95 = -0.13 N

(d)

In magnetic field radius r = m*v/(q*B)

charge q = m*v/(r*B)

q = -5.7*10^-8*2.1*10^3/(0.95*1.2)

charge q = -105*10^-6 C = -105 uC

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