A charged particle of mass m = 6.9X10 -8 kg, moving with constant velocity in th

Question

A charged particle of mass m = 6.9X10-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 1.6T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.73 m, 0) and leaves the region at (x,y) = 0, 0.73 m a time t = 639 ?s after it entered the region.

1) With what speed v did the particle enter the region containing the magnetic field? m/s

2) What is Fx, the x-component of the force on the particle at a time t1 = 213 ?s after it entered the region containing the magnetic field. N

3) What is Fy, the y-component of the force on the particle at a time t1 = 213 ?s after it entered the region containing the magnetic field.    N    

4) What is q, the charge of the particle? Be sure to include the correct sign.         ?C

5)

Increase B by a factor of 2

Increase B by less than a factor of 2

Decrease B by less than a factor of 2

Decrease B by a factor of 2

There is no change that can be made to B to keep the trajectory the same

A charged particle of mass m = 6.9X10-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 1.6T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.73 m, 0) and leaves the region at (x,y) = 0, 0.73 m a time t = 639 ?s after it entered the region. 1) With what speed v did the particle enter the region containing the magnetic field? m/s 2) What is Fx, the x-component of the force on the particle at a time t1 = 213 ?s after it entered the region containing the magnetic field. N 3) What is Fy, the y-component of the force on the particle at a time t1 = 213 ?s after it entered the region containing the magnetic field. N 4) What is q, the charge of the particle? Be sure to include the correct sign. ?C 5) Increase B by a factor of 2 Increase B by less than a factor of 2 Decrease B by less than a factor of 2 Decrease B by a factor of 2 There is no change that can be made to B to keep the trajectory the same

Explanation / Answer

a)

V=s/t =0.73*(pi/2)/(639*10-6)

V=1794.5 m/s

b)

angle

o=Wt =(V/d)t =(1794.5/0.73)*(213*10-6)

o=0.5235 rad

Fx=-(mV2/d)Cos(o)=-[(6.9*10-8)*(1794.5)2/0.73]*Cos(0.5236)

Fx=-0.2636 N

c)

Fy=-(mV2/d)Sin(o)==-[(6.9*10-8)*(1794.5)2/0.73]*sin(0.5236)

Fy=-0.1522 N

d)

Centripeal force =magnetic force

mV2/d =qVB

=>q=mV/dB =(6.9*10-8)*1794.5/0.73*1.6

q=1.06*10-4 C =106 uC

e)

Increase B by a factor of 2

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