A charged particle of mass m = 4.7X10 -8 kg, moving with constant velocity in th

Question

A charged particle of mass m = 4.7X10-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 2.3T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.79 m, 0) and leaves the region at (x,y) = 0, 0.79 m a time t = 428 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 = 142.7 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 = 142.7 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)

If the velocity of the incident charged particle were doubled, how would B have to change (keeping all other parameters constant) to keep the trajectory of the particle the same?

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

It followed a path of a circle with radius as distance d.
Centripetal Force = Magnetic Force
m*v^2/d = q*v*B
m*v/d = q*B ------1

Distance = Speed time
Speed = (*d)/2 / 428 us
V = (*0.79)/2 /( 428 * 10^-6 )
V = 2.9 * 10^3 m/s

Charge on the particle,
Substituing Values in above eq,
(4.7*10^-8 * 2.9 * 10^3)/0.79 = q * 2.3
q = 75 * 10^-6 C
Charge will be negative, as the Field is outwards and Force is to the Left.
q = - 75 uC

Distance travelled by particle in time, t = 142.7 us
S = v*t
S = (2.9*10^3) * 142.7 * 10^-6 m
S = 0.414 m

= S/r
= 0.414/0.79
= 0.524 * (180/)
= 30 o

Horizontal Component of Force, Fx = F*cos(30)
Fx = (q*v*B) * cos(30)
Fx = (75*10^-6 * 2.9*10^3 * 2.3) * cos(30)
Fx = 0.433 N

Vertical Component of Force, Fy = F*sin(30)
Fy = (75*10^-6 * 2.9*10^3 * 2.3) * sin(30)
Fy = 0.250 N

To keep the trajectory the same, d should remain same,
d = m*v/q*B
If we double the velocity, B should also be increased by a factor of 2.

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