Question

A charged particle of mass m = 7.3X10^-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 1.8T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.5 m, 0) and leaves the region at (x,y) = 0, 0.5 m a time t = 463 μs 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 F_x, the x-component of the force on the particle at a time t₁ = 154.3 μs after it entered the region containing the magnetic field.(N)

3. What is F_y, the y-component of the force on the particle at a time t₁ = 154.3 μs 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. (μC)

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?

a. Increase B by a factor of 2

b. Increase B by less than a factor of 2

c. Decrease B by less than a factor of 2

d. Decrease B by a factor of 2

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

Answer 1

radius of circular path r = 0.5 m

(1)

distance travelled in magnetic field d = (1/4)*2*pi*r = pi*r/2 = pi*0.5/2 = pi/4

time it was in magneitc field t = 463*10^-6 s

speed of particle v = d/t = (pi/4)/(463*10^-6) = 1696.3 m/s

-----------------------

(2)

at t = 154.3 *10^-6 s

distanc travelled d = v*t = 1696.3*154.3*10^-6 = 0.262 m

angle subtented at origin theta = d/r = 0.524 rad = 30 degrees

Fx = -(m*v^2/r)*cos30

Fx = -(7.3*10^-8*1696.3^2*cos30)/0.5

Fx = -0.364 N

(3)

Fy = -(m*v^2/r)*sin30

Fy = -(7.3*10^-8*1696.3^2*sin30)/0.5

Fx = -0.21 N

(4)

In magneitc field Fnet = m*ac

ac = centripetal acceleration = v^2/r

Fb = magneitc force = q*v*B*sin90

q*v*B = m*v^2/r

q = m*v/(r*B)

charge q = 7.3*10^-8*1696.3/(0.5*1.8) = 137.6*10^-6 C

by right hand rule the charge is negative

charge q = -137.6*10^-6 C = -137.6 uC <<<-----------ANSWER

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r = mv/(qB)

if velocity is doubled B will increase by a factor of 2

OPTION (a) Increase B by a factor of 2