1) With what speed v did the particle enter the region containing the magnetic f
ID: 2144795 • Letter: 1
Question
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 = 177.7 (u)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 = 177.7 (u)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. ___________uC
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 charged particle of mass m = 7.8X10^ -8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 2T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.51 m, 0) and leaves the region at (x,y) = 0, 0.51 m a time t = 533 (u)s after it entered the region. With what speed v did the particle enter the region containing the magnetic field? m/s What is Fx, the x-component of the force on the particle at a time t1 = 177.7 (u)s after it entered the region containing the magnetic field. N What is Fy, the y-component of the force on the particle at a time t1 = 177.7 (u)s after it entered the region containing the magnetic field. N What is q, the charge of the particle? Be sure to include the correct sign. uC 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? ncrease 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. Velocity = {pie d/2}/ 2t = 3.14*(0.51/2)/533*10^-6 = 1502.25 m/sec
b. Angle = vt'/d =1502.25*177.7*10^-6/0.51 = pie/6 = 30 degree
hence X component = mv^2/r * cos30
= {4.8 *10^-8 *(2551.94)^2 cos 30}/0.46
= 0.5885 N
c.
hence X component = mv^2/r * sin30
= {7.8*10^-8 *(1502.25)^2 cos 30}/0.255
= 0.59781867729 N
d. q = mv/Bd
= 7.8*10^-8*1502.25/2*0.51
= 0.00011487794 C
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