A charged particle of mass m2 = 4.1 × 10-8 kg, initially moving with constant ve
ID: 1774361 • Letter: A
Question
A charged particle of mass m2 = 4.1 × 10-8 kg, initially moving with constant velocity in the y-direction, enters a region containing a constant magnetic field B2.1 T aligned with the positive z-axis as shown below. The particle enters the region at (z,y) = (d,0) and leaves the region at (z, y) = (0, d), d = 0.52 m, at a time t = 590 s after it entered the region. Figure 1 (a) With what speed v did the particle enter the region containing the magnetic field? (3pt) b) What are F and Fy, the r- and y-components of the force on the particle at a time = 1967 s after it entered the region containing the magnetic field? Be careful with radians vs degrees here. (9pt) (c) What is q, the charge of the particle? Be sure to include the correct sign. (5pt) (d) If the velocity of the incident particle were doubled, how would B have to change (keeping all other parameters constant) to keep the trajectory (meaning the value of d) of the particle the same? (3pt)Explanation / Answer
(a)
speed = distance/time
distance = length of semicircle = 2pir/4 = pi*r/2
r = radius = d = 0.52 m
v = (pi*r/2)/t
v = pi*d/(2t)
v = pi*0.52/(2*590*10^-6)
v = 1384.43 m/s
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(b)
after t1 time
distance travelled s = v*t1 = 1384.43*196.7*10^-6 = 0.272 m
distance s = d*theta
theta = s/d = 0.523 rad
angle rotated = 0.523*(360/(2pi)) = 30 degrees
Fx = -m*v^2*cos30/d = -4.1*10^-8*1384.43^2*cos30/0.52 = -0.131 N
Fy = -m*v^2*sin30/d = -4.1*10^-8*1384.43^2*sin30/0.52 = -0.076 N
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centripetal force = magnetic force
Fc = Fb
m*v^2/d = q*v*B
q = mv/(dB)
q = (4.12*10^-8*1384.43)/(0.52*2.1)
q = 5.22*10^-5 C
charge q = -5.22*10^-5 C <<<<---ANSWER
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B = mv/(qd)
B1 = mv1/(dq)
v1 = 2*v
B1 = m*2*v/(dq)
B1 = 2*B
B has to be doubled
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