Part A, B, and C: A tiny dust particle (m = 2.50 Times 10^5 kg) has a charge of
ID: 1450133 • Letter: P
Question
Part A, B, and C:
A tiny dust particle (m = 2.50 Times 10^5 kg) has a charge of 2.05 Times 10^-4 C when it enters an MRI machine with a speed of 7.12 m/s. The machine's strong magnetic field (B = 2.50 T) and the particle are shown in the diagram. The field points out of your screen. Which path will the particle take while inside the field? (straight through) (curl to the left) (curl to the right) (curl out of the screen - not pictured) (curl into the screen - not picture) What is the magnitude of the force (F) acting on particle? What would be the radius of curvature (r) of the particle as it travels through the magnetic field?Explanation / Answer
A) magnetic force ; F = q ( v X B)
q = +ve
v = along j
B = k
F = + ( j x k) = +i
path -> curl to the left = C
B) Fm = 2.05 x 10^-4 x (7.12 x 2.50) = 3.65 x 10^-3 N
c) charge will move in circular path.
Fm = m v^2 / r (for circular motion)
3.65 x 10^-3 = 2.50 x 10^-5 x 7.12^2 / r
r = 0.347 m