Part A, B, and C A tiny dust particle (m = 2.50 times 10^-5 kg) has a charge of
ID: 1451319 • 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 = 2.53 T) and the particle are shown in the diagram. The field points out of your screen. Answer the three questions below, using three significant figures. Part A: Which path will the particle take while inside the field? A (straight through) B (curt to the left) C (curl to the right) D (curl out of the screen - not pictured) E (curl into the screen - not picture) Part B: What is the magnitude of the force (F) acting on particle? F = Part C: What would be the radius of curvature (r) of the particle as it travels through the magnetic field?Explanation / Answer
givendata:
m=2.50*10^-5 kg
q=2.05*10^-4C
v=7.12m/s
B=2.53T
A)
To find the direction of the force, use the right-hand rule.
Let the fingers of your right hand point in the direction of v. Orient the palm of your hand, so that as you curl your fingers, you can sweep them over to point into the direction of B. Your thumb points in the direction of the vector product v × B. If q is positive then this is the direction of F. If q is negative, your thumb points opposite to the direction of F.
answer is =curl towards the right
B) magnetic force=?
from the Lorentz force F=qvB
F=2.05*10^-4*7.12*2.53 =0.00369N
C) radius of curvature
Centripetal force = Lorentz force
mv2/r = qvB,
r = (mv)/(qB) =(2.50*10^-5*7.12)/(2.05*10^-4*2.53)
r=0.3432 m