A tiny dust particle (m = 2.90 x 10^-5)....... A tiny dust particle (m = 2.90 ti

Question

A tiny dust particle (m = 2.90 x 10^-5).......

A tiny dust particle (m = 2.90 times 10^-5 kg) has a charge of 2.25 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 Which path will the particle take while inside the field? What is the magnitude of the force (F) acting on particle? F = What would be the radius of curvature (r) of the particle as it travels through the magnetic field? r =

Explanation / Answer

A) C) curl to the right (according to right hand rule)

B) F = q*v*B*sin(90)

= 2.25*10^-4*7.12*2.53*1

= 4.05*10^-3 N

C) Apply, F = q*v*B*sin(90)

m*a_rad = q*v*B

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

r = m*v/(B*q)

= 2.9*10^-5*7.12/(2.53*2.25*10^-4)

= 0.363 m

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