tiny dust particle (m = 2.50 10% kg) has a charge of 2.25 10-4 C when it enters

Question

tiny dust particle (m = 2.50 10% kg) has a charge of 2.25 10-4 C when it enters an MRI machine with a speed of 7.02 m/s. The machine's strong magnetic field (B2.50 T) and the particle are shown in the diagram. The field points out of your screen Map Answer the three questions below, using three significant figures Part A: Which path will the particle take while inside the field? O A (straight through) O B (curl to the left) O C (curl to the right) O 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? Number Part C: What would be the radius of curvature (r) of the particle as it travels through the magnetic field? Number Previous Next Save And Exit

Explanation / Answer

i = right

j = up

k = out of page

F = q *( v x B )

q = charge of the particle

v = speed of the particle = 7.02 j m/s

B = magnetic field = 2.5 k

F = (j x k) = i

C (curl to right)

++++++++++

F = 2.25*10^-4*( 7.02 j x 2.5 k)

F = 2.25*10^-4*7.02*2.5

F = 0.003945 N i

magnitude = 0.003945 N

+++++++++++++

C)

Fb = Fc

Fc = centrtipetal force = mv^2/r

r = mv^2/Fb

r = (2.5*10^-5*7.02^2)/0.003945

r = 0.3123 m <<<<<------answer

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