A tiny dust particle (m 2.90 x 105 kg) has a charge of 2.05 x 04 C when it enter

Question

A tiny dust particle (m 2.90 x 105 kg) has a charge of 2.05 x 04 C when it enters an MRI machine with a speed of 7.02 m/s. The machine's strong field points out (B 2.50 and the particle are shown in the diagram. The T magnetic field of your screen. 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) O E (curl into the screen not picture) Part B: What is the magnitude of the force (F) acting on particle? Number

Explanation / Answer

This dust particle is under the force field F = q (V X B) . According to cross product direction property, the particle will take the C direction (curl to the right) (option C ) . |F| = q (V X B) = 2.05* 10-4 *(7.02 * 2.50) = 3.597 * 10-3 N (Ans)

As it will move in a circle under the influence of force so centripetal force = lorentz force => mv2/r = |F| => r = (2.9*10-5 * (7.02)2 ) / 3.6 * 10-3 = 0.397 m (Ans)

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