A tiny dust particle (m = 2.50 times 10^-5 kg) has a charge of 2.05 times 10^-4

Question

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 7.22 m/s. The machine's strong magnetic field (B = 2.56 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. Which path will the particle take while inside the field? A (straight through) B (curl to the left) C (curl to the right) D (curl out of the screen - not pictured) E (curl into the screen - not picture) What is the magnitude of the force (F) acting on particle? F = N What would be the radius of curvature (r) of the particle as it travels through the magnetic field?

Explanation / Answer

mass of charged particle(m)=2.5x10^-5kg

charge(q)=2.05x10^-4c

magnetic field(B)=2.56T

velocity 0f charged particle(v)=7.22m/s

when charged particle enters the magnetic field it experience magnetic force and that magnetic force acts towards right

part a

option c is correct

part b

magnetic force(F)=qvB

=(2.05X10^-4X7.22X2.56)N

=3.789x10^-3N

part c

the charged particle moves in a circular path

r=mv/qB=(2.50x10^-5x7.22)/(2.05x10^-4x2.56)

r =0.343m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.