Problem 22.9 A 0.40 uC particle moves with a speed of 20 m/s through a region wh

Question

Problem 22.9 A 0.40 uC particle moves with a speed of 20 m/s through a region where the magnetic field has a strength of 0.99 T Part A At what angle to the field is the particle moving if the force exerted on it is 4.8 x 10 N? Express your answer using two significant figures. 61 Submit Answers Give U incorrect; Try Again 5 attempts remaining Part B At what angle to the field is the particle moving if the force exerted on it is 3.0 x 10 ON? Express your answer using two significant figures. Submit My Answers Give Up

Explanation / Answer

F = q(vxB)

F = qvB sin(theta)

4.8*10^-6 =0.40*20*0.99*10^-6 sin(theta)

Theta= 37.30 degree

B)

F=qvBsin(theta)

3*10^-3 = 0.4*20.99*10^-6 sin(theta)

Theta = 22.25 degree

C)

1*10^-7 = 0.4×20×.99*10^-6sin(theta)

Theta = 0.723 degree

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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