A particle with charge 94.5 nC is moving in the region where there is a uniform

Question

A particle with charge 94.5 nC is moving in the region where there is a uniform magnetic field of magnitude 0.45 T pointing in the negative x-direction. At a particular instant of time the velocity of the particle has components: Vx= -1.68 X 10^4 m/s; Vy= -3.11 X10^4 m/s; Vz= 5.85 X 10^4 m/s.

a.) What is the magnetic field vector in vector notation

b.) what is the velocity vector of the particle in vector notation

c.) what is the force on the particle at this time in vector form

d.) what are the components of the force on the particle at this time

e.) at this instant, what is the angle between the velocity of the particle and the magnetic field

f.) at this instant, what is the direction of force on the particle with respect to its direction of motion

Explanation / Answer

a) B = -0.45 i + 0 j + o k

b) v = -1.68X E4 i - 3.11 X E4 j + 5.85 X E4k

c) F = q v X B

= 0 i + 2.48 E-3 j + 1.32 E-3 k

d) Fx = 0

Fy = 2.48 E-3 N

Fz = 1.32 E-3 N

e)|v| = 6.83 E4 m/s

angle with x-y plane is the angle between velocity and magnetic field

theta = 90-arccos (5.85/6.83) = 58.93 degrees

f) |F| = 2.81 E-3 N

F.v = 1.0 E-3 Nm/s

cos(alpha) = F.v/(|F||v|) = 0.52 -5

alpha = 89.99 degrees

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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