At a given instant, a particle with a mass of 4.20×10 3 kg and a charge of 3.80×

Question

At a given instant, a particle with a mass of 4.20×103 kg and a charge of 3.80×108 C has a velocity with a magnitude of 2.40×105 m/s in the +y direction. It is moving in a uniform magnetic field that has magnitude 0.800 T and is in the -xdirection.

A) What is the direction of the magnetic force on the particle ?

B) What is the magnitude of the magnetic force on the particle ?

C) What is the direction of the resulting acceleration of the particle?

D) What is the magnitude of the resulting acceleration of the particle?

Explanation / Answer

m = 4.2*10-3 kg, q = 3.8*10^-8 C

v =2.4*10^5 m/s, B = 0.8 T

(a) F = q(vXB)

F = +(+y X -x)

F = +z direction

(b) F = qvB

F = 3.8*10^-8*2.4*10^5*0.8

F = 0.0073 N

(c) a is along F direction

i.e a is along z direction

(d) F = ma

0.0073 = 4.2*10^-3*a

a = 1.74 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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