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(a) What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56 * 104 V/m and a magnetic field of 4.62 * 10-3 T, with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors v, E, and B. (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

Short Answer

Expert verified
  1. The speed of the beam electrons is3.88×106m/s
  2. The relative orientation of vectors is

The radius of the orbit is4.17×10-3 and period of the orbit is7.74×10-9s

Step by step solution

01

Magnetic force

Magnetic force is given by

FB=q(v×B)

02

Determine the speed of beam of the electrons

(a)

Magnetic force is given by

FB=q(v×B)

Therefore, the speed of the beam is

v=EB=1.56×104V/m4.62×103T=3.38×106m/sV=3.38×106m/s

03

Determine the relative orientation of the vectors

The sketch of the vector’s velocity electric field and magnetic field is

04

Determine the radius of the orbit and period of orbit

There is 90-degree angle between the velocity and magnetic field

The force acting on the ion is

FB=|q|vBsin(ϕ)=|q|vBsin(90)=|q|vBAndtheequationisF=|q|vB=ma|q|vB=mv2R

And the radius is

R=mv|q|B=(9.11×1031kg)(3.38×106m/s)(1.60×1010C)(4.62×103T)=4.17×103m

Therefore, the radius of the orbit is4.17×10-3m

And the period of orbit is

T=2πω=2πRV=2π(4.17×103m)(3.38×106m/s)

Therefore, the period of the orbit is7.74×10-9s

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