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Chapter 28: Q 8P (page 829)

Question: An electric field of1.50kV/mand a perpendicular magnetic field of 0.400Tact on a moving electron to produce no net force. What is the electron’s speed?

Short Answer

Expert verified

The electron’s speed is |v|=3.75x103m/s.

Step by step solution

01

Given

ϕ=90.0∘E=1.50kVm=1.50x103V/mB=0.400TF=0

02

Determining the concept

Find the velocity of an electron by inserting the given values in the formula for electromagnetic force.

Formulae are as follow:

F=eE+evBsinϕ

Where, F is force, v is velocity, E is electric field, B is magnetic field, e is charge of particle.

03

Determining the speed of electron

The electromagnetic force experienced by an electron is,

F=eE+evBsinϕ

Inserting given values,

0=-1.6×10-19×1.50×103+-1.6×10-19×v×0.400×sin90∘0=-1.6×10-19×1.50×103-1.6×10-19×v×0.400×sin90∘v=1.50×1030.400v=3.75×103m/s

Hence, the electron’s speed is v=3.75×103m/s.

Therefore, the velocity of an electron can be determined by inserting the given values in the formula for electromagnetic force.

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Most popular questions from this chapter

An electron follows a helical path in a uniform magnetic field given by B⇀=(20i^-50j^-30k^)mT . At time t = 0, the electron’s velocity is given by v⇀=(20i^-30j^+50k^)m/s.

(a)What is the angleϕbetweenv⇀ andB⇀ The electron’s velocity changes with time.

(b) Do its speed change with time?

(c) Do the angleϕchange with time?

(d) What is the radius of the helical path?

Question: At one instantv→=(-2.00i^+4.00j^-6.00k^)m/s, is the velocity of a proton in a uniform magnetic fieldB→=(2.00i^-4.00j^+8.00k^)mTAt that instant, what are (a) the magnetic force acting on the proton, in unit-vector notation, (b) the angle betweenv→ and F→, and (c) the angle betweenv→ and B→?

In Figure, a metal wire of mass m = 24.1 mg can slide with negligible friction on two horizontal parallel rails separated by distance d = 2.56 cm. The track lies in a vertical uniform magnetic field of magnitude 56.3 mT. At time t = 0, device Gis connected to the rails, producing a constant current i = 9.13 mA in the wire and rails (even as the wire moves). Att = 61.1 ms, (a) what is the wire’s speed? (b) What is the wire’s direction of motion (left or right)?

An electron is moving at 7.20×106m/sin a magnetic field of strength 83.0mT. What is the (a) maximum and (b) minimum magnitude of the force acting on the electron due to the field? (c) At one point the electron has an acceleration ofmagnitude role="math" localid="1662987933736" 4.90×1014m/s2.What is the angle between the electron’s velocity and the magnetic field?

(a) In Fig. 28-8, show that the ratio of the Hall electric field magnitude E to the magnitude Ecof the electric field responsible for moving charge (the current) along the length of the strip is

EEC=Bneρ

where ρ is the resistivity of the material and nis the number density of the charge carriers. (b) Compute this ratio numerically for Problem 13. (See Table 26-1.)
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