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An electron moves at 0.100 cas shown in Fig. Find the magnitude and direction of the magnetic field this electron produces at the following points, each 2.00 mm from the electron:

(a) points Aand B;

(b) point C;

(c) point D.

Short Answer

Expert verified
  1. BA=6.00×10-8TandBB=6.00×10-8T
  2. BC=1.20×10-7T
  3. role="math" BD=0

Step by step solution

01

Solving part (a) or the problem.

Consider an electron that moves with a speed of v = 0.100 c as shown in the figure.

First, we need to find the magnitude and direction of the magnetic field that this electron produces at A where the distance from this point to the electron is r= 2.00 um. The magnetic field due to this moving electron at the nucleus is given by,

B=μoqv×r^4πr2 (1)

where v is the direction of the position vector (from the charge to the point that we want to find the field at it) we can see that the angle between the velocity and the position vector is, so we can write,

B=μo4πqvsin(ϕ)r2

substitute with the givens we get,

BA=μ04πqvsinϕr2

=4π×10-7T.m/A1.60×10-19C3.00×107m/ssin30°2.00×10-6m2=6.00×10-8TBA=6.00×10-8T

according to the right-hand rule, the direction of the magnetic fieldBqv×r^ is out of the page. Now we need to find the magnitude and direction of the magnetic field that this

electron produces at B where the distance from this point to the electron is r= 2.00 μm. We get,

BB=μ04πqvsinϕr2=4π×10-7T.m/A4π1.60×10-19C3.00×107m/ssin30°2.00×10-6m2=6.00×10-8TBA=6.00×10-8T

according to the right-hand rule, the direction of the magnetic fieldBqv×r^ is out of the page.

02

Solving part (b) or the problem.

Now we need to find the magnitude and direction of the magnetic field that this electron produces at C where the distance from this point to the electron is r= 2.00, we can see that the angle between the velocity and the position vector is . We get,

BC=μ04πqvsinϕr2=4π×10-7T.m/A4π1.60×10-19C3.00×107m/ssin30°2.00×10-6m2=1.20×10-7TBC=1.20×10-7T

according to the right-hand rule, the direction of the magnetic fieldBqv×r^ is out of the page.

03

Solving part (c) or the problem.

Finally, we need to find the magnitude and direction of the magnetic field that this electron produces at D where the distance from this point to the electron is r= 2.00 μm, we can see that the angle between the velocity and the position vector is ϕ=180°So we get

BD=0

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