Problem: An electron moves at 0.150 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron. Find the magnitude of the magnetic field this electron produces at the pointno title providedA) Find the magnitude and direction of the magnetic field this electron produces at the point A.B) Find the magnitude and direction of the magnetic field this electron produces at the point B.C) Find the magnitude and direction of the magnetic field this electron produces at the point C.D) Find the magnitude and direction of the magnetic field this electron produces at the point D.

FREE Expert Solution

The magnetic field is expressed as:

$\boxed{\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{q}\mathbf{(}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{\times }\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}}$

The magnitude of the magnetic field is then:

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{q}\mathbf{v}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{4}\mathbf{\pi }{\mathbf{r}}^{\mathbf{2}}}$

But, $\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{q}\mathbf{v}}{\mathbf{4}\mathbf{\pi }{\mathbf{r}}^{\mathbf{2}}}\mathbf{=}{\mathit{B}}_{\mathbf{0}}$

Therefore,

$\boxed{\mathbf{B}\mathbf{=}{\mathbf{B}}_{\mathbf{0}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}$

We'll also use the left-hand rule to determine the direction of the magnetic field. 

This is because the electron is negatively charged and we use the left-hand rule for negative charges.

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