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A +7.00-μC point charge is moving at a constant 8.50×106 m/s in the +y-direction, relative to a reference frame. At the instant when the point charge is at the origin of this reference frame, find the magnetic-field vector B⃗ it produces at the following points.

Answer :

The points are:

(a) x = 0.500 m, y = 0, z = 0 (b) x = 0, y = 0, z = +0.500 m

Answer:

a) [tex] B = -2.38\times10^{-5} \hat k[/tex]

b) [tex] B = 2.38\times10^{-5} \hat i [/tex]

Explanation:

The magnetic field can be calculated as follows:

[tex]B= \frac{\mu_o}{4\pi}\frac{q (\vec v\times \hat r)}{r^2}[/tex]

a) The velocity vector is in +y direction and the point is on x-axis

Because [tex]\hat j \times \hat i = -\hat k[/tex],

[tex]B= \frac{\mu_o}{4\pi}\frac{q (\vec v\times \hat r)}{r^2}[/tex]

[tex]B= 1\times 10^{-7}\times 7\times 10^{-6}\times \frac{8.50\times 10^6 \hat j\times \hat i}{0.500^2}\\B = -2.38\times10^{-5} \hat k[/tex]

b) The velocity vector is in +y direction and the point is on z-axis

Because [tex]\hat j \times \hat k = \hat i[/tex],

[tex]B= \frac{\mu_o}{4\pi}\frac{q (\vec v\times \hat r)}{r^2}[/tex]

[tex]B= 1\times 10^{-7}\times 7\times 10^{-6}\times \frac{8.50\times 10^6 \hat j\times \hat k}{0.500^2}\\B = 2.38\times10^{-5} \hat i[/tex]

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