Physics: 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.

I believe the answer would be C...

A +7.00-μC point charge is moving at a constant

marenh04938

21.01.2022

The points are:

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

a) $B = -2.38\times10^{-5} \hat k$

b) $B = 2.38\times10^{-5} \hat i$

Explanation:

The magnetic field can be calculated as follows:

$B= \frac{\mu_o}{4\pi}\frac{q (\vec v\times \hat r)}{r^2}$

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

Because $\hat j \times \hat i = -\hat k$ ,

$B= \frac{\mu_o}{4\pi}\frac{q (\vec v\times \hat r)}{r^2}$

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

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

Because $\hat j \times \hat k = \hat i$ ,

$B= \frac{\mu_o}{4\pi}\frac{q (\vec v\times \hat r)}{r^2}$

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

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