- #1

Physicslearner500039

- 124

- 6

- Homework Statement
- A 6.00 uC point charge is moving at a constant 8 * 10^6 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, what is the magnetic-field vector it produces at the following points x = 0, y = -0.500 m, z = +0.500 m?

- Relevant Equations
- B = μo/4##\pi \frac {q \vec v \times \vec r} {r^2} ##

The problem is simple, but have one confusion, if i substitute the values given, I get

##

B = \frac {10^{-7}(6*10^{-6})[(8*10^6 \vec j) \times (-0.5\vec j + 0.5 \vec k)]} {r^2} ##

## B = 48\mu T\vec i##

First thing the answer does not match. I don't see the angle in calculations between ##\vec v , \vec r## which i assume is ##135 Deg##. What is the mistake? Please advise.

##

B = \frac {10^{-7}(6*10^{-6})[(8*10^6 \vec j) \times (-0.5\vec j + 0.5 \vec k)]} {r^2} ##

## B = 48\mu T\vec i##

First thing the answer does not match. I don't see the angle in calculations between ##\vec v , \vec r## which i assume is ##135 Deg##. What is the mistake? Please advise.