Magnetic field from moving charge cross product

[Physicslearner500039]

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.

 

[TSny]

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

There is a mistake in this equation. Check to see if you have the correct the power of $r$ in the denominator, or if the vector $\vec r$ in the numerator should be a unit vector.

I don't see the angle in calculations between $\vec v , \vec r$ which i assume is $135 Deg$.

Since you are working with unit vector representation of the vectors, you don't need to know the angle between $\vec v$ and $\vec r$.

 

[Physicslearner500039]

Ok understood. Thank you, yes the formula i wrongly understood. The answer is
$B = \frac {10^{-7} * 6 * 8\vec j * (-\frac {1} {\sqrt2} \vec j + \frac {1} {\sqrt2} \vec k)} {0.05}$
$B = 6.78\mu T \vec i$

 

[TSny]

Looks good except for the placement of the decimal point in the denominator. I believe your final answer is correct.