What is the mistake in calculating the magnetic field in this problem?

[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.