Hi there!
We can use Biot-Savart's Law for a moving particle:
![B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }](https://tex.z-dn.net/?f=B%3D%20%5Cfrac%7B%5Cmu_0%20%7D%7B4%5Cpi%7D%5Cfrac%7Bq%5Cvec%7Bv%7D%5Ctimes%20%5Cvec%7Br%7D%7D%7Br%5E2%20%7D)
B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)
q = charge of particle (1.6 × 10⁻¹⁹ C)
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
r = distance from particle (2.10 μm)
There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:
![B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }](https://tex.z-dn.net/?f=B%3D%20%5Cfrac%7B%5Cmu_0%20%7D%7B4%5Cpi%7D%5Cfrac%7Bq%5Cvec%7Bv%7D%20%5Cvec%7Br%7Dsin%5Ctheta%7D%7Br%5E2%20%7D)
Where 'θ' is the angle between the velocity and radius vectors.
a)
To find the angle between the velocity and radius vector, we find the complementary angle:
θ = 90° - 60° = 30°
Plugging 'θ' into the equation along with our other values:
![B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }](https://tex.z-dn.net/?f=B%3D%20%5Cfrac%7B%5Cmu_0%20%7D%7B4%5Cpi%7D%5Cfrac%7Bq%5Cvec%7Bv%7D%20%5Cvec%7Br%7Dsin%5Ctheta%7D%7Br%5E2%20%7D%5C%5C%5C%5CB%3D%20%5Cfrac%7B%284%5Cpi%20%2A10%5E%7B-7%7D%29%7D%7B4%5Cpi%7D%5Cfrac%7B%281.6%2A10%5E%7B-19%7D%29%283.9%2A10%5E%7B7%7D%29%20%5Cvec%7Br%7Dsin%2830%29%7D%7B%282.1%2A10%5E%7B-5%7D%29%5E2%20%7D)
![B = \boxed{7.07 *10^{-10} T}](https://tex.z-dn.net/?f=B%20%3D%20%5Cboxed%7B7.07%20%2A10%5E%7B-10%7D%20T%7D)
b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.
c)
In this instance, the radius vector and the velocity vector are perpendicular so
'θ' = 90°.
![B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}](https://tex.z-dn.net/?f=B%3D%20%5Cfrac%7B%284%5Cpi%20%2A10%5E%7B-7%7D%29%7D%7B4%5Cpi%7D%5Cfrac%7B%281.6%2A10%5E%7B-19%7D%29%283.9%2A10%5E%7B7%7D%29%20%5Cvec%7Br%7Dsin%2890%29%7D%7B%282.1%2A10%5E%7B-5%7D%29%5E2%20%7D%20%3D%20%5Cboxed%7B1.415%20%2A%2010%5E%7B-9%7DT%7D)
d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.
Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.
![\boxed{B = 0 T}](https://tex.z-dn.net/?f=%5Cboxed%7BB%20%3D%200%20T%7D)