9.2.8. a) Using the formula given in problem 9.2.1, determine the magnetic induction created by a charge $q$ moving with velocity $v$ at a distance $r$ from this charge. The radius vector $\vec{r}$ forms an angle $\alpha$ with the velocity $v$.
b) Determine the magnetic induction of a straight wire of length $l$ carrying a current $I$ at a distance $r$ from the wire if $l \ll r$. The radius vector $\vec{r}$ forms an angle $\alpha$ with the wire.
Solution
a) Let's write down the formula given in the problem 9.2.1, $$\vec{B} = \mu_0 \varepsilon_0 [\vec{v} \times \vec{E}] \quad(1)$$ We will find the current flow rate from 9.2.3$$v = \frac{I}{\lambda}\quad(2)$$ Considering that the charge $q = l\lambda$, the electric field strength $$E = \frac{l\lambda}{4\pi\varepsilon_0r^2}\quad(3)$$ We substitute $(2)$ and $(3)$ into $(1)$, and we get the expression $$\vec{B} = \mu_0 \varepsilon_0\left[\vec{\left(\frac{I}{\lambda}\right)} \times \frac{l\lambda}{4\pi\varepsilon_0} \cdot \frac{\vec{r}}{r^3}\right] \quad(4)$$$$\vec{B} = \frac{\mu_0 q}{4\pi r^3}[\vec{v} \times \vec{r}]$$ The modulus of this vector product is $$B = \frac{\mu_0 q}{4\pi r^3}vr\cdot \sin\alpha \Rightarrow \boxed{B = \frac{\mu_0 qv}{4\pi r^2}\cdot \sin\alpha}$$
b) Since $l \ll r$ then the approximation of the thread as a point charge is valid $$B = \mu_0 \varepsilon_0 Ev\cdot \sin\alpha$$ We substitute the expressions $v$ and $E$ from $(2)$ and $(3)$, respectively $$B = \mu_0 \varepsilon_0 \frac{l\lambda}{4\pi\varepsilon r^2} \cdot \frac{I}{\lambda}\sin\alpha \Rightarrow \boxed{B=\frac{\mu_0Il}{4\pi r^2}\sin\alpha}$$
