Escaping the magnetic field of a wire

We set up everything in $3$ -dimensional co-ordinates first. Take the axis of the wire to be the $z$ axis, and let the direction of the electron be the $x$ axis. The magnetic field in the $xz$ plane is then directed along the $y$ axis. Let $B$ be the magnetic field on the electron. Then, Ampere's law gives: $B= \frac{ \mu_0 I}{2 \pi x}$ However, this field is perpendicular to the velocity of the electron, and hence does no work. This forces the kinetic energy of the electron to be constant. Let $v_0$ be the initial velocity of the electron, and let $v_x$ and $v_z$ be the components of the velocity of the electron on the $x$ and $z$ axis respectively (note that the electron has no motion along the $y$ axis). Then, we have: $v_0^2= v_x^2 + v_z^2$ $\implies v_z=-\sqrt{v_0^2-v_x^2}$ We take the negative value because the electron moves along the negative $z$ axis. According to Lorentz formula, we get: $\vec{F}= q \vec{v} \times \vec{B}$ $\implies m \frac{dv}{dt}= ev_zB= -e\sqrt{v_0^2-v_x^2}\frac{\mu_0I}{2 \pi x}$ Note that $x$ is maximized when $v_x= 0$ . Also note that $\frac{dv_x}{dt}= v_x \frac{dv_x}{dx}$ . All what remains is to substitute this value in the equation above. Integrating, we can show that $x_{max}= R e^{\frac{2v_0 \pi m}{\mu_0Ie}}$ Plugging the values, we obtain $v_{max}= \boxed{0.0234 \ m}$

Let us denote the perpendicular distance from the wire by $x$ . Let us denote the axes by $X$ (along the perpendicular ) and $Y$ (perpendicular to $X$ in the plane of motion ) .

Since magnetic force is perpendicular to velocity , it does no work and the speed always remains constant.

Let $\theta$ be the angle which $\vec{v}$ makes with the $X$ axis.

$F_{x} = - evBsin\theta$

Here, $v\sin\theta = v_{y} = \sqrt{{v_{0}}^2 - {v_{x}}^2}$

Hence , $F_{x} = ma_{x} = -e(\frac{\mu_{0} i}{2 \pi x}) \sqrt{{v_{0}}^2 - {v_{x}}^2}$

Putting $a_{x} = v_{x}\frac{d(v_{x})}{dx}$ , we get ,

$\frac{-v_{x} d(v_{x})}{\sqrt{{v_{0}}^2 - {v_{x}}^2}} = \frac{\mu_{0} i e}{2 \pi m} \frac{dx}{x}$

We know that x would maximum when $v_{x} = 0$

$\int_{v_{0}}^0 \frac{-v_{x} d(v_{x})}{\sqrt{{v_{0}}^2 - {v_{x}}^2}} = \int_{ R}^{x_{max}} \frac{\mu_{0} i e}{2 \pi m} \frac{dx}{x}$

$\Rightarrow x_{max} = R e^{\frac{2 \pi m_{e} v_{0}}{\mu_{0} i e}} = \fbox{0.0234}$

${Distance}_{max}=Re^{\frac{2{\pi}m_ev_0}{u_0Ie}}$ By substituting the values given in the problem and taking exponent, $e=2.718281828$ , we will get ${Distance}_{max}=0.0234$ .

The magnetic field at a distance $r$ from the axis of the wire is given as $B = \dfrac{\mu_0 I } {2\pi r}$ . The force on the electron is the Magnetic-Lorentz. A quick check shows that the force is in the direction that is opposite to the motion of the electron (hence the question!). The force is given by $F = -evB$ since the angle between the velocity vector and the magnetic field vector is $90^{\circ}$ at every instant. $v$ is the magnitude of the velocity of electron and the negative sign is because $v$ and $F$ are anti-parallel.

Writing the god equation:

$F_{net} = ma = -evB$

This gives $a = \dfrac{dv}{dt} = v\dfrac{dv}{dr} = -\dfrac{e}{m}vB$ .

So, $dv = -\dfrac{e}{m}B dr = -\dfrac e m \left(\dfrac{\mu_0 I} {2\pi}\right) \dfrac{ dr} { r}$

Now, the maximum distance occurs when the velocity of the particle becomes zero (and it starts returning to the wire surface).

So, integrating from limits $v_i = 30, v_f = 0$ and $r_i = 0.01, r_f = R$ gives: $-30 = -\dfrac{\mu_0 I e} {2\pi m} \ln \left( \dfrac{ R} { 0.01} \right)$

Plugging in the values and solving for $R$ gives $R = 0.023 m$ .