Question

An infinitely long charged wire having linear charge density $\lambda$ is placed along z-axis & a particle of mass m, charge Q is projected from (R, 0, 0) with velocity $v_0$ along (+ve) y- axis in gravity free space Then.

• A. Speed of particle when its distance from wire grows to 2R is $\sqrt{v_0^2 + \frac{\lambda Q}{\pi \epsilon_0 m}ln2}$
• B. Speed of particle when its distance from wire grows to 2R is $\sqrt{v_0^2 + \frac{\lambda Q}{2 \pi \epsilon_0 m}ln2}$
• C. Velocity component of particle along the radial line (joining the line charge to particle) at an instant its distance becomes 2R is $\sqrt{\frac{v_0^2}{4} + \frac{\lambda Q}{2 \pi \epsilon_0 m}ln2}$
• D. Velocity component of particle along the radial line (joining the line charge to particle) at an instant its distance becomes 2R is $\sqrt{\frac{3v_0^2}{4} + \frac{\lambda Q}{\pi \epsilon_0 m}ln2}$

Answer 1

Answer: (A), (D)

(A), (D)

Answer 2

The electric field due to an infinitely long charged wire with linear charge density $\lambda$ at a perpendicular distance $r$ is given by $\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \hat{r}$, where $\hat{r}$ is the unit vector in the radial direction from the wire. The force on a particle with charge Q is $\vec{F} = Q\vec{E} = \frac{Q\lambda}{2\pi\epsilon_0 r} \hat{r}$. This force is conservative.

The initial position of the particle is $(R, 0, 0)$, so the initial distance from the wire is $r_i = R$. The initial velocity is $\vec{v}_0 = v_0 \hat{j}$. The initial speed is $v_0$. The motion is in the xy-plane since the initial velocity is in the xy-plane and the force is also in the xy-plane (radially outwards from the z-axis).

The potential energy of the particle at a distance $r$ from the wire is given by $U(r) = Q V(r)$, where $V(r)$ is the electric potential. The potential difference between two points at distances $r_1$ and $r_2$ is $V(r_2) - V(r_1) = -\int_{r_1}^{r_2} \vec{E} \cdot d\vec{r} = -\int_{r_1}^{r_2} \frac{\lambda}{2\pi\epsilon_0 r} dr = -\frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_2}{r_1}\right) = \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_1}{r_2}\right)$. The change in potential energy as the particle moves from $r_i$ to $r_f$ is $\Delta U = U(r_f) - U(r_i) = Q(V(r_f) - V(r_i)) = \frac{Q\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_i}{r_f}\right)$.

By conservation of mechanical energy, $\Delta KE + \Delta U = 0$, so $\Delta KE = -\Delta U$. $KE_f - KE_i = - \frac{Q\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_i}{r_f}\right) = \frac{Q\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_f}{r_i}\right)$. The initial kinetic energy is $KE_i = \frac{1}{2} m v_0^2$. Let the speed of the particle when its distance from the wire is $r_f = 2R$ be $v_f$. The final kinetic energy is $KE_f = \frac{1}{2} m v_f^2$. Substituting $r_i = R$ and $r_f = 2R$: $\frac{1}{2} m v_f^2 - \frac{1}{2} m v_0^2 = \frac{Q\lambda}{2\pi\epsilon_0} \ln\left(\frac{2R}{R}\right) = \frac{Q\lambda}{2\pi\epsilon_0} \ln(2)$. $\frac{1}{2} m v_f^2 = \frac{1}{2} m v_0^2 + \frac{Q\lambda}{2\pi\epsilon_0} \ln(2)$. $m v_f^2 = m v_0^2 + \frac{Q\lambda}{\pi\epsilon_0} \ln(2)$. $v_f^2 = v_0^2 + \frac{Q\lambda}{\pi\epsilon_0 m} \ln(2)$. $v_f = \sqrt{v_0^2 + \frac{Q\lambda}{\pi\epsilon_0 m} \ln(2)}$. This matches option (A).

Now consider the velocity component along the radial line. The radial velocity component is $v_r = \frac{dr}{dt}$. The velocity vector in polar coordinates in the xy-plane is $\vec{v} = v_r \hat{r} + v_\theta \hat{\theta}$, where $v_\theta = r \frac{d\theta}{dt}$ is the tangential velocity component. The force is purely radial, $\vec{F} = F_r \hat{r}$. There is no tangential force, $F_\theta = 0$. The angular momentum about the z-axis is conserved because the torque about the z-axis is zero: $\vec{\tau} = \vec{r} \times \vec{F} = r\hat{r} \times F_r \hat{r} = 0$. The angular momentum about the z-axis is $L_z = m r^2 \dot{\theta} = m r v_\theta$. Initial position is $(R, 0, 0)$, so $r_i = R$. Initial velocity is $\vec{v}_0 = v_0 \hat{j}$. At the initial position $(R, 0)$, the radial direction is along the +x-axis ($\hat{r} = \hat{i}$) and the tangential direction is along the +y-axis ($\hat{\theta} = \hat{j}$). So, initial velocity $\vec{v}_0 = v_0 \hat{j} = v_0 \hat{\theta}_i$. Initial radial velocity $v_{ri} = \vec{v}_0 \cdot \hat{r}_i = v_0 \hat{j} \cdot \hat{i} = 0$. Initial tangential velocity $v_{\theta i} = \vec{v}_0 \cdot \hat{\theta}_i = v_0 \hat{j} \cdot \hat{j} = v_0$. The initial angular momentum about the z-axis is L_z_i = m r_i v_{\theta i} = m R v_0.

When the distance from the wire is $r_f = 2R$, let the tangential velocity component be $v_{\theta f}$. By conservation of angular momentum about the z-axis: L_z_f = L_z_i $m r_f v_{\theta f} = m R v_0$. $m (2R) v_{\theta f} = m R v_0$. $2 v_{\theta f} = v_0 \implies v_{\theta f} = \frac{v_0}{2}$.

The speed of the particle at $r_f = 2R$ is $v_f$. The speed is related to the radial and tangential velocity components by $v_f^2 = v_{rf}^2 + v_{\theta f}^2$. We found $v_f^2 = v_0^2 + \frac{Q\lambda}{\pi\epsilon_0 m} \ln(2)$ and $v_{\theta f} = \frac{v_0}{2}$. So, $v_f^2 = v_{rf}^2 + \left(\frac{v_0}{2}\right)^2 = v_{rf}^2 + \frac{v_0^2}{4}$. Substitute the expression for $v_f^2$: $v_0^2 + \frac{Q\lambda}{\pi\epsilon_0 m} \ln(2) = v_{rf}^2 + \frac{v_0^2}{4}$. $v_{rf}^2 = v_0^2 - \frac{v_0^2}{4} + \frac{Q\lambda}{\pi\epsilon_0 m} \ln(2)$. $v_{rf}^2 = \frac{3v_0^2}{4} + \frac{Q\lambda}{\pi\epsilon_0 m} \ln(2)$. The radial velocity component is $v_{rf} = \sqrt{\frac{3v_0^2}{4} + \frac{Q\lambda}{\pi\epsilon_0 m} \ln(2)}$. This matches option (D).