Question

A charged particle (charge: +q, mass: m) enters a region of magnetic field $B_0 (-\hat{k})$ with velocity $v_0 \hat{i}$ at the origin.

The magnetic field exists from x = 0 to $x = \frac{3mv_0}{5qB_0}$. Then

• A. The y-co-ordinate of the point at which the particle exits the magnetic field is $\frac{2mv_0}{5qB_0}$
• B. The y-co-ordinate of the point at which the particle exits the magnetic field is $\frac{mv_0}{5qB_0}$
• C. The time spent by the particle in the magnetic field is $\frac{53\pi m}{180qB_0}$
• D. The magnitude of impulse imparted by the magnetic force on the particle is $mv_0 \sqrt{\frac{2}{5}}$

Answer 1

Answer: (B), (D)

B, D

Answer 2

The particle enters the magnetic field region ($0 \le x \le \frac{3mv_0}{5qB_0}$) at the origin (0,0) with velocity $\vec{v} = v_0 \hat{i}$. The magnetic field is $\vec{B} = -B_0 \hat{k}$.

The magnetic force on the particle is $\vec{F} = q(\vec{v} \times \vec{B})$.

Initially, at the origin, $\vec{F} = q(v_0 \hat{i} \times -B_0 \hat{k}) = -qv_0 B_0 (\hat{i} \times \hat{k}) = -qv_0 B_0 (-\hat{j}) = qv_0 B_0 \hat{j}$.

The force is perpendicular to the velocity, so the particle moves in a circular path in the x-y plane. The magnitude of the velocity remains constant, $v = v_0$.

The radius of the circular path is $R = \frac{mv}{qB} = \frac{mv_0}{qB_0}$.

The initial velocity is in the +x direction, and the initial force is in the +y direction. This means the center of the circle must be located at a point (x_c, y_c) such that the radius vector from the center to the initial position (0,0) is opposite to the initial force direction. The initial force is in the +y direction, so the radius vector from the center to (0,0) must be in the -y direction. Thus, the center is at (0, R).

The equation of the circular path is $x^2 + (y-R)^2 = R^2$.

The particle exits the magnetic field when its x-coordinate reaches $x_{exit} = \frac{3mv_0}{5qB_0}$.

Let $R = \frac{mv_0}{qB_0}$. The exit x-coordinate is $x_{exit} = \frac{3}{5}R$.

To find the y-coordinate at the exit point, we substitute $x = \frac{3}{5}R$ into the equation of the circle:

$(\frac{3}{5}R)^2 + (y_{exit} - R)^2 = R^2$

$\frac{9}{25}R^2 + (y_{exit} - R)^2 = R^2$

$(y_{exit} - R)^2 = R^2 - \frac{9}{25}R^2 = \frac{16}{25}R^2$

$y_{exit} - R = \pm \sqrt{\frac{16}{25}R^2} = \pm \frac{4}{5}R$

$y_{exit} = R \pm \frac{4}{5}R$

The two possible y-coordinates on the circle at $x = \frac{3}{5}R$ are $y_1 = R + \frac{4}{5}R = \frac{9}{5}R$ and $y_2 = R - \frac{4}{5}R = \frac{1}{5}R$.

The particle starts at (0,0) and the initial force is in the +y direction, so the particle moves into the region $y>0$. As the x-coordinate increases from 0, the y-coordinate initially increases.

The velocity components are $v_x = \frac{v_0}{R}(R-y)$ and $v_y = \frac{v_0}{R}x$. Since $x$ increases from 0 in the region, $v_y > 0$, confirming that y increases initially.

The particle path starts at (0,0) and moves towards increasing x and y.

The point $( \frac{3}{5}R, \frac{9}{5}R)$ corresponds to an angle $\phi = \arccos(-\frac{4}{5}) \approx 143.1^\circ$ from the -y axis (using $x=R\sin\phi, y=R-R\cos\phi$).

The point $( \frac{3}{5}R, \frac{1}{5}R)$ corresponds to an angle $\phi = \arccos(\frac{4}{5}) \approx 36.9^\circ$ from the -y axis.

The motion starts at $\phi=0$ (at (0,0)). As x increases from 0 to $\frac{3}{5}R$, the angle $\phi$ increases from 0. The y-coordinate is $y = R(1-\cos\phi)$. For $0 \le \phi \le \pi$, $1-\cos\phi$ increases from 0 to 2.

The x-coordinate is $x = R \sin\phi$. For $0 \le \phi \le \pi/2$, x increases from 0 to R. For $\pi/2 \le \phi \le \pi$, x decreases from R to 0.

The particle exits at $x = \frac{3}{5}R$. This x-coordinate is reached at two possible angles $\phi_1$ and $\phi_2$ in the range $[0, \pi]$, where $\sin \phi = \frac{3}{5}$. These angles are $\phi_1 = \arcsin(\frac{3}{5})$ (in the first quadrant, $\approx 36.9^\circ$) and $\phi_2 = \pi - \arcsin(\frac{3}{5})$ (in the second quadrant, $\approx 143.1^\circ$).

The y-coordinate is $y = R(1-\cos\phi)$.

If $\sin\phi = \frac{3}{5}$, then $\cos\phi = \pm \sqrt{1 - (\frac{3}{5})^2} = \pm \frac{4}{5}$.

For $\phi_1 = \arcsin(\frac{3}{5})$, $\cos\phi_1 = +\frac{4}{5}$. The y-coordinate is $y_1 = R(1 - \frac{4}{5}) = \frac{1}{5}R$. This corresponds to the point $(\frac{3}{5}R, \frac{1}{5}R)$.

For $\phi_2 = \pi - \arcsin(\frac{3}{5})$, $\cos\phi_2 = -\frac{4}{5}$. The y-coordinate is $y_2 = R(1 - (-\frac{4}{5})) = \frac{9}{5}R$. This corresponds to the point $(\frac{3}{5}R, \frac{9}{5}R)$.

Since the particle starts at (0,0) and moves towards increasing x, the first time it reaches $x = \frac{3}{5}R$ will be at the smaller angle $\phi_1$.

Thus, the y-coordinate at which the particle exits the magnetic field is $y_{exit} = \frac{1}{5}R = \frac{mv_0}{5qB_0}$.

Option (A) is incorrect. Option (B) is correct.

(C) The time spent by the particle in the magnetic field is the time taken to travel from the entry point (0,0) to the exit point $(\frac{3}{5}R, \frac{1}{5}R)$.

The particle moves along the arc of the circle. The angle subtended by this arc at the center (0,R) is the change in the angle $\phi$.

At the entry point (0,0), $x=0, y=0$. $0 = R\sin\phi_{entry}$, $0 = R(1-\cos\phi_{entry})$. $\sin\phi_{entry}=0, \cos\phi_{entry}=1$. This corresponds to $\phi_{entry} = 0$.

At the exit point $(\frac{3}{5}R, \frac{1}{5}R)$, $x=\frac{3}{5}R, y=\frac{1}{5}R$. $\frac{3}{5}R = R\sin\phi_{exit}$, $\frac{1}{5}R = R(1-\cos\phi_{exit})$. $\sin\phi_{exit}=\frac{3}{5}$, $\cos\phi_{exit}=1-\frac{1}{5}=\frac{4}{5}$. This corresponds to an angle $\phi_{exit} = \arcsin(\frac{3}{5})$ (since $\sin>0, \cos>0$, it's in the first quadrant). Let $\alpha = \arcsin(\frac{3}{5})$. The angle is $\alpha$.

The angle swept by the radius vector from the center is $\Delta \phi = \phi_{exit} - \phi_{entry} = \alpha - 0 = \alpha$.

The angular speed of the particle around the center is $\omega = \frac{v_0}{R} = \frac{v_0}{mv_0/(qB_0)} = \frac{qB_0}{m}$.

The time taken is $t = \frac{\Delta \phi}{\omega} = \frac{\alpha}{qB_0/m} = \frac{m \alpha}{qB_0}$.

Here $\alpha = \arcsin(\frac{3}{5})$. In radians, $\alpha \approx 0.6435$ radians.

The time is $t = \frac{m \arcsin(3/5)}{qB_0}$.

Let's check the value in option (C): $\frac{53\pi m}{180qB_0}$.

$\frac{53\pi}{180}$ radians corresponds to 53 degrees. $\arcsin(3/5) \approx 36.87^\circ$.

So option (C) is incorrect.

(D) The impulse imparted by the magnetic force on the particle is $\vec{J} = \int \vec{F} dt = \Delta \vec{p} = m(\vec{v}_{exit} - \vec{v}_{entry})$.

$\vec{v}_{entry} = v_0 \hat{i}$.

At the exit point $(\frac{3}{5}R, \frac{1}{5}R)$, the velocity vector is tangent to the circle. The velocity vector is perpendicular to the radius vector from the center (0,R) to the exit point $(\frac{3}{5}R, \frac{1}{5}R)$.

The radius vector is $(\frac{3}{5}R - 0, \frac{1}{5}R - R) = (\frac{3}{5}R, -\frac{4}{5}R)$.

The velocity vector is proportional to $(-(-\frac{4}{5}R), \frac{3}{5}R) = (\frac{4}{5}R, \frac{3}{5}R)$ or its negative.

The magnitude of the velocity is $v_0$. So $\vec{v}_{exit} = v_0 \frac{(\frac{4}{5}R, \frac{3}{5}R)}{\sqrt{(\frac{4}{5}R)^2 + (\frac{3}{5}R)^2}} = v_0 \frac{(\frac{4}{5}R, \frac{3}{5}R)}{R} = v_0 (\frac{4}{5}, \frac{3}{5}) = v_0 (\frac{4}{5}\hat{i} + \frac{3}{5}\hat{j})$.

Let's check the direction of motion. The particle moves from $(0,0)$ to $(\frac{3}{5}R, \frac{1}{5}R)$. This is in the first quadrant. The velocity vector should have positive components. Our derived velocity vector has positive components, which is consistent.

$\vec{v}_{exit} = \frac{4}{5}v_0 \hat{i} + \frac{3}{5}v_0 \hat{j}$.

The impulse is $\vec{J} = m(\vec{v}_{exit} - \vec{v}_{entry}) = m((\frac{4}{5}v_0 \hat{i} + \frac{3}{5}v_0 \hat{j}) - v_0 \hat{i}) = m(-\frac{1}{5}v_0 \hat{i} + \frac{3}{5}v_0 \hat{j})$.

The magnitude of the impulse is $|\vec{J}| = m v_0 \sqrt{(-\frac{1}{5})^2 + (\frac{3}{5})^2} = m v_0 \sqrt{\frac{1}{25} + \frac{9}{25}} = m v_0 \sqrt{\frac{10}{25}} = m v_0 \sqrt{\frac{2}{5}}$.

Option (D) states the magnitude is $mv_0 \sqrt{\frac{2}{5}}$. This is correct.