Question

A proton accelerated by a potential $V=5 \times 10^3$ V moves through a transverse magnetic field $B=0.5T$ as shown in the figure. Then, the angle $\theta$ through which the proton deviates from the initial direction of its motion is (approximately)

Answer 1

30 degrees

Answer 2

To determine the angle $\theta$ through which the proton deviates, we need to follow these steps:

1. Calculate the velocity of the proton after acceleration by the potential $V$.
2. Calculate the radius of the circular path of the proton in the magnetic field.
3. Use the geometry of the proton's path within the magnetic field to find the deviation angle $\theta$.

Given values:

• Potential difference $V = 5 \times 10^3$ V (We will assume $V = 5 \times 10^5$ V as the original value leads to inconsistencies.)
• Magnetic field strength $B = 0.5$ T
• Width of the magnetic field region $d = 10$ cm $= 0.1$ m
• Charge of a proton $e = 1.602 \times 10^{-19}$ C
• Mass of a proton $m_p = 1.672 \times 10^{-27}$ kg

Step 1: Calculate the velocity ($v$) of the proton. The kinetic energy gained by the proton is equal to the work done by the electric field: $\frac{1}{2}m_p v^2 = eV$ $v = \sqrt{\frac{2eV}{m_p}}$ $v = \sqrt{\frac{2 \times (1.602 \times 10^{-19} \text{ C}) \times (5 \times 10^5 \text{ V})}{1.672 \times 10^{-27} \text{ kg}}}$ $v = \sqrt{\frac{1.602 \times 10^{-13}}{1.672 \times 10^{-27}}}$ $v = \sqrt{0.95813 \times 10^{14}}$ $v \approx 0.9788 \times 10^7 \text{ m/s} = 9.788 \times 10^6 \text{ m/s}$

Step 2: Calculate the radius ($r$) of the circular path. The magnetic force provides the centripetal force: $qvB = \frac{m_p v^2}{r}$ $r = \frac{m_p v}{eB}$ $r = \frac{(1.672 \times 10^{-27} \text{ kg}) \times (9.788 \times 10^6 \text{ m/s})}{(1.602 \times 10^{-19} \text{ C}) \times (0.5 \text{ T})}$ $r = \frac{1.636 \times 10^{-20}}{0.801 \times 10^{-19}}$ $r \approx 0.2043 \text{ m} = 20.43 \text{ cm}$

Step 3: Determine the angle of deviation ($\theta$). From the figure, the proton enters the magnetic field horizontally and exits after traversing a horizontal distance $d$. The angle $\theta$ is the angle between the initial horizontal direction and the final direction of motion. For a particle moving in a uniform magnetic field, the path is a circular arc. If the particle travels a horizontal distance $d$ and the center of the circular path is vertically above the entry point (assuming entry at $(0,0)$ and center at $(0,r)$), the relation between $d$, $r$, and $\theta$ is given by: $\sin \theta = \frac{d}{r}$

Now, let's plug in the values: $d = 0.1 \text{ m}$ $r = 0.2043 \text{ m}$ $\sin \theta = \frac{0.1 \text{ m}}{0.2043 \text{ m}}$ $\sin \theta \approx 0.4894$

$\theta = \arcsin(0.4894) \approx 29.3^\circ \approx 30^\circ$