Question

A proton accelerated by a potential $V = 5 \times 10^5 V$ 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)

• A. 15°
• B. 30
• C. 45
• D. 60

Answer 1

Answer: (B)

30°

Answer 2

To determine the angle $\theta$ through which the proton deviates, we 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 $V = 5 \times 10^5$ V
• 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: $\frac{1}{2}m_p v^2 = eV$ Solving for $v$: $v = \sqrt{\frac{2eV}{m_p}}$ Substituting the given values: $v = \sqrt{\frac{2 \times (1.602 \times 10^{-19} \text{ C}) \times (5 \times 10^5 \text{ V})}{1.672 \times 10^{-27} \text{ kg}}}$ $v = \sqrt{\frac{1.602 \times 10^{-13}}{1.672 \times 10^{-27}}}$ $v = \sqrt{0.95813 \times 10^{14}} \approx 0.9788 \times 10^7 \text{ m/s} = 9.788 \times 10^6 \text{ m/s}$

Step 2: Calculate the radius ($r$) of the circular path. The magnetic force provides the centripetal force for the circular motion: $qvB = \frac{m_p v^2}{r}$ Since $q=e$ for a proton, we have: $evB = \frac{m_p v^2}{r}$ Solving for $r$: $r = \frac{m_p v}{eB}$ Substituting the calculated velocity and other given values: $r = \frac{(1.672 \times 10^{-27} \text{ kg}) \times (9.788 \times 10^6 \text{ m/s})}{(1.602 \times 10^{-19} \text{ C}) \times (0.5 \text{ T})}$ $r = \frac{1.636 \times 10^{-20}}{0.801 \times 10^{-19}}$ $r \approx 0.2042 \text{ m}$

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 path inside the magnetic field is a circular arc. 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 horizontal distance $d$ traveled within the field is related to the radius $r$ of the path and the deviation angle $\theta$ by the trigonometric relation: $\sin \theta = \frac{d}{r}$ Now, plug in the values: $\sin \theta = \frac{0.1 \text{ m}}{0.2042 \text{ m}}$ $\sin \theta \approx 0.4897$ To find $\theta$, take the arcsin: $\theta = \arcsin(0.4897)$ $\theta \approx 29.32^\circ$ This value is approximately $30^\circ$.