Question

A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude \(\dfrac{\pi}{2} \times 10^{-3} \text{T}\). The angle between the direction of magnetic field and velocity of proton is 60°. The pitch of the helical path taken by the proton is \_\_\_\ cm.

• A. 65 cm
• B. 40 cm
• C. 50 cm
• D. 80 cm

Answer 1

The Correct Option is B

To solve the problem of finding the pitch of the helical path taken by a proton in a uniform magnetic field, we need to understand the relationship between the proton's motion, kinetic energy, magnetic force, and helical motion. The pitch is the distance the proton travels parallel to the magnetic field during one complete rotation around the field lines.

1. The kinetic energy \( K \) of the proton is given as 2.0 eV. First, we need to convert this into joules for standard unit usage: \(K = 2.0 \text{ eV} = 2.0 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-19} \text{ J}\).
2. The velocity \( v \) of the proton can be found using the kinetic energy formula: \(K = \dfrac{1}{2} m v^2\), where \( m \) is the mass of proton \((1.67 \times 10^{-27} \text{ kg})\). Solving for \( v \): \(v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 3.2 \times 10^{-19}}{1.67 \times 10^{-27}}} = 6.19 \times 10^5 \text{ m/s}\).
3. The velocity can be decomposed into components parallel \((v_{\parallel})\) and perpendicular \((v_{\perp})\) to the magnetic field:
   • Parallel component: \(v_{\parallel} = v \cos \theta = 6.19 \times 10^5 \times \cos(60^\circ) = 3.095 \times 10^5 \text{ m/s}\).
   • Perpendicular component: \(v_{\perp} = v \sin \theta = 6.19 \times 10^5 \times \sin(60^\circ) = 5.36 \times 10^5 \text{ m/s}\).
4. The cyclotron frequency (angular frequency for circular motion in a magnetic field) \( \omega \) is: \(\omega = \dfrac{qB}{m}\), where \( q \) is the charge of the proton \((1.6 \times 10^{-19} \text{ C})\) and \( B = \dfrac{\pi}{2} \times 10^{-3} \text{ T}\). \(\omega = \dfrac{1.6 \times 10^{-19} \times \dfrac{\pi}{2} \times 10^{-3}}{1.67 \times 10^{-27}} = 1.508 \times 10^{7} \text{ rad/s}\).
5. The time period \( T \) of the circular motion is given by: \(T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{1.508 \times 10^{7}}\).
6. The pitch \( p \) of the helical path is the distance traveled in one complete cycle along the direction of the magnetic field: \(p = v_{\parallel} \times T = 3.095 \times 10^5 \times \dfrac{2\pi}{1.508 \times 10^{7}} = 0.40 \text{ m} = 40 \text{ cm}\).

Thus, the pitch of the helical path taken by the proton is 40 cm.