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

Given: \[ B = \frac{\pi}{2} \times 10^{-3} \, \text{T}, \quad \text{K.E.} = 2.0 \, \text{eV}, \quad m = 1.67 \times 10^{-27} \, \text{kg}, \quad \text{Charge on proton} = 1.6 \times 10^{-19} \, \text{C} \] From the kinetic energy formula \( K.E. = \frac{1}{2} m v^2 \), we can solve for the velocity \(v\) as follows: \includegraphics[width=0.75\linewidth]{6.png} \[ v = \sqrt{\frac{2 K.E.}{m}} \] The pitch of the helical path is given by: \[ \text{Pitch} = v \cos 60^\circ \times \text{time period of one rotation} \] \[ v = v \cos 60^\circ = \frac{2\pi r}{e B} \] \[ = \sqrt{\frac{2 \times 2 \times 10^{-19}}{1.6 \times 10^{-27}}} \times \cos 60^\circ = \frac{2 \pi \times 1.6 \times 10^{-19}}{1.6 \times 10^{-27} \times \frac{\pi}{2} \times 10^{-3}} \] \[ = 2 \times 10^4 \times \frac{1}{2} \times 4 \times 10^{-5} = 4 \times 10^1 \text{ cm} \] Thus, the pitch of the helical path is 40 cm.