Question

The figure shows a region of length 'l' with a uniform magnetic field of 0.3T in it and a proton entering the region with velocity $4 \times 10^5 ms^{-1}$ making an angle $60^\circ$ with the field. If the proton completes 10 revolution by the time it cross the region shown,' l ' is close to (mass of proton = $1.67 \times 10^{-27}kg$, charge of the proton = $1.6 \times 10^{-19}C$)

• A. 0.11m
• B. 0.88m
• C. 0.44m
• D. 0.22m

Answer 1

Answer: (C)

0.44m

Answer 2

The motion of a charged particle in a uniform magnetic field is helical if the velocity of the particle is not parallel or perpendicular to the magnetic field. The velocity component parallel to the magnetic field, $v_{||} = v \cos \theta$, remains constant, while the velocity component perpendicular to the magnetic field, $v_{\perp} = v \sin \theta$, results in uniform circular motion in a plane perpendicular to the magnetic field.

The time period of the circular motion is given by $T = \frac{2\pi m}{qB}$. This is the time taken for one complete revolution. The distance traveled by the particle along the direction of the magnetic field during one revolution is called the pitch of the helix, $p$. The pitch is given by $p = v_{||} T$.

In this problem, the proton enters the magnetic field region of length $l$ with velocity $v = 4 \times 10^5 \, ms^{-1}$ at an angle $\theta = 60^\circ$ with the magnetic field $B = 0.3 \, T$. The mass of the proton is $m = 1.67 \times 10^{-27} \, kg$ and the charge is $q = 1.6 \times 10^{-19} \, C$. The proton completes 10 revolutions by the time it crosses the region of length $l$. This means that the total distance traveled along the direction of the magnetic field is $l$, and this distance is covered in the time taken for 10 revolutions.

The time period of one revolution is: $T = \frac{2\pi m}{qB} = \frac{2\pi \times 1.67 \times 10^{-27} \, kg}{1.6 \times 10^{-19} \, C \times 0.3 \, T}$ $T = \frac{2\pi \times 1.67 \times 10^{-27}}{0.48 \times 10^{-19}} \, s$ $T = \frac{2\pi \times 1.67}{0.48} \times 10^{-8} \, s$

The velocity component parallel to the magnetic field is: $v_{||} = v \cos \theta = (4 \times 10^5 \, ms^{-1}) \times \cos 60^\circ = (4 \times 10^5) \times 0.5 \, ms^{-1} = 2 \times 10^5 \, ms^{-1}$.

The pitch of the helix is: $p = v_{||} T = (2 \times 10^5 \, ms^{-1}) \times \frac{2\pi \times 1.67}{0.48} \times 10^{-8} \, s$ $p = \frac{4\pi \times 1.67}{0.48} \times 10^{-3} \, m$

The proton completes 10 revolutions by the time it crosses the region of length $l$. Therefore, the length $l$ is 10 times the pitch: $l = 10 \times p = 10 \times \frac{4\pi \times 1.67}{0.48} \times 10^{-3} \, m$ $l = \frac{40\pi \times 1.67}{0.48} \times 10^{-3} \, m$

Using $\pi \approx 3.14159$: $l = \frac{40 \times 3.14159 \times 1.67}{0.48} \times 10^{-3} \, m$ $l = \frac{125.6636 \times 1.67}{0.48} \times 10^{-3} \, m$ $l = \frac{209.998572}{0.48} \times 10^{-3} \, m$ $l \approx 437.497 \times 10^{-3} \, m$ $l \approx 0.4375 \, m$

Comparing this value with the given options, the calculated value 0.4375 m is closest to 0.44 m.