Question

A proton of mass $1.67 \times {10^{ - 27}}kg$ and charge $1.68 \times {10^{ - 19}}C$ is projected with a speed of $2 \times {10^6}m{s^{ - 1}}$ at an angle of ${60^ \circ }$to the X-axis. If a uniform magnetic field of $0.10T$ is applied along Y-axis, the path of proton is:
A. a circle of radius $0.2m$and time period $\pi \times {10^{ - 7}}s$
B. a circle of radius $0.1m$and time period $2\pi \times {10^{ - 7}}s$
C. a helix of radius $0.1m$and time period $2\pi \times {10^{ - 7}}s$
D. a helix of radius $0.2m$and time period $4\pi \times {10^{ - 7}}s$

Answer 1

It is given that the uniform magnetic field is applied along Y-axis and the electron is projected along X-axis with a certain velocity and at a certain angle. Therefore, the magnetic field is parallel to the velocity. The parallel magnetic field is responsible for the charged particle to move in a curved helical path.

Formula used:
The radius of curved path of a charged particle is: $r = \dfrac{{mv\cos \theta }}{{qB}}$
The time period is the period taken by the charged particle to go around the curved path is: $T = \dfrac{{2\pi r}}{v}$

Complete step by step solution: The values of mass, charge and projection speed of protons is given in the problem
The mass of proton =$m = 1.67 \times {10^{ - 27}}kg$
Charge of proton = $q = 1.68 \times {10^{ - 19}}C$
Velocity of proton = $v = 2 \times {10^6}m{s^{ - 1}}$
Magnetic field = $B = 0.10T$
The magnetic force is always perpendicular to the velocity of the charged particle. Therefore, the
a charged path follows a curved path in the magnetic field. The magnetic force produces a centripetal
force, that is: ${F_c} = \dfrac{{m{v^2}}}{r}$. On the other hand, the magnitude of the magnetic force
is: $F = qvB$. Because this centripetal force is provided by the magnetic force, $\dfrac{{m{v^2}}}{r} = qvB$. Therefore, the radius of the curved path will be
$r = \dfrac{{m{v^2}}}{{qvB}} = \dfrac{{mv}}{{qB}}$
The magnetic field is applied along Y-axis and the velocity of electron is along X-axis at an angle of ${60^ \circ }$. The component of velocity parallel to the magnetic field is ${v_{parallel}} = v\cos \theta$.
Thus, $\theta = {60^ \circ }$. Therefore, the radius will be $r = \dfrac{{mv\cos \theta }}{{qB}}$

\to (1)$$ Substituting the values for mass, velocity, charge and magnetic field in equation (1) $$r = \dfrac{{mv}}{{qB}} = \dfrac{{1.67 \times {{10}^{ - 27}} \times 2 \times {{10}^6} \times \cos ({{60}^ \circ })}}{{1.68 \times {{10}^{ - 19}} \times 0.10}} = \dfrac{{1.67 \times {{10}^{ - 27}} \times 2 \times {{10}^6} \times \dfrac{1}{2}}}{{1.68 \times {{10}^{ - 19}} \times 0.10}} \approx 10 \times {10^{ \- 2}}m = 0.1m$$ The time period is the period taken by the charged particle to go around the curved path. Thus, the time period $$T$$ is the distance covered by the charged particle (or the circumference of the curved path) divided by velocity. $$T = \dfrac{{2\pi r}}{v} = \dfrac{{2\pi \left( {10 \times {{10}^{ - 2}}} \right)}}{{2 \times {{10}^6} \times \cos ({{60}^ \circ })}} = \dfrac{{2\pi \left( {10 \times {{10}^{ - 2}}} \right)}}{{2 \times {{10}^6} \times \dfrac{1}{2}}} = 2\pi \times {10^{ - 7}}s$$ If velocity of the charged particle is not perpendicular to the magnetic field, then each component of velocity is compared to the magnetic field. The parallel magnetic field is responsible for the charged particle to move in a form of helix. Thus, a helix of radius $$0.1m$$ and time period $$2\pi \times {10^{ - 7}}s$$ is formed. **Hence, option (C) is the correct answer.** **Note:** The magnetic force is always perpendicular to velocity. Therefore, no work is done on the charged particle. The Kinetic energy of the particle and the speed will always remain constant (provided if there is a constant magnetic field), but the direction may change due to parallel magnetic field forming a helical path. When the magnetic field is perpendicular to the velocity of the charged particle, then it ends up forming a circular path.