Question

A beam of proton moving with velocity $4 \times 10^5$ m/sec enters in a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. Calculate radius of helical path and pitch of helix.

Answer 1

Radius of helical path ≈ 0.01205 m, Pitch of helix ≈ 0.04372 m

Answer 2

A charged particle moving in a uniform magnetic field experiences a force that is perpendicular to both its velocity and the magnetic field. When the velocity of the particle is at an angle to the magnetic field, the motion is helical. This helical motion can be decomposed into two independent motions:

1. Uniform circular motion in a plane perpendicular to the magnetic field, caused by the velocity component perpendicular to the field ($v_{\perp}$).
2. Uniform linear motion along the direction of the magnetic field, caused by the velocity component parallel to the field ($v_{\parallel}$).

Given:

Velocity of proton, $v = 4 \times 10^5$ m/s Magnetic field, $B = 0.3$ T Angle to the magnetic field, $\theta = 60^\circ$

Constants for proton:

Mass of proton, $m_p = 1.67 \times 10^{-27}$ kg Charge of proton, $q_p = 1.6 \times 10^{-19}$ C

1. Calculate the components of velocity:

The velocity component parallel to the magnetic field is:

$v_{\parallel} = v \cos \theta = (4 \times 10^5 \text{ m/s}) \times \cos 60^\circ = (4 \times 10^5) \times 0.5 = 2 \times 10^5 \text{ m/s}$

The velocity component perpendicular to the magnetic field is:

$v_{\perp} = v \sin \theta = (4 \times 10^5 \text{ m/s}) \times \sin 60^\circ = (4 \times 10^5) \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \times 10^5 \text{ m/s}$

Using $\sqrt{3} \approx 1.732$, $v_{\perp} \approx 2 \times 1.732 \times 10^5 = 3.464 \times 10^5 \text{ m/s}$

2. Calculate the radius of the helical path ($r$):

The magnetic force provides the centripetal force for the circular motion:

$q_p v_{\perp} B = \frac{m_p v_{\perp}^2}{r}$

$r = \frac{m_p v_{\perp}}{q_p B}$

Substitute the values:

$r = \frac{(1.67 \times 10^{-27} \text{ kg}) \times (2\sqrt{3} \times 10^5 \text{ m/s})}{(1.6 \times 10^{-19} \text{ C}) \times (0.3 \text{ T})}$

$r = \frac{1.67 \times 2\sqrt{3} \times 10^{-22}}{0.48 \times 10^{-19}}$

$r = \frac{1.67 \times 2 \times 1.732}{0.48} \times 10^{-3}$

$r = \frac{5.78368}{0.48} \times 10^{-3}$

$r \approx 12.049 \times 10^{-3} \text{ m}$

$r \approx 0.01205 \text{ m}$

3. Calculate the time period of the circular motion ($T$):

The time period is given by:

$T = \frac{2\pi m_p}{q_p B}$

Substitute the values:

$T = \frac{2\pi \times 1.67 \times 10^{-27} \text{ kg}}{1.6 \times 10^{-19} \text{ C} \times 0.3 \text{ T}}$

$T = \frac{2\pi \times 1.67 \times 10^{-27}}{0.48 \times 10^{-19}}$

Using $\pi \approx 3.14159$:

$T = \frac{2 \times 3.14159 \times 1.67}{0.48} \times 10^{-8}$

$T = \frac{10.4933}{0.48} \times 10^{-8}$

$T \approx 21.861 \times 10^{-8} \text{ s}$

$T \approx 2.186 \times 10^{-7} \text{ s}$

4. Calculate the pitch of the helix ($p$):

The pitch is the distance traveled along the magnetic field direction in one time period:

$p = v_{\parallel} T$

Substitute the values:

$p = (2 \times 10^5 \text{ m/s}) \times (2.186 \times 10^{-7} \text{ s})$

$p = 4.372 \times 10^{-2} \text{ m}$

$p = 0.04372 \text{ m}$

The angular frequency mentioned in the question text before question 9 is $\omega = \frac{qB}{m}$.

$\omega = \frac{1.6 \times 10^{-19} \times 0.3}{1.67 \times 10^{-27}} = \frac{0.48 \times 10^{-19}}{1.67 \times 10^{-27}} = \frac{0.48}{1.67} \times 10^8 \approx 0.2874 \times 10^8 \text{ rad/s} = 2.874 \times 10^7 \text{ rad/s}$

Explanation of the solution:

1. Decompose the proton's velocity into components parallel ($v_{\parallel}$) and perpendicular ($v_{\perp}$) to the magnetic field.
2. Calculate the radius of the circular path using the formula $r = \frac{m v_{\perp}}{qB}$, where $m$ is the mass, $q$ is the charge, and $B$ is the magnetic field strength.
3. Calculate the time period of one revolution using $T = \frac{2\pi m}{qB}$.
4. Calculate the pitch of the helix using $p = v_{\parallel} T$, which is the distance traveled along the magnetic field direction during one revolution.