Question

A proton, an electron, a neutron, and an alpha particle are entering a region of uniform magnetic field with same velocities. The tracks of these particles are labelled.

1. Identify the tracks of each particle.

2. Write the expression for the force experienced by a charged particle in the magnetic field in vector form.

3. If the proton is moving at 90° to the uniform magnetic field what will be the change in kinetic energy of the proton? Give reason.

4. An electron with energy 1 keV is entering a uniform magnetic field of 0.04 T at an angle 60° with the field. Predict the path of the electron and find the characteristics of the path.

Answer 1

1. Identify the tracks of each particle:

• Path 1 – proton
• Path 2 – alpha particle
• Path 3 – neutron
• Path 4 – electron

2. 

\(\overrightarrow F=q(\overrightarrow V\times \overrightarrow B)\)

3. Zero. Since the force is perpendicular to the direction of velocity work done is zero.

4. KE = 1 keV = 1 × 10³ × 1.6 × 10^-19 = 1.6 × 10^-16 

\(\frac{1}{2}mv^2\) = 1.6 x 10^-16 J

v² = \(\frac{1}{m}\times1.6\times10^{-16}\times2 \)  = \(\frac{3.2\times10^{-16}}{9.1\times10^{-31}}\)

= 0.352 x 10¹⁵ = 3.52 x 10¹⁴

V = \( \sqrt{3.52\times10^{14}}\) = 1.88 x 10⁷ ms^-1

Radius of the path = \(\frac{mvsinθ}{qB}\)

= \(\frac{9.1\times10^{-31}\times1.88\times10^7\times\sqrt{\frac{3}{2}}}{1.6\times10^{-19}\times 0.04}\) = 2.31 x 10^-3m

Pitch = \(\frac{2\times\pi\times mvcos θ}{qB}\)

 = \(\frac{2\times\pi\times9.1\times10^{-31}\times1.88\times10^7\times0.5}{1.6\times10^{19}\times0.04}\)

= 8.39 × 10^-3