The problem is asking to determine the momentum of a proton in a uniform magnetic field. Let's break it down step by step.

Given:

• Magnetic induction $B = 20 \, \text{mT} = 20 \times 10^{-3} \, \text{T}$
• Radius of the proton's circular path $r = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m}$
• Charge of the proton $q = 1.6 \times 10^{-19} \, \text{C}$

Formula:

In a uniform magnetic field, the momentum $p$ of a charged particle moving in a circular path can be found using the relationship between magnetic field, charge, velocity, and radius:

$p = qBr$

Substitution:

Substitute the given values into the formula:

$p = (1.6 \times 10^{-19} \, \text{C}) \times (20 \times 10^{-3} \, \text{T}) \times (5 \times 10^{-2} \, \text{m})$

$p = 1.6 \times 10^{-19} \times 20 \times 10^{-3} \times 5 \times 10^{-2}$

$p = 1.6 \times 10^{-19} \times 10^{-3}$

$p = 1.6 \times 10^{-22} \, \text{kg} \cdot \text{m/s}$

Answer:

The momentum of the proton is $p = 1.6 \times 10^{-22} \, \text{kg} \cdot \text{m/s}$.

Would you like more details on this solution, or have any questions?

Related questions:

1. What is the relationship between force and momentum in a magnetic field?
2. How would the momentum change if the magnetic field strength increased?
3. How does the radius of the circular path affect the proton’s speed?
4. What is the velocity of the proton in this magnetic field?
5. How does the direction of the magnetic field affect the proton's motion?

Tip:

In problems involving charged particles in magnetic fields, always check if the particle's mass is needed for calculations like velocity but not directly for momentum.