A beam of protons moves at 459005 m/s through a uniform 1.3 T-magnetic field directed along the positive z-axis. The velocity of each proton lies in the xz-planes and is directed at 40 rad to the +z-axis. Find the force on a proton.

To find the force on a proton moving in a magnetic field, we can use the following formula for the magnetic force on a charged particle:

\[
F = q v B \sin(\theta)
\]

where:
- \( q \) is the charge of the proton (\( 1.6 \times 10^{-19} \, \text{C} \)),
- \( v \) is the velocity of the proton,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field direction.

### Given Data
- \( v = 459005 \, \text{m/s} \)
- \( B = 1.3 \, \text{T} \)
- \( \theta = 40^\circ \) (we will use this angle in radians directly, as it's provided as 40 radians).

**Note:** However, an angle in the context of a physical system is typically expressed in degrees or radians within the range \([0, 2\pi]\) radians. Since 40 radians is far beyond this, let's proceed assuming you meant \(40^\circ\) for practical accuracy. Thus:

1. **Convert the angle to radians if needed** (assuming it's 40 degrees):
   \[
   \theta = 40^\circ = \frac{40 \times \pi}{180} \approx 0.698 \, \text{radians}
   \]

2. **Calculate the force:**
   \[
   F = (1.6 \times 10^{-19}) \times (459005) \times (1.3) \times \sin(0.698)
   \]

Let’s compute this explicitly:

\[
F \approx 1.6 \times 10^{-19} \times 459005 \times 1.3 \times 0.6428
\]
\[
F \approx 6.131 \times 10^{-14} \, \text{N}
\]

### Answer
The force on a proton in the given conditions is approximately:
\[
F \approx 6.131 \times 10^{-14} \, \text{N}
\]

Would you like more details on any part of this calculation?

---

1. What happens to the force if the angle changes?
2. How would the force change if the magnetic field strength increased?
3. What if the proton’s velocity was doubled?
4. How is the direction of the force determined?
5. What role does the proton’s charge play in this formula?

**Tip:** Remember, the force is always perpendicular to both the magnetic field and the velocity direction, following the right-hand rule.

Learn how to calculate the magnetic force on a proton moving through a magnetic field. This example involves using the formula F = q v B sin(θ) and provides a detailed step-by-step solution for finding the force with given velocity, magnetic field strength, and angle of motion. Suitable for high school students in Grades 11-12.

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