Math Problem Statement

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.

Solution

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=qvBsin(θ)F = q v B \sin(\theta)

where:

  • qq is the charge of the proton (1.6×1019C1.6 \times 10^{-19} \, \text{C}),
  • vv is the velocity of the proton,
  • BB is the magnetic field strength,
  • θ\theta is the angle between the velocity vector and the magnetic field direction.

Given Data

  • v=459005m/sv = 459005 \, \text{m/s}
  • B=1.3TB = 1.3 \, \text{T}
  • θ=40\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 [0,2π][0, 2\pi] radians. Since 40 radians is far beyond this, let's proceed assuming you meant 4040^\circ for practical accuracy. Thus:

  1. Convert the angle to radians if needed (assuming it's 40 degrees): θ=40=40×π1800.698radians\theta = 40^\circ = \frac{40 \times \pi}{180} \approx 0.698 \, \text{radians}

  2. Calculate the force: F=(1.6×1019)×(459005)×(1.3)×sin(0.698)F = (1.6 \times 10^{-19}) \times (459005) \times (1.3) \times \sin(0.698)

Let’s compute this explicitly:

F1.6×1019×459005×1.3×0.6428F \approx 1.6 \times 10^{-19} \times 459005 \times 1.3 \times 0.6428 F6.131×1014NF \approx 6.131 \times 10^{-14} \, \text{N}

Answer

The force on a proton in the given conditions is approximately: F6.131×1014NF \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.

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Math Problem Analysis

Mathematical Concepts

Magnetic Force
Vector Calculations
Trigonometry

Formulas

F = q v B sin(θ)

Theorems

Right-Hand Rule for Magnetic Force

Suitable Grade Level

Grades 11-12