Calculating Deviation of a Bullet's Trajectory

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SUMMARY

The discussion focuses on calculating the deviation of a bullet's trajectory influenced by Earth's magnetic field. A bullet weighing 3.40g, traveling at a velocity of 160m/s, and possessing a charge of 13.5 x 10^-9 C, experiences a force due to the magnetic field of 5 x 10^-5 T. The radius of the bullet's circular motion is calculated using the formula R = mv/qB, resulting in a radius of 8.05 x 10^11 m. Despite this large radius, the bullet's deviation after traveling 1.00 km is minimal but can be quantified.

PREREQUISITES
  • Understanding of Lorentz force: F = q(v × B)
  • Knowledge of circular motion and radius of curvature
  • Familiarity with basic physics concepts such as mass, velocity, and charge
  • Ability to perform calculations involving scientific notation
NEXT STEPS
  • Research the derivation of the Lorentz force equation
  • Learn about the effects of magnetic fields on charged particles
  • Study the principles of circular motion in physics
  • Explore methods for calculating small deviations in projectile motion
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Physics students, educators, and anyone interested in the effects of magnetic fields on charged projectiles will benefit from this discussion.

davidbenari
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Homework Statement


A bullet (3.40g) has a velocity of 160m/s perpendicular to Earth's magnetic field 5x10^-5 T . The bullet has a charge of 13.5 x 10^-9 C.With what distance will it be deviated from its trajectory after it has traveled 1.00km?

Homework Equations


##\vec{F}=q\vec{v}\times\vec{B}##

The Attempt at a Solution


The way I visualise this problem. The bullet travels in circular motion.
It's radius of motion will be given by ##R=\frac{mv}{qB}=8.05\times 10^{11} m ##
With a radius this big, after it has traveled 1km it's almost perfectly on top of the line of its initial trajectory. Because this are my answers to the problem, I'm almost 100% sure I'm not doing it right.
 
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davidbenari said:
With a radius this big, after it has traveled 1km it's almost perfectly on top of the line of its initial trajectory.
It's certainly a small deviation, but it is not zero and can be calculated. The trick is to calculate it in such a way that you do not completely lose precision.
Please post an equation for the deviation.
 

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