Magnitude of electric and magnetic fields question. Help,

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SUMMARY

The discussion centers on calculating the speed of a charged ink drop (q = 3 × 10-9 C) moving in an electric field (E = 8 × 105 N/C) and a magnetic field (B = 1.3 T) such that the electric and magnetic forces are equal. The equation F = q × v × B is used to relate the forces, but the initial calculation of speed resulted in an incorrect value of 2.05 × 1014 m/s. The error was identified as a misunderstanding of the relationship between electric field strength and force, emphasizing that the electric field does not directly equate to force.

PREREQUISITES
  • Understanding of electric fields and magnetic fields
  • Familiarity with the equation F = q × v × B
  • Knowledge of force units (Newtons) and electric field units (N/C)
  • Basic algebra for solving equations
NEXT STEPS
  • Review the concept of electric force: FE = q × E
  • Study the relationship between electric and magnetic forces in charged particle motion
  • Learn about the Lorentz force law and its applications
  • Practice solving problems involving charged particles in electric and magnetic fields
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of charged particles in electromagnetic fields.

astru025
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Homework Statement



An ink drop with charge q = 3 ✕ 10-9 C is moving in a region containing both an electric field and a magnetic field. The strength of the electric field is 8E5 N/C and the strength of the magnetic field is 1.3 T. At what speed must the particle be moving perpendicular to the magnetic field so that the magnitudes of the electric and magnetic forces are equal?

Homework Equations


F= q x v x B



The Attempt at a Solution



Right now I know the magnitude of the electric field is 8E5 so I can take that and set it equal to my other numbers... 8 x 10^5 = 3E-9 C x 1.3 T x ( ? M/s)
Solving for m/s, I get 2.05E14 which turns out to be incorrect.
Not sure what i am doing wrong any help would be nice!
 
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##q \vec{v}\times\vec{B}## gives you the magnetic force on the charge. ##\vec{E}## is the electric field, which is not a force, so you can't compare them directly. What's the force on the charge due to the electric field?
 
Hint: What are the units on the left and right hand sides of your equation, and what should they be?
 
N/C= C x T x m/s. those are my units.. Which I now realize I have done something wrong. Any more help?!
 
At what speed must the particle be moving perpendicular to the magnetic field so that the magnitudes of the electric and magnetic forces are equal?

What's the unit of force?
 

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