Crossed Electric and Magnetic fields problem

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SUMMARY

The discussion centers on calculating the net force acting on a positive 1.8 µC charge moving at 3.2 x 106 m/s in crossed electric and magnetic fields. The electric field has a magnitude of 5.1 x 103 N/C, and the magnetic field has a magnitude of 1.2 x 10-3 T, both pointing in the same direction. The correct approach to find the net force involves recognizing that the electric force is parallel to the electric field while the magnetic force is perpendicular to both the magnetic field and the velocity of the charge. The net force must be calculated as a vector sum of these two forces.

PREREQUISITES
  • Understanding of electric fields and forces (Coulomb's Law)
  • Knowledge of magnetic fields and forces (Lorentz Force Law)
  • Familiarity with vector addition in physics
  • Basic concepts of charge and motion in electromagnetic fields
NEXT STEPS
  • Study the Lorentz Force Law in detail to understand the interaction of electric and magnetic fields.
  • Learn about vector addition techniques for forces in physics.
  • Explore the concept of electromagnetic waves and their properties.
  • Investigate practical applications of crossed electric and magnetic fields in devices like cathode ray tubes.
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Students and professionals in physics, electrical engineering, and anyone interested in understanding the dynamics of charged particles in electromagnetic fields.

futron
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"A magnetic field has a magnitude of 1.2 x 10-3 T, and an electric field has a magnitude of 5.1 x 10^3 N/C. Both fields point in the same direction. A positive 1.8 µC charge moves at a speed of 3.2 x 10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge."

Using the crossed magnetic and electric field equation, wouldn't the net force be F=qE+qvB? I resolved it to 0.01609N, which isn't the correct answer. Any ideas?

~Futron
 
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futron said:
"A magnetic field has a magnitude of 1.2 x 10-3 T, and an electric field has a magnitude of 5.1 x 10^3 N/C. Both fields point in the same direction. A positive 1.8 µC charge moves at a speed of 3.2 x 10^6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge."

Using the crossed magnetic and electric field equation, wouldn't the net force be F=qE+qvB? I resolved it to 0.01609N, which isn't the correct answer. Any ideas?

~Futron

Did you consider the direction of the forces? I did not check your answer, but that is a likely source of error.
 
Wouldn't Fnet be the sum if they are both pointed the same direction?

~Futron
 
futron said:
Wouldn't Fnet be the sum if they are both pointed the same direction?

~Futron

The fields are in the same direction. Electric force is parallel to the electric field. What about the magnetic force?
 
It is perpendicular relative to the magnetic field. How would I then calculate that into the Fnet equation? Thanks.

~Futron
 
futron said:
It is perpendicular relative to the magnetic field. How would I then calculate that into the Fnet equation? Thanks.

~Futron

Not only is it perpendicular to the magnetic field, it is perpendicular to the direction of the velocity of the charge. Of imporatance here is that since the electric force is parallel to the electric field, and the magnetic force is perpendicuar to that, the two forces involved are perpendicular. They must be added as vectors.
 
Got it. Thanks for your help!

~Futron
 

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