Magnetic Force on Electric Currents

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SUMMARY

The discussion centers on calculating the electric current in a wire subjected to a magnetic field. Using the formula F = ILB, where F is the magnetic force (0.810 N), L is the length of the wire (2.00 m), and B is the magnetic field strength (0.030 T), the correct calculation yields a current (I) of 0.074 A. The orientation of the wire relative to the magnetic field is crucial, as the formula assumes a 90-degree angle for accurate results. Clarity in algebraic representation is emphasized to avoid confusion in problem-solving.

PREREQUISITES
  • Understanding of the formula F = ILB for magnetic forces
  • Knowledge of basic physics concepts related to electric currents and magnetic fields
  • Familiarity with unit conversions in physics (e.g., Newtons, Teslas)
  • Ability to interpret and solve algebraic equations
NEXT STEPS
  • Study the implications of wire orientation in magnetic fields
  • Learn about the Lorentz force law and its applications
  • Explore the effects of varying magnetic field strengths on current-carrying wires
  • Investigate the relationship between current, voltage, and resistance in circuits
USEFUL FOR

Students and educators in physics, electrical engineers, and anyone interested in understanding the principles of electromagnetism and electric currents in magnetic fields.

Mikala
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1. A current in a wire which has a length of 2.00m in a magnetic Field of 0.030 Teslas, feels a magnetic force of 0.810 Newtons. What is the current in the wire?2. F=IlB, Find I3. F= 0.810, l= 2.00m, B= 0.030 T, I=? (A)
I= l x B I= 2.00m x 0.030 T I= 0.074 A
F 0.810
Not really sure if the setup is correct.
 
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The question is poorly worded ... it should be more that 2m of a wire carrying a current passes through a 0.030T uniform magnetic field and experiences a force of 0.810N... what is the current? (Missing: orientation of the wire to the field.)

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html
You appear to have started with the correct approach: F=ILB
You know F, L, and B: so solve for I and plug the numbers in.
(assuming the current is oriented 90deg to the field.)

I don't know why you wrote I=lxBI or how you got I=0.074A
Because of the way you have written your algebra, your thinking is unclear: this would cost you marks.
 

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