How Is the Potential Energy of a Bobbin Calculated in a Magnetic Field?

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SUMMARY

The potential energy of a rectangular bobbin with 50 turns, dimensions of 5.00 cm by 8.00 cm, and carrying a current of 1.75 A in a uniform magnetic field of 1.5 T is calculated using the formula dU = -μB sin(θ) dθ. The magnetic moment (μ) is determined to be 0.35 Am². When the bobbin is oriented at an angle of 37º to the magnetic field, the calculated potential energy is -0.315 J, indicating that the energy decreases as the angle increases from zero.

PREREQUISITES
  • Understanding of magnetic fields and forces
  • Familiarity with the concept of magnetic moment
  • Knowledge of calculus for integration
  • Basic principles of electromagnetism
NEXT STEPS
  • Study the derivation of the potential energy formula in magnetic fields
  • Learn about the effects of varying angles on magnetic potential energy
  • Explore the applications of magnetic moments in different geometries
  • Investigate the relationship between current, turns, and magnetic field strength
USEFUL FOR

Physics students, electrical engineers, and anyone studying electromagnetism and magnetic field interactions with current-carrying conductors.

txoricillo
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Homework Statement


A rectangular bobbin of 50 turns is got sides of 5.00 and 8.00 cm and carries a current of 1.75 A. I'ts orientated as shown in figure 37º and turns around z axis. Determine the potential energy of the bobbin when its inside a uniform magnetic field of 1.5T in the y-axis (potential energy is zero when angle is zero)

Homework Equations


dU=-muBsinthetadtheta
mu=NIA=0.35 Am2
B=1.5T
theta= angle between magneticfield and normal vector to the bobbin

The Attempt at a Solution


Integrate dU. Solution -0.315 J
 
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txoricillo said:

The Attempt at a Solution


Integrate dU. Solution -0.315 J
Can you show your work? And what exactly is your question?
 

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