Current in a rectangle on a hinge

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SUMMARY

The discussion focuses on calculating the torque acting on a rectangular coil of wire, consisting of 15 turns, measuring 10 cm by 5 cm, carrying a current of 0.90 A, and positioned at a 30° angle in a uniform magnetic field of 0.50 T. The torque is determined using the formula τ = μ × B, where μ = NiA. The calculated torque is 0.16875 Nm, but there is a need to reassess the angle due to the direction of the magnetic moment vector μ.

PREREQUISITES
  • Understanding of electromagnetic principles, specifically torque on current-carrying coils.
  • Familiarity with the formula for magnetic moment (μ = NiA).
  • Knowledge of trigonometric functions, particularly sine, in relation to angles in physics.
  • Ability to interpret vector directions in magnetic fields.
NEXT STEPS
  • Review the concept of torque on a current loop in a magnetic field.
  • Study the implications of angle adjustments in torque calculations.
  • Learn about the magnetic moment vector and its orientation in magnetic fields.
  • Explore the effects of varying current and magnetic field strength on torque.
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Students studying electromagnetism, physics educators, and anyone involved in solving problems related to torque in magnetic fields.

lodovico
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29_46.gif

Homework Statement


Figure 29-36 shows a rectangular, 15-turn coil of wire, 10 cm by 5.0 cm. It carries a current of 0.90 A and is hinged along one long side. It is mounted in the xy plane, at an angle of 30° to the direction of a uniform magnetic field of 0.50 T. Find the magnitude and direction of the torque acting on the coil about the hinge line.

Homework Equations



\tau=μ × B
μ=NiA

The Attempt at a Solution



I don't know how to approach this. I tried to plug in numbers into ^ formula
 
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If you'll show more details of how you tried the calculation, we will be able to see where you're having trouble.
 
τ=μ × B
μ=NiA

τ=NiABsin30
τ=(15)(.9)(.1*.05)(.5)sin30

τ=.16875 Nm
 
lodovico said:
τ=μ × B
μ=NiA

τ=NiABsin30
τ=(15)(.9)(.1*.05)(.5)sin30

τ=.16875 Nm

See if you can figure out why the 30o is not correct here. You'll need to think about the direction of \vec{\mu}
 

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