Magnetic Field - Torque on a Current Loop

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SUMMARY

The discussion focuses on calculating the current in a circular coil with 720 turns and an area of 0.6 m² placed in a uniform magnetic field of 0.63 T, where the maximum torque is 0.0026 N·m. The equation used is T = IABn, which allows for the determination of current in the coil. The second part of the problem involves recalculating the current when the coil is transformed into a single-turn coil with a larger area, where the relationship A2 = (n²)A1 is crucial for finding the new area and subsequently the current.

PREREQUISITES
  • Understanding of magnetic torque equations, specifically T = IABn
  • Knowledge of circular coil properties and magnetic fields
  • Ability to manipulate algebraic equations for problem-solving
  • Familiarity with the concept of turns in coils and their effect on magnetic properties
NEXT STEPS
  • Study the derivation and applications of the torque equation T = IABn
  • Explore the relationship between coil turns and magnetic field strength
  • Learn about the effects of area changes on torque in magnetic fields
  • Investigate practical applications of current loops in electromagnetic devices
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Students studying electromagnetism, physics educators, and anyone interested in the practical applications of magnetic fields and current loops in engineering and technology.

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Homework Statement



A circular coil of 720 turns and area 0.6 m^2 is in a uniform magnetic field of 0.63 T. The maximum torque exerted on the coil by the field is 0.0026 N · m.

a) Calculate the current in the coil.

b) Assume the 720 turns of wire are used to form a single-turn coil with the same shape but much larger area. What is the current if the maximum torque exerted on the coil by the field is 0.0026 N · m?

NOTE: Sigfigs do not matter

Homework Equations



T = IABn

The Attempt at a Solution



Part A:
I solved by substitution using the equation T = IABn, getting T = 9.5532e-6 N · m (answer is correct)

Part B:
This is the part I'm having trouble with. I thought I could fine the new Area doing the following (I'm sure this approach is wrong):

.6 = pi*r^2, r*720 = 314.7
A = pi*314.7^2, plugging in the new area into T = IAB gives the wrong answer.

I'd be very grateful for any help.
 
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I got 1.32644e-8 N-m
 
rl.bhat said:
I got 1.32644e-8 N-m
yeah, Thanks. I got the same answer or thereabouts.

After some algebra (and some help) I found that the new area A2 = (n^2)*A1.

Plugging that into T = IAB gives the correct answer.
 

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