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

AI Thread Summary
To calculate the potential energy of a rectangular bobbin in a magnetic field, the formula dU = -μB sin(θ) dθ is used, where μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between the magnetic field and the bobbin's normal vector. For a bobbin with 50 turns, sides of 5.00 cm and 8.00 cm, and a current of 1.75 A, the magnetic moment (μ) is calculated as 0.35 Am². Given a uniform magnetic field of 1.5 T along the y-axis, the potential energy is determined by integrating dU, resulting in a value of -0.315 J. The potential energy is defined as zero when the angle is zero. This calculation is crucial for understanding the interaction between magnetic fields and 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?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

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