Linear acceleration in a uniform magnetic field due to torque

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SUMMARY

The discussion focuses on calculating the maximum linear acceleration of a square loop in a uniform magnetic field, specifically when the loop is hinged along one side. The loop has a perimeter of 4L, a mass per unit length of 0.1000 kg/m, and carries a current of 5.000 A in a magnetic field of 10.00 mT. The key equations involved are torque (τ = Iα) and the relationship between linear acceleration (a) and angular acceleration (α) (a = rα). The maximum torque occurs when the angle between the magnetic field and the current is optimal.

PREREQUISITES
  • Understanding of torque and angular acceleration in physics
  • Familiarity with the concept of moment of inertia
  • Knowledge of magnetic fields and their effects on current-carrying conductors
  • Basic principles of rotational dynamics
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  • Calculate the moment of inertia for a square loop about the hinge axis
  • Explore the relationship between torque and angular acceleration in rotational systems
  • Investigate the conditions for maximum torque in magnetic fields
  • Learn about the effects of varying current and magnetic field strength on linear acceleration
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Physics students, educators, and anyone interested in understanding the dynamics of current-carrying loops in magnetic fields.

astralboy15
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Here the extra credit question I'm stuck on:


A square loop (perimeter of 4L and hinged along one side) is made of a wire that has a mass per unit length of 0.1000 kg/m and carries a current of 5.000 A. A uniform magnetic field of 10.00 mT directed perpendicular to the hinged side exists in the region. Determine the maximum linear acceleration magnitude, (a t) max , of the side of the loop opposite the hinged side.


My Question:
My teacher gave me a hint that torque = I*alpha, where I is the moment of inertia. How do I find the moment of inertia for this box? Then how do I work with angular acceleration to get the answer I'm after?

Thanks!
 
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What is the axis of rotation of the loop.

To find out I of the loop about this axis, you will need to find out the I's for each side of the loop about the axis and sum them.

[tex]\tau = I \alpha[/tex]

[tex]a = r \alpha[/tex]

You can see that a depends upon the torque and it is max when the torque is max.

When is the torque max?
 

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