Question about magnetic fields

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SUMMARY

The discussion centers on calculating the required current in a conductor suspended by flexible wires to achieve zero tension in a magnetic field of 2.60 T. The mass per unit length of the conductor is 0.0800 kg/m. Using the formula for magnetic force (F = IL x B) and equating it to the weight of the conductor (W = mg), the necessary current is determined to be 302 mA. This calculation ensures that the magnetic force counteracts the weight of the conductor, resulting in no tension in the supporting wires.

PREREQUISITES
  • Understanding of magnetic force calculations using F = IL x B
  • Knowledge of weight calculations using W = mg
  • Familiarity with the concept of mass per unit length
  • Basic grasp of the right-hand rule for current direction in magnetic fields
NEXT STEPS
  • Study the application of the right-hand rule in electromagnetic contexts
  • Learn about the principles of magnetic force on current-carrying conductors
  • Explore the effects of varying magnetic field strengths on current requirements
  • Investigate the relationship between mass per unit length and tension in flexible supports
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This discussion is beneficial for physics students, electrical engineers, and anyone interested in the practical applications of electromagnetism in suspended systems.

andrew410
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A conductor suspended by two flexible wires as in the figure below has a mass per unit length of 0.0800 kg/m.

What current must exist in the conductor in order for the tension in the supporting wires to be zero when the magnetic field is 2.60 T into the page?

Figure: http://east.ilrn.com/books/sepsp06t/pse6e.29.14p.e.jpg

I know you have to use the formula F=IL x B to get the current but I don't know how to put it all together...
I think the force is the tension so you can make that zero. I don't know how to get the length because it only gives you the mass per unit length.
Maybe I'm doing this wrong...I don't know...
Please help me! Thanks in advance! :)
 
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The magnetic force (F) per unit length (L) is:
F/L = BI

It's weight (W) per unit length is:
W/L = mg/L
= (m/L)*g
= 0.08*9.8 N/m
= 0.784 N/m

To have zero tension in the supporting wires, the conductor's magnetic force must cancel out it's weight (i.e. W=F)

BI = 0.784 N
-> I = 0.784/2.6 A
= 302 mA (current must go from left to right, using the right-hand rule)
 
thx for the help and the quick response! :)
 

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