# The magnitude and direction of the minimum magnetic field

## Homework Statement:

A .76 meter long wire runs horizontally and carries a current of 28 A from left to right. What would be the magnitude and direction of the minimum magnetic field to suspend the wire in mid air if the wire has a mass of 46.6 g/m

## Relevant Equations:

F=ILB
F=ILB
F=(48)(.76)B
F/36.48=B
I am stuck at how to find F, is there a different formula I am missing?

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kuruman
Homework Helper
Gold Member
How about a different force? If the current were turned off, would the wire still be suspended in mid air?

• themountain
Ok, so F=ma so F=(46.6)(-9.8) so the force is 456.68. Using my past calculations, 456.68/36.48=B so B=12.52 T and the direction is perpendicular to the wire, so direction is 90 degrees. Is this correct? Thanks!

kuruman
Homework Helper
Gold Member
Ok, so F=ma so F=(46.6)(-9.8) so the force is 456.68. Using my past calculations, 456.68/36.48=B so B=12.52 T and the direction is perpendicular to the wire, so direction is 90 degrees. Is this correct? Thanks!
Not correct.
1. The mass 46.6 is incorrect. It has no units and it's the wrong number.
2. The direction of the B field is ambiguous, 90 degrees relative to what? Use the directions of gravity and the current to specify the direction of the field or, even better, use unit vectors.

I would recommend getting an expression in terms of symbols and then put in the numbers at the very end.

• themountain
46.6 x .76 =35.416g
so F=(35.416)(-9.8) which means F=347.07
347.07/36.48= 9.514 T and it is towards bottom of the page

kuruman