Physics Question about Magnetic Force & parallel wires

[tnbstudent]

I have a homework question that I am having some trouble with. This is my first post - so be easy on me.

Question:
A weight-lifter is able to hold a barbell with a maximum total mass of 205 kg above his head. A physicist uses him to conduct a classroom demonstration, asking him to hold one insulated conducting rod (without any weights on it) above his head, while a second, very long rod is placed under his feet, parallel to the first rod. The rod that he is "lifting" has a length of 2.00 m and, together with its electrical leads, a mass of 10.0 kg. The rods are separated by a distance of 2.15 m. What is the maximum current that can be passed through the rods, in the same direction, before he is forced to yield? The same amount of current flows through both rods.

I started by using the basic force equation: F=m*a and then the formula for Force of parallel wires: F=μI1I2/2∏r

My answer:
I tried to set the force of what the lifter could lift equal to the force of the wire.
I got 2009N as the force he could lift (2009N=205kg*9.8m/s^2)
Then I found the force of the wire to be 98N (98N=10kg*9.8m/s^2)
Then I subtracted the 98 from 2009 to get 1911N as the force created by the current in the wire
Since I1 and I2 are the same I used 2I and my equation looks like:
1911N = (4∏X10^-7)*(2I)*(2.0m)/(2∏*2.15m)
When I solve I get I=5.14 e ^9

 

[gneill]

I have a homework question that I am having some trouble with. This is my first post - so be easy on me.

Question:
A weight-lifter is able to hold a barbell with a maximum total mass of 205 kg above his head. A physicist uses him to conduct a classroom demonstration, asking him to hold one insulated conducting rod (without any weights on it) above his head, while a second, very long rod is placed under his feet, parallel to the first rod. The rod that he is "lifting" has a length of 2.00 m and, together with its electrical leads, a mass of 10.0 kg. The rods are separated by a distance of 2.15 m. What is the maximum current that can be passed through the rods, in the same direction, before he is forced to yield? The same amount of current flows through both rods.

I started by using the basic force equation: F=m*a and then the formula for Force of parallel wires: F=μI1I2/2∏r

My answer:
I tried to set the force of what the lifter could lift equal to the force of the wire.
I got 2009N as the force he could lift (2009N=205kg*9.8m/s^2)
Then I found the force of the wire to be 98N (98N=10kg*9.8m/s^2)
Then I subtracted the 98 from 2009 to get 1911N as the force created by the current in the wire
Since I1 and I2 are the same I used 2I and my equation looks like:
1911N = (4∏X10^-7)*(2I)*(2.0m)/(2∏*2.15m)
When I solve I get I=5.14 e ^9

Your expression: μI1I2/2∏r gives the force per unit length of the parallel wires, not the total force. Check the units --- they should work out to N/m.

Also, since when is I*I = 2*I?

 

[tnbstudent]

Thanks - I had a typo. I actually have my formula as
F=μ*I1*I2*l/2∏r where l is the length of the wire
Once, I clear up my brain lapse and use I^2 instead of 2I I get 101,349A.
I only have one try at this one, so I'd like to make sure this is right before I submit it

Thanks again for the help

 

[gneill]

Thanks - I had a typo. I actually have my formula as
F=μ*I1*I2*l/2∏r where l is the length of the wire
Once, I clear up my brain lapse and use I^2 instead of 2I I get 101,349A.
I only have one try at this one, so I'd like to make sure this is right before I submit it

Thanks again for the help

That result looks good. You might want to consider expressing your answer using the appropriate significant figures.