Minimum force in a single turn square loop

[Cheezay]

Homework Statement

A single-turn square loop carries a current of 18.0 A. The loop is 17.1 cm on a side and has a mass of 0.0360 kg. Initially the loop lies flat on a horizontal tabletop. When a horizontal magnetic field is turned on, it is found that only one side of the loop experiences an upward force. Calculate the minimum magnetic field, Bmin, necessary to start tipping the loop up from the table.

Homework Equations

B=mg/2IL

The Attempt at a Solution

So basically i am using the above equation, and I'm dividing the weight of the loop by 2, because only half the loop is feeling force. So plugging in my numbers, i get .5435 T and it's incorrect, any suggestions? Thanks in advance!

 

[rl.bhat]

Initially the force is acting on one side of the loop. Other end is on the table. This force acts on the loop until the loop stands vertically on the table. Find the work done by this force on the loop. equate it to the potential energy of the loop. And find Bmin.

 

[nlightner]

I would like to clarify a few things before providing a response. Firstly, I would like to confirm that the loop is a perfect square and not a rectangle, as this would affect the calculations. Secondly, I would like to know the orientation of the magnetic field with respect to the loop. Is it parallel or perpendicular? This will also impact the calculations.

Assuming the loop is a perfect square and the magnetic field is perpendicular to the loop, your calculation of Bmin=0.5435 T is correct. However, if the magnetic field is parallel to the loop, the minimum force required would be zero as there would be no torque acting on the loop.

If the loop is a rectangle, the calculation would be slightly different and would depend on the orientation of the magnetic field. I would suggest double-checking your numbers and equations to ensure accuracy. If you are still getting an incorrect answer, it would be helpful to see your work and calculations to identify any errors.