Minimum force in a single turn square loop

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

B=mg/2IL

3. 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

Homework Helper
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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.
 

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