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A single-turn square loop carries a current of 18 A. The loop is 14 cm on...

A single-turn square loop carries a current of 18 A. The loop is 14 cm on a side and has a mass of 3.7×10−2 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.

Find the minimum magnetic field, Bmin, necessary to start tipping the loop up from the table. Express your answer using two significant figures.

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Answer #1

Given

I = 18A

L = 0.14 m

m = 0.037 kg

Only one side of the loop experiences an upward force, so tipping the loop, not raising the whole thing

F = (m/2)*g.

F = B*I*L

B = F/(I*L)

B = 0.0185*9.81/(18*0.14)

B = 0.072 T

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