I=P/A=1/2CE_0^2=.5*C/U_0*B? I understand that the current must be rotating clock

Question

I=P/A=1/2CE_0^2=.5*C/U_0*B?

I understand that the current must be rotating clockwise.
I=? and what are the units?

As shown in the figure, a square loop of wire, with a mass of 400 g, is free to pivot about a horizontal axis through one of its sides. A 0.40 T horizontal magnetic field is directed as shown. What current I in the loop is needed to hold the loop steady in a horizontal plane? I=P/A=1/2CE_0^2=.5*C/U_0*B? I understand that the current must be rotating clockwise. I=? and what are the units?

Explanation / Answer

a = 0.10 m, B = 0.4 T, mass for each side = m = 400/4 g = 100 g = 0.100 kg
find current I
magnetic force on the rightmost side = F = IBa
magnetic torque = F*a = IBa2

torque of gravity = mg*a/2 + mg*a + mg*a/2 = 2mga

so IBa2 = 2mga

I = 2mg/(Ba) = 49 A

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