A square loop of wire of side 3.80 cm carries 2.10 A of current. A uniform magnetic field of magnitude 0.730 T makes an angle of 37.0° with the plane of the loop. What is the magnitude of the torque on the loop?
Added by Garrett F.
Step 1
The area (A) of a square is given by the formula A = side^2. So, A = (3.80 cm)^2 = 14.44 cm^2. But we need the area in m^2, so we convert cm^2 to m^2 by multiplying by 10^-4. So, A = 14.44 * 10^-4 m^2 = 0.001444 m^2. Show more…
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