A circular current loop of radius 2.0 cm carries a current of 2.0 mA. (a) What is the magnitude of its magnetic dipole moment? (b) If the dipole is oriented at 30 degrees to a uniform magnetic field of magnitude 0.50 T, what is the magnitude of the torque it experiences and what is its potential energy?
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0 mA = 2.0 x 10^-3 A, radius = 2.0 cm = 0.02 m magnetic dipole moment = 2.0 x 10^-3 A * π * (0.02 m)^2 magnetic dipole moment = 2.5 x 10^-6 A m^2 Show more…
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