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A square coil of side 0.10 m has 50 turns and carries a current of…
Question
A square coil of side 0.10 m has 50 turns and carries a current of 2.0 A. It is placed in a uniform magnetic field of 0.50 T with the plane of the coil parallel to the field. Find the magnitude of the torque (in N*m) acting on the coil.
✓ Verified answer: 0.5checked by our engine — not a guess
Step-by-step solution
Torque tau = N I A B sin(phi), where phi is the angle between the magnetic moment and the field.
With the coil plane parallel to B, the moment is perpendicular to B, so phi = 90 deg and sin(phi) = 1.
A = (0.10)^2 = 0.010 m^2.
tau = 50 x 2.0 x 0.010 x 0.50 = 0.50 N*m.
Final answer0.5
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