The torque expression of a current carrying conductor is(a) T = BIA cos θ(b) T = BA cos θ(c) T = BIA sin θ(d) T = BA sin θThe question was posed to me in unit test.Query is from Magnetization in section Magnetic Forces and Materials of Electromagnetic Theory

The torque expression of a current carrying conductor is(a) T = BIA co

1.	The torque expression of a current carrying conductor is(a) T = BIA cos θ(b) T = BA cos θ(c) T = BIA sin θ(d) T = BA sin θThe question was posed to me in unit test.Query is from Magnetization in section Magnetic Forces and Materials of Electromagnetic Theory
Answer» Right ANSWER is (c) T = BIA sin θ Easy explanation: The torque is given by the product of the flux density, magnetic MOMENT IA and the sine ANGLE of the conductor held by the FIELD. This gives T = BIA sin θ.

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The torque expression of a current carrying conductor is(a) T = BIA cos θ(b) T = BA cos θ(c) T = BIA sin θ(d) T = BA sin θThe question was posed to me in unit test.Query is from Magnetization in section Magnetic Forces and Materials of Electromagnetic Theory

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