A long cylindrical conductor of radius a = 2.00 cm and length L = 2.00 m carries

Question

A long cylindrical conductor of radius a = 2.00 cm and length L = 2.00 m carries a current I = 2.00 A uniformly over its cross-sectional area. The conductor is made of a material with resistivity 90.0 nano-Ohm-m. A long cylindrical conductor of radius a = 2.00 cm and length L = 2.00 m carries a current I = 2.00 A uniformly over its cross-sectional area. The conductor is made of a material with resistivity 90.0 nano-Ohm-m. What is the magnitude of the Poynting vector at the surface of the conductor? What is the direction of the Poynting vector at the surface of the cylinder? What is the total power delivered by the Poynting vector to the cylinder? I have the equation S=1/mu0 ExB but can't figure out how to apply it using the given information.

Explanation / Answer

The radius of the cylindrical conductor is a = 2.00 cm = 2.00 * 10^-2 m The length of the cylindrical conductor is L = 2.00 m The conductor carries a current of I = 2.00 A The conductor is made of a material with resistivity ? = 90.0 nano-Ohm-m = 90.0 * 10^-9 Ohm-m The resistance of the conductor is R = (?L/A) where A = pa2 From Ohm's law,we have V = I * R or E * a = I * R or E = (I * R/a) = (I * ?L/pa3) The magnetic field around the conductor is B = (µoI/2pa) The magnitude of the Poynting vector at the surface of the conductor is S = (1/µo) * (E X B) = (1/µo) * E * B * sin? or S = (1/µo) * (I * ?L/pa3) * (µoI/2pa) * sin? = (I^2 * ?L/2p^2a^4) * sin? where ? = 90o The direction of the poynting vetor is perpendicular to the surface of the cylinder. The total power delivered by the Poynting vector to the cylinder is P = S * 2paL or P = (I^2 * ?L/2p^2a^4) * sin? * 2paL = (I^2 * ?L^2/pa^3) * sin? From Ohm's law,we have V = I * R or E * a = I * R or E = (I * R/a) = (I * ?L/pa3) The magnetic field around the conductor is B = (µoI/2pa) The magnitude of the Poynting vector at the surface of the conductor is S = (1/µo) * (E X B) = (1/µo) * E * B * sin? or S = (1/µo) * (I * ?L/pa3) * (µoI/2pa) * sin? = (I^2 * ?L/2p^2a^4) * sin? where ? = 90o The direction of the poynting vetor is perpendicular to the surface of the cylinder. The total power delivered by the Poynting vector to the cylinder is P = S * 2paL or P = (I^2 * ?L/2p^2a^4) * sin? * 2paL = (I^2 * ?L^2/pa^3) * sin?

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