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A straight wire of circular cross-section carries a direct current I, as shown in figure below. If R is the resistance per unit length of the wire, then the Poynting vector at the surface of the wire will be—
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RI2 → n 2∏r -
RI2 → (−n) 2∏r -
RI2 → n 2∏ -
RI2 → (−n) 2∏
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Correct Option: B
R is the resistance per unit length and I is the current flowing. Therefore, power loss occurring in the conductor per unit length is I2R. This power is furnished on E and H fields at the surface of the wire. According to Poynting theorem E × H is the power flow per unit area. Since the power is fed from outside through the cylindrical surface of the conductor, E × H at every point on the surface is radial and directed into the surface. Therefore, Poynting vector is
| = | (−n) | ||
| surface area | 2∏r × 1 |
where n is the unit radial vector directed outward.
