Question

Consider current I passing through a resistor of radius r, length L, and resistance R. Show that the flux of the Poynting vector (i.e., the integral of S→⋅dA→\overrightarrow{S}\cdot d\overrightarrow{A}) over the surface of the resistor is I2R. Then give an interpretation of this result.

Accepted Answer

Step 1: Recall the Poynting vector, $$ \overrightarrow{S} $$, which represents the power per unit area flowing through a surface due to electromagnetic fields. It is given by $$ \overrightarrow{S} = \frac{1}{\mu_0} \overrightarrow{E} 	imes \overrightarrow{B} $$, where $$ \overrightarrow{E}  is $$the electric field and $$ \overrightarrow{B}  is $$the magnetic field. Step 2: For a resistor, the electric field $$ \overrightarrow{E} $$ inside the resistor is related to the current density $$ \overrightarrow{J}  by $$Ohm's law: $$ \overrightarrow{E} = ho \overrightarrow{J} $$, where $$ ho  is $$the resistivity of the material. The current density $$ \overrightarrow{J}  is $$given by $$ \overrightarrow{J} = \frac{I}{A} $$, where $$ A = \pi r^2  is $$the cross-sectional area of the resistor. Step 3: The magnetic field $$ \overrightarrow{B} $$ around the resistor can be determined using Ampère's law: $$ \oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_0 I . $$For a cylindrical resistor, the magnetic field at a distance $$ r $$ from the axis is $$ B = \frac{\mu_0 I}{2 \pi r} . $$Step 4: The Poynting vector $$ \overrightarrow{S}  at $$the surface of the resistor is then calculated as $$ \overrightarrow{S} = \frac{1}{\mu_0} \overrightarrow{E} 	imes \overrightarrow{B} . $$Substituting $$ \overrightarrow{E} $$ and $$ \overrightarrow{B} $$, and noting that the cross product is perpendicular to the surface, the magnitude of $$ \overrightarrow{S}  is$$ $$ S = \frac{E B}{\mu_0} . $$Step 5: To find the total power flow, integrate $$ \overrightarrow{S} \cdot d\overrightarrow{A} $$ over the cylindrical surface of the resistor. This yields $$ P = \int \overrightarrow{S} \cdot d\overrightarrow{A} = I^2 R $$, which matches the power dissipated in the resistor due to Joule heating. This result shows that the electromagnetic energy flow into the resistor is entirely converted into thermal energy, consistent with energy conservation.

Consider current I passing through a resistor of radius r, length L, and resistance R. Show that the flux of the Poynting vector (i.e., the integral of $\vec{S} \cdot d \vec{A}$ ) over the surface of the resistor is I²R. Then give an interpretation of this result.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Poynting Vector

Resistance and Ohm's Law

Surface Integral