Discussion
A long straight wire, having radius \(a\) and resistance per unit length \(r\), carries a current \(I\). The magnitude and direction of the Poynting vector on the surface of the wire is
Answer: A
The Poynting vector has magnitude \(I^{2}r/2\pi a\), perpendicular to the axis and pointing inwards.
At the surface, the electric field driving the current is along the wire, \(E = Ir\) (the voltage drop per unit length), and the magnetic field circles the wire with \(H = I/(2\pi a)\). They are perpendicular, so
\[ S = EH = Ir \cdot \frac{I}{2\pi a} = \frac{I^{2}r}{2\pi a} \]
With \(\vec{E}\) along the current and \(\vec{H}\) circling it by the right-hand rule, \(\vec{E} \times \vec{H}\) points radially inward. This is the energy flowing in from the field to be dissipated as heat: multiplying by the surface area per unit length, \(2\pi a\), gives \(I^{2}r\), the Joule heating per unit length. An outward flux would have the wire supplying energy, and \(I^{2}r/\pi a\) would deliver twice the heat produced.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development