Discussion
What is the magnitude of the Poynting vector at the surface of a long cylindrical wire of radius R, length L, carrying current I, when its ends are kept at a potential difference V ?
Answer: D
The Poynting vector has magnitude \(\frac{VI}{2\pi RL}\).
Along the wire the electric field is \(E = V/L\), and at its surface the magnetic field is \(H = \frac{I}{2\pi R}\), at right angles to E. The Poynting vector \(\bar{S} = \bar{E} \times \bar{H}\) points radially into the wire, with magnitude
\[ \begin{aligned} S = EH &= \frac{V}{L}\cdot\frac{I}{2\pi R} \\ &= \frac{VI}{2\pi RL} \end{aligned} \]
Check: multiplying by the curved surface area \(2\pi RL\) gives the power \(VI\), exactly the Joule heat in the wire, so the energy flows in through the sides. The options with \(\pi R^{2}L\) or the end faces \(2\pi R^{2}\) divide by the wrong area, and it is not zero, because both fields are present.
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