Question

Choose the correct answer from the alternatives given.
An infinitely long thin wire carrying a uniform linear static charge density $\lambda$ is placed along the z-axis. The wire is set into motion along its length with a uniform velocity $\overrightarrow{v}=v\hat{k}z$.Find the poynting vector.

• A. $\frac{-{\lambda }^{2}v}{4{\pi }^2{\epsilon }_0{a}^2}\hat{k}$
• B. $\frac{{\lambda }^{2}v}{4\pi {\epsilon }_0{a}^2}\hat{i}$
• C. $\frac{{\lambda }^{2}v}{5\pi {\epsilon }_0{a}^2}\hat{k}$
• D. $\frac{-{\lambda }^{2}v}{3\pi {\epsilon }_0{a}^2}\hat{k}$

Answer 1

The correct option is A $\frac{-{\lambda }^{2}v}{4{\pi }^2{\epsilon }_0{a}^2}\hat{k}$

Electric field due to the infinite long wire, $\overrightarrow{E}=\frac{\lambda }{2\pi {ϵ}_{0}a}\hat{j}$

Magnetic field due to the motion of the wire, $\overrightarrow{B}=\frac{{\mu }_{0}i}{2\pi a}\hat{i}=\frac{{\mu }_0\lambda v}{2\pi a}\hat{i}$

Poynting vector, $\overrightarrow{S}=\frac{1}{{\mu }_0}(\overrightarrow{E}\times \overrightarrow{B})=\frac{1}{{\mu }_0}(\frac{\lambda }{2\pi {ϵ}_{0}a}\hat{j}\times \frac{{\mu }_0\lambda v}{2\pi a}\hat{i})=\frac{-{\lambda }^{2}v}{4{\pi }^2{ϵ}_0{a}^2}\hat{k}$

Hence option $\left(A\right)$ is correct.