Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 34 - Electromagnetic Fields and Waves - Exercises and Problems - Page 1030: 33

Answer

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Work Step by Step

We need to use Galilean fields transformation equations, $$\left.\begin{matrix} &\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A \\ & \vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}\\ \end{matrix}\right\}\tag 1$$ Assuming that B is the mosquito's frame. We need the magnetic field through this frame to be zero, so $\vec B_B=0$. Now we need to find $B_A$ and $E_A$, so we can use the second formula from (1) to find the mosquito's velocity. Recalling that the magnetic field for a very long curing carrying wire is given by $$\vec B_A=\dfrac{\mu_0I}{2\pi r}\hat r$$ And since the current is moving to the right, the direction of the magnetic field above the wire is out of the page. See the left figure below (view from the left side). So above the wire, the magnetic field is $$\vec B_A=\dfrac{\mu_0I}{2\pi r}\hat i\tag1$$ We can see that the electric field is radially outward at any point. $$\vec E_A=\dfrac{1}{4\pi\epsilon_0}\dfrac{2\lambda}{ r}\hat r$$ So, at a point above the wire, its direction is upward. $$\vec E_A=\dfrac{1}{4\pi\epsilon_0}\dfrac{2\lambda}{ r}\hat j\tag 2$$ According to the left figure below, the current direction is into the page $(-\hat k)$. Now we can use the second formula from (2) to find the velocity of the mosquito. $$ \vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}$$ $$0=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}$$ $$ \vec B_A=\dfrac{\vec v_{BA}\times \vec E_A}{c^2}$$ It is obvious that $(\vec v_{BA}\times \vec E_A)$ must have the same direction of $\vec B_A$ which is in the positive $x$-direction. And we can see that the electric field at this point is upward $$\hat i= \hat ?\times \hat j=-\hat k\times \hat j$$ Hence, the direction of the mosquito at this point is into page in the direction of the current which is in the negative $z$-direction. Note that we are focusing on the position above the wire between the wire and the plastic insulation as shown in the left figure below. Now we know the direction of the velocity and we need to find its magnitude. Solving for $\vec v_{BA}$; $$\vec v_{BA}=\dfrac{ c^2\vec B_A}{\vec E_A} $$ Plugging from (1) and (2), $$\vec v_{BA}= 4\pi\epsilon_0 c^2 \dfrac{\mu_0I}{ 2\pi\color{red}{\bf\not} r} \dfrac{ \color{red}{\bf\not} r}{2\lambda}$$ $$\vec v_{BA}= 4\pi\epsilon_0 c^2 \dfrac{\mu_0I}{ 4\pi \lambda } $$ Plug the known; $$\vec v_{BA}= (9\times 10^9)^{-1} (3\times 10^8)^2 \cdot \dfrac{(4\pi\times 10^{-7})(2.5)}{ 4\pi (250\times 10^{-9})} $$ $$\vec v_{BA}=(-\color{red}{\bf 1\times 10^7}\;\hat k)\;\rm m/s$$ The mosquito needs to move at a speed of 10$^7$ m/s in a direction that is parallel to the current.
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