# Why don't stationary charge feel force from a current carrying wire?

The current carrying wire doesn't apply any magnetic force on nearby charge $$q$$( positive stationary charge) because it has 0 velocity in lab frame. We found that there is no force on q by wire. But if we take a moving frame then, q is in relative motion and hence a current carrying wire applies a magnetic force on q. Let's denote this magnetic force in moving frame by $$F_B$$. Since net force on charge is still 0 there is some force needed to cancel out this $$F_B$$ force.

This was answered by length contraction. I have seen many videos referring to this as solution but i don't think if it works. I need a calculation which can show how length contraction actually helps in cancelling out $$F_B$$.

For calculation you are going to do i would like to refer some sign but you can take numerical value if you wish.

Area of cross-section of wire, $$A$$; length of wire in lab frame, $$L$$; electron density of wire in lab frame, $$n$$ ; the average velocity of electrons in lab frame, $$v_d$$; the moving frame is moving in opposite direction of electrons motion as seen from lab frame and the velocity of this frame relative to lab frame is $$v_F$$. The charge q is placed at $$r$$ distance from center of wire that. Moving frame is parallel to straight wire.

• @MariusLadegårdMeyer i removed that part to only focus on answering how there would be no net force on q in any frame. Feb 17 at 0:06
• Blah. : ) I would much rather calculate force F between test charge and wires negative charges. And then force -F between test charge and wire's positive charges. Feb 19 at 1:12
• @stuffu you can enter the chat of me and frobenius. Its link is in his answer's comments Feb 19 at 7:24


I post below the Figure 5.22 and a paragraph extracted from the book $$''$$Electricity and Magnetism$$''$$ by Edward M. Purcell & David J. Morin, 3rd Edition.

$$\hebl$$

Figure 5.22. A test charge $$\:q\:$$ moving parallel to a current in a wire. (a) In the lab frame, the wire, in which the positive charges are fixed, is at rest. The current consists of electrons moving to the right with speed $$\:\nu_0$$. The net charge on the wire is zero. There is no electric field outside the wire. (b) In a frame in which the test charge is at rest, the positive ions are moving to the left with speed $$\:\nu\:$$, and the electrons are moving to the right with speed $$\:\nu'_0$$. The linear density of positive charge is greater than the linear density of negative charge. The wire appears positively charged, with an external field $$\:E'_r$$, which causes a force $$\:qE'_r$$ on the stationary test charge $$\:q$$. (c) That force transformed back to the lab frame has the magnitude $$\:qE'_r/\gamma$$, which is proportional to the product of the speed $$\:\nu\:$$ of the test charge and the current in the wire, $$\m \lambda_0\nu_0$$.

$$\hebl$$

The case above is not exactly an answer to your question. But it gives some hints :

(1) First, since you suppose that the stationary charge in the lab system doesn't feel electric force, $$(\mathbf E\e\bl0)$$, the wire must have zero net charge, so we must have its configuration as in (a) of the Figure-01, that is ions at rest with linear charge density $$\:\lambda_0\:$$ and electrons with linear charge density $$\:\m\lambda_0\:$$ moving to the right with velocity $$\:\nu_0\e v_d$$. The current of moving electrons produces a magnetic field $$\:\mathbf B\neq\bl0\:$$ but zero magnetic force $$\:\mathbf u_q\x\mathbf B\e\bl0\:$$ on the stationary charge $$q\:(\mathbf u_q\e\bl0)$$. So, in the lab frame we have zero Lorentz force $$$$\mathbf f\e q\plr{\mathbf E\p\mathbf u_q\x\mathbf B}\e\bl0 \tl{01}$$$$

(2) Second, in a frame moving to the right the ions are moving to the left and the linear density of positive charge is greater than the linear density of negative charge. The wire appears positively charged, with an external electric field $$\:\mathbf E'\neq\bl0$$, which causes an electric force $$\:q\mathbf E'\:$$ on the test charge $$\:q$$. This electric force is compensated by the magnetic force $$\:q\plr{\mathbf u'_q\x\mathbf B'}$$ on the moving charge $$\:q$$. So, in the moving to the right frame we have also zero Lorentz force $$$$\mathbf f'\e q\plr{\mathbf E'\p\mathbf u'_q\x\mathbf B'}\e\bl0 \tl{02}$$$$ A first quick proof of equation \eqref{02} is based on the Lorentz transformation of the electromagnetic field, see the Figure and equations (04a),(04b) in my answer therein Is it a typo in David Tong's derivation of spin-orbit interaction? repeated herein for convenience \begin{align} \mathbf E' & \e\gamma \mathbf E\m\dfrac{\gamma^2}{c^2 \plr{\gamma\p1}}\plr{\mathbf E\bl\cdot \bl\upsilon}\bl\upsilon\,\p\,\gamma\plr{\bl\upsilon\x\mathbf B} \tl{03a}\\ \mathbf B' & \e\gamma \mathbf B\m\dfrac{\gamma^2}{c^2 \plr{\gamma\p1}}\plr{\mathbf B\bl\cdot \bl\upsilon}\bl\upsilon\,\m\,\dfrac{\gamma}{c^2}\plr{\bl\upsilon\x\mathbf E} \tl{03b} \end{align} If at a point in the unprimed system (lab system) we have $$\:\mathbf E\e\bl0\:$$ then in the primed system moving with velocity $$\:\bl\upsilon$$ $$$$\mathbf E' \e\bl\upsilon\x\mathbf B' \tl{04}$$$$ see equation (08) and its proof in aforementioned linked answer. But in the new frame we have for the velocity $$\:\mathbf u'_q\:$$ of the test charge $$$$\mathbf u'_q \e \m\bl\upsilon \tl{05}$$$$ Equations \eqref{04},\eqref{05} imply equation \eqref{02}.

Note that based on the Lorentz transformation of the electromagnetic field, equations \eqref{03a} & \eqref{03b}, the Lorentz 3-vector force applied on a point charge $$\:q\:$$ moving with velocity $$\:\mathbf u\:$$, that is $$$$\mathbf f\e q\plr{\mathbf E\p\mathbf u\x\mathbf B} \tl{06}$$$$ under a Lorentz boost with velocity $$\:\bl\upsilon\:$$ is transformed as follows $$$$\mathbf f'\e \dfrac{\mathbf f\p\dfrac{\gamma^2}{c^2 \plr{\gamma\p 1}}\plr{\mathbf f\bl\cdot \bl\upsilon}\bl\upsilon\m\gamma \bl\upsilon\plr{\dfrac{\mathbf f\bl\cdot\mathbf u}{c^{2}}}}{\gamma \plr{1\m\dfrac{\bl\upsilon\bl\cdot\mathbf u}{c^{2}\vphantom{\tfrac{a}{b}}}}} \tl{07}$$$$ see equation (11) in my answer therein Are magnetic fields just modified relativistic electric fields?. So, if $$\:\mathbf f\e\bl 0\:$$ in an inertial frame $$\:\rm S\:$$ then $$\:\mathbf f'\e\bl 0\:$$ in any other inertial $$\:\rm S'$$.

$$\hebl$$

By this ADDENDUM A we respond to the OP's demand to have $$''$$...a calculation which can show how length contraction actually helps in cancelling out $$F_B$$$$''$$. First we must have in mind that

Length Contraction (LC) and Time Dilation (TD) are consequences of the Lorentz $$\,$$ Trans- formation (LT). Their use to get results is unsafe. In our calculations we use LT to get safe results while we use LC and TD to interpret these results.

This will be done in the following for our case of the infinite straight wire. We will determine the electromagnetic field $$\:\plr{\mb E,\mb B}\:$$ in the inertial frame $$\:\mr S$$, the rest frame of the wire and the charge $$\:q$$, from the distribution of electric charges and electric currents directly by the Maxwell equations. Using a Lorentz boost transformation, equations \eqref{03a} & \eqref{03b}, we will determine the electromagnetic field $$\:\plr{\mb E',\mb B'}\:$$ in an inertial frame $$\:\mr S'\:$$ moving in the direction of the straight wire. By Lorentz Contraction and relativistic addition of velocities we will interpret the results determining the distribution of electric charges and electric currents in the frame $$\:\mr S'$$.

In Figure-02 we have an infinitely long wire with :

(1) ions of linear charge density $$\:\lambda_{\bf i}\e\lambda_0\gr 0\:$$ at rest so producing zero electric current $$\:\mb I_{\bf i}\e\mr I_{\bf i}\mb e_{\bl z}\e\bl0\:$$ and

(2) electrons of linear charge density $$\:\lambda_{\bf e}\e\m\lambda_0\:$$ in motion with velocity $$\:\mb v\e \mr v \,\mb e_{\bl z} , \mr v \bl\in \plr{\m c,\p c}$$ so producing electric current $$\:\mb I_{\bf e}\e\mr I_{\bf e}\mb e_{\bl z}\e\lambda_{\bf e}\,\mb v\e \m\lambda_0\mr v \,\mb e_{\bl z}$$.

So in the wire we have :

(a) zero linear charge density $$\:\lambda\e\lambda_{\bf i}\p\lambda_{\bf e}\e0\:$$ and

(b) electric current $$\:\mb I\e \mb I_{\bf i}\p\mb I_{\bf e}\e \m\lambda_0\mr v \,\mb e_{\bl z}$$.

Although in many textbooks and the web we give in Figure-03 the electromagnetic field $$\:\plr{\mb E,\mb B}\:$$ produced by an infinitely long straight wire with constant uniform linear charge density $$\:\lambda\:$$ and constant electric current $$\:\mb I\e \mr I\,\mb e_{\bl z}$$. Note that $$\:\mr I\:$$ is not the magnitude of the vector $$\:\mb I\:$$, it's a real number, that is it could take negative values. Because of the rotational around and translational along the $$\:z\m$$axis symmetries we use cylindrical coordinates $$\:\rho,\phi,z\:$$ so we have $$$$\mb E\e\dfrac{\lambda}{2\pi\epsilon_0\rho}\mb e_{\bl\rho}=\dfrac{\lambda}{2\pi\epsilon_0\rho^2}\bl\rho\,,\qquad \mb B=\dfrac{\mu_0\mr I}{2\pi\rho}\mb e_{\bl\phi}\e\dfrac{\mu_0}{2\pi\rho^2}\plr{\mb I\bl{\times\rho}} \tl{A-01}$$$$ where $$\:\bl\rho\e \rho\,\mb e_{\bl\rho}\e \plr{\rho\cos\phi,\rho\sin\phi,0}$$.

In equation \eqref{A-01} the magnitude of the vector $$\:\mb E\:$$ is determined from the Maxwell equation $$$$\bl{\nabla\cdot}\mb E\e\dfrac{\varrho}{\epsilon_0}\quad \texttt{where } \varrho\e\texttt{volume charge density} \tl{A-02}$$$$ by volume integration in a cylinder of radius $$\:\rho\:$$ and height $$\:\mr L\:$$ $$$$\begin{split} \iiint\limits_V\bl{\nabla\cdot}\mb E\,\mr dV & \e\iiint\limits_V\dfrac{\varrho}{\epsilon_0}\,\mr dV\:\bl\Longrightarrow\:\iint\limits_S \mb E\bl\cdot\mr d\bl S\e\dfrac{\lambda\mr L}{\epsilon_0}\:\bl\Longrightarrow\\ \mr E\cdot 2\pi\rho \,\mr L & \e \dfrac{\lambda\mr L}{\epsilon_0}\:\bl\Longrightarrow\: \mr E\e\dfrac{\lambda}{2\pi\epsilon_0\rho}\\ \end{split} \tl{A-03}$$$$ that is essentially by Gauss Law.

Also the magnitude of the vector $$\:\mb B\:$$ is determined from the Maxwell equation $$$$\bl{\nabla\times}\mb B\e \mu_0 \mb j \p\dfrac{1}{c^2}\dfrac{\partial\mb E}{\partial t}\quad \texttt{where } \mb j\e\texttt{electric current density} \tl{A-04}$$$$ by surface integration on a circular disk of radius $$\:\rho\:$$ (note that $$\:\mb E\:$$ is constant in time) $$$$\begin{split} \iint\limits_S\plr{\bl{\nabla\times}\mb B}\bl\cdot\mr d\bl S & \e\iint\limits_S\mu_0 \mb j \bl\cdot\mr d\bl S\:\bl\Longrightarrow\:\oint\limits_C\mb B\bl\cdot\mr d\bl \ell\e \mu_0\,\mr I\:\bl\Longrightarrow\\ \mr B\cdot 2\pi\rho & \e\mu_0\,\mr I\:\bl\Longrightarrow\:\mr B\e\dfrac{\mu_0\mr I}{2\pi\rho} \\ \end{split} \tl{A-05}$$$$ Note that the electromagnetic field $$\:\plr{\mb E,\mb B}\:$$ in equation \eqref{A-01} as derived directly from Maxwell equations is exactly (not approximately) relativistic.

Using above general solution for our case here with zero linear electric charge density $$\:\lambda\e0\:$$ and electric current $$\:\mb I\e \m\lambda_0\mr v \,\mb e_{\bl z}$$., that is for the configuration of Figure-02, we have in the rest frame $$\:\mr S\:$$ of the wire $$$$\mb E\e\bl 0\:(\texttt{everywhere)}\,,\qquad \mb B\e\m\dfrac{\mu_0\lambda_0\mr v}{2\pi\rho}\mb e_{\bl\phi} \tl{A-06}$$$$ as shown in Figure-04.

In an inertial frame $$\:\mr S'\:$$ moving with velocity $$\:\bl\upsilon\e \upsilon\,\mb e_{\bl z}\:$$ with respect to the rest frame $$\:\mr S\:$$ of the wire the electromagnetic field $$\:\plr{\mb E',\mb B'}\:$$ will be derived from that of \eqref{A-06} via the Lorentz boost transformation, equations \eqref{03a} & \eqref{03b}, that is $$$$\mb E'\e\gamma_{\bl \upsilon}\plr{\bl{\upsilon\times}\mathbf B}\,,\qquad \mb B'\e\gamma_{\bl \upsilon}\mb B\qquad \texttt{ where }\gamma_{\bl \upsilon}\e 1\bigg/\sqrt{1\m\dfrac{\upsilon^2}{c^2}} \tl{A-07}$$$$ since $$\:\mb E\e\bl 0\:$$ and $$\:\plr{\mb B\bl{\cdot\upsilon }}\e 0$$.

Inserting the expression \eqref{A-06} of $$\:\mb B\:$$ in \eqref{A-07} we have $$$$\mb E'\e\gamma_{\bl\upsilon}\plr{\bl{\upsilon\times}\mb B}\e \m\gamma_{\bl\upsilon}\dfrac{\mu_0\lambda_0\upsilon\,\mr v}{2\pi\rho}\plr{\mb e_{\bl z}\bl\times\mb e_{\bl\phi}}\e \m\gamma_{\bl\upsilon}\dfrac{\mu_0\lambda_0\upsilon\,\mr v}{2\pi\rho'}\plr{\mb e'_{\bl z}\bl\times\mb e'_{\bl\phi}} \tl{A-08}$$$$ so $$$$\mb E'\e \dfrac{\gamma_{\bl\upsilon}\mu_0\lambda_0\upsilon\,\mr v}{2\pi\rho'}\mb e'_{\bl\rho}\,,\qquad \mb B'\e\m\dfrac{\gamma_{\bl\upsilon}\mu_0\lambda_0\mr v}{2\pi\rho'}\mb e'_{\bl\phi} \tl{A-09}$$$$ as shown in Figure-05.

We bring these results in expressions similar to the general solution \eqref{A-01} as follows $$$$\mb E'\e\dfrac{\lambda'}{2\pi\epsilon_0\rho'}\mb e'_{\bl\rho}=\dfrac{\lambda'}{2\pi\epsilon_0\rho'^{\,2}}\bl\rho'\,,\qquad \mb B'=\dfrac{\mu_0\mr I'}{2\pi\rho'}\mb e'_{\bl\phi}\e\dfrac{\mu_0}{2\pi\rho'^{\,2}}\plr{\mb I'\bl{\times\rho'}} \tl{A-10}$$$$ where $$$$\boxed{\:\:\lambda'\e \dfrac{\gamma_{\bl\upsilon}\upsilon\,\mr v}{c^2}\lambda_0\vp\:\:}\quad \texttt{and}\quad \boxed{\:\;\mr I'\e \m\gamma_{\bl\upsilon}\lambda_0\,\mr v\e \gamma_{\bl\upsilon} \mr I\vp\:\:} \tl{A-11}$$$$ So it looks like on the $$\:z'\m$$axis of the moving frame $$\:\mr S'\:$$ there exists a linear electric charge density $$\:\lambda'\:$$ that produces the electric field $$\:\mb E'\:$$ and an electric current $$\:\mr I'\:$$ that produces the magnetic field $$\:\mb B'$$. But how these facts are explained and especially the values of $$\:\lambda', \mr I'\:$$ in equation \eqref{A-11} ??? These are explained in ADDENDUM B by interpretations using the length contraction and the relativistic addition of velocities.

$$\hebl$$

In the inertial frame $$\:\mr S'\:$$ for the linear electric charge density $$\:\lambda'\:$$ we have $$$$\begin{split} \lambda'& \e\lambda'_{\bf i}\p\lambda'_{\bf e}\\ \texttt{where}\quad\lambda'_{\bf i} & \e \texttt{charge density of the ion array}\\ \texttt{and }\quad\lambda'_{\bf e} & \e \texttt{charge density of the electron array}\\ \end{split} \tl{B-01}$$$$

The ion array is at rest in frame $$\:\mr S$$, the rest frame of the wire, but it moves with velocity $$\:\plr{\m\bl\upsilon}\:$$ in frame $$\:\mr S'$$. Take a straight segment of length $$\:\mr L_{\bf i}\:$$ on the ion array in frame $$\:\mr S$$. Because of length contraction for its length $$\:\mr L'_{\bf i}\:$$ in frame $$\:\mr S'\:$$ we have $$$$\mr L'_{\bf i}\e \dfrac{\mr L_{\bf i}}{\gamma_{\bl\upsilon}} \tl{B-02}$$$$ Because of the Lorentz invariance of the electric charge we have by equating the charge between $$\:\mr L_{\bf i}\:$$ and $$\:\mr L'_{\bf i}\:$$ $$$$\lambda'_{\bf i}\mr L'_{\bf i}\e \lambda_{\bf i}\mr L_{\bf i}\:\bl\Longrightarrow\: \lambda'_{\bf i}\e \dfrac{\mr L_{\bf i}}{\mr L'_{\bf i}}\lambda_{\bf i}\e\gamma_{\bl\upsilon}\lambda_{\bf i} \tl{B-03}$$$$ so $$$$\boxed{\:\:\lambda'_{\bf i}\e\gamma_{\bl\upsilon}\lambda_0\vp\:\:} \tl{B-04}$$$$

Now, in order to find the linear charge density $$\:\lambda'_{\bf e}\:$$ of the electron array in the frame $$\:\mr S'\:$$ we must determine :

(1) First, the velocity $$\:\mb v'\:$$ of the electron array in the frame $$\:\mr S'\:$$ and

(2) Second, how the length of straight segments of the electron array are contracted between the inertial frames.

For (1) we have the following equation for the Lorentz transformation of the velocity 3-vector $$\:\mb v\:$$ under a Lorentz boost with velocity $$\:\bl\upsilon$$ $$$$\mb v'\e\dfrac{\mb v\p\dfrac{\gamma^2_{\bl\upsilon}\plr{\bl{\upsilon\cdot}\mb v}}{c^2 \plr{\gamma_{\bl\upsilon}\p 1}}\bl\upsilon\m\gamma_{\bl\upsilon}\bl\upsilon}{\gamma_{\bl\upsilon}\plr{1\m\dfrac{\bl{\upsilon\cdot}\mb v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}}} \tl{B-05}$$$$ which essentially gives the relativistic addition of the velocities $$\:\mb v\:$$ and $$\:\m\bl\upsilon$$.

For (2) we have the following important relation between the $$\:\gamma\m$$factors of the velocities $$\:\bl\upsilon, \mb v, \mb v'\:$$ of equation \eqref{B-05} $$$$\dfrac{\gamma_{\mb v'}}{\gamma_{\mb v}}\e\gamma_{\bl\upsilon}\plr{1\m\dfrac{\bl{\upsilon\cdot}\mb v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}}\quad \texttt{where }\gamma_{\bl \alpha}\e 1\bigg/\sqrt{1\m\dfrac{\alpha^2}{c^2}}\:\plr{\bl\alpha\e \bl\upsilon,\mb v,\mb v'} \tl{B-06}$$$$ For the derivations of equations \eqref{B-05} and \eqref{B-06} see my answer here Transformation of 4-velocity, equations (08) and (14) respectively.

For our case the velocities $$\:\bl\upsilon, \mb v, \mb v'\:$$ are collinear on the same $$\:z\m$$axis so \eqref{B-05} yields $$$$\mb v'\e\dfrac{\mb v\m\bl\upsilon}{1\m\dfrac{\bl{\upsilon\cdot}\mb v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tl{B-07a}$$$$ expressed also between scalars $$$$\mr v'\e\dfrac{\mr v\m\upsilon}{1\m\dfrac{\upsilon\,\mr v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tl{B-07b}$$$$ while equation \eqref{B-06} yields $$$$\dfrac{\gamma_{\mb v'}}{\gamma_{\mb v}}\e\gamma_{\bl\upsilon}\plr{1\m\dfrac{\upsilon\,\mr v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tl{B-08}$$$$

Consider now a straight segment $$\:\mr L^0_{\bf e}\:$$ on the electron array in its rest frame. Because of length contraction for the lengths $$\:\mr L_{\bf e},\mr L'_{\bf e}\:$$ in the inertial frames $$\:\mr S,\mr S'\:$$ respectively we have $$$$\gamma_{\mb v}\mr L_{\bf e}\e\gamma_{\mb v'}\mr L'_{\bf e}\e \mr L^0_{\bf e}\e \texttt{Lorentz invariant} \tl{B-09}$$$$ Because of the Lorentz invariance of the electric charge we have by equating the charge between $$\:\mr L_{\bf e}\:$$ and $$\:\mr L'_{\bf e}\:$$ $$$$\lambda'_{\bf e}\mr L'_{\bf e}\e \lambda_{\bf e}\mr L_{\bf e}\:\bl\Longrightarrow\: \lambda'_{\bf e}\e \dfrac{\mr L_{\bf e}}{\mr L'_{\bf e}}\lambda_{\bf e}\e \dfrac{\gamma_{\mb v'}}{\gamma_{\mb v}}\lambda_{\bf e} \e\gamma_{\bl\upsilon}\plr{1\m\dfrac{\upsilon\,\mr v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}}\lambda_{\bf e} \tl{B-10}$$$$ so $$$$\boxed{\:\:\lambda'_{\bf e}\e\m\gamma_{\bl\upsilon}\plr{1\m\dfrac{\upsilon\,\mr v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}}\lambda_0\vp\:\:} \tl{B-11}$$$$ Adding \eqref{B-04}, \eqref{B-11} side by side we have $$$$\lambda'\e\lambda'_{\bf i}\p\lambda'_{\bf e}\e\gamma_{\bl\upsilon}\lambda_0\m\gamma_{\bl\upsilon}\plr{1\m\dfrac{\upsilon\,\mr v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}}\lambda_0\e\dfrac{\gamma_{\bl\upsilon}\upsilon\,\mr v}{c^2}\lambda_0 \tl{B-12}$$$$ so verifying the first of equations \eqref{A-11}.

For the electric current of the ions in frame $$\:\mr S'\:$$ we have $$$$\mb I'_{\bf i}\e \lambda'_{\bf i}\plr{\m\bl\upsilon}\e \m \gamma_{\bl\upsilon}\lambda_0\bl\upsilon \tl{B-13}$$$$ so $$$$\boxed{\:\:\mr I'_{\bf i}\e\m \gamma_{\bl\upsilon}\lambda_0\upsilon\vp\:\:} \tl{B-14}$$$$ while for the electric current of the electrons in frame $$\:\mr S'\:$$ we have $$$$\mb I'_{\bf e}\e \lambda'_{\bf e}\mb v'\e \m\gamma_{\bl\upsilon}\plr{1\m\dfrac{\upsilon\,\mr v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}}\lambda_0\dfrac{\mb v\m\bl\upsilon}{\plr{1\m\dfrac{\upsilon\,\mr v}{c^2}\vphantom{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}}}\e \m\gamma_{\bl\upsilon}\lambda_0\plr{\mb v\m\bl\upsilon} \tl{B-15}$$$$ that is $$$$\boxed{\:\:\mr I'_{\bf e}\e \m\gamma_{\bl\upsilon}\lambda_0\plr{\mr v\m\upsilon} \vp\:\:} \tl{B-16}$$$$ Adding \eqref{B-14}, \eqref{B-16} side by side we have $$$$\mr I'\e\mr I'_{\bf i}\p\mr I'_{\bf e}\e\m \gamma_{\bl\upsilon}\lambda_0\upsilon \m\gamma_{\bl\upsilon}\lambda_0\plr{\mr v\m\upsilon}\e \m \gamma_{\bl\upsilon}\lambda_0\mr v \tl{B-17}$$$$ so verifying the second of equations \eqref{A-11}.

$$\hebl$$

Important Note :

From the expression of the linear electric charge density $$\:\lambda'$$, see the first of equations \eqref{A-11}
$$$$\boxed{\:\:\lambda'\e \dfrac{\gamma_{\bl\upsilon}\upsilon\,\mr v}{c^2}\lambda_0\vp\:\:} \nonumber$$$$ since $$\gamma_{\bl\upsilon}\gr 0, \lambda_0\gr 0$$ we have $$$$\begin{split} \lambda' & \gr 0 \quad \texttt{if } \upsilon\,\mr v\gr 0\\ \lambda' & \les 0 \quad \texttt{if } \upsilon\,\mr v\les 0\\ \end{split} \tl{B-18}$$$$ that is : the neutral in $$\:\mr S\:$$ wire appears in the moving system $$\:\mr S'\:$$ positively charged if the velocities $$\:\bl\upsilon, \mb v\:$$ point to the same direction of the $$\:z\m$$axis and negatively charged if these velocities point to opposite directions of the $$\:z\m$$axis.

• Feb 18 at 9:50
• @James Webb : Ok. Mar 30 at 7:53
• there is an eq. (02a) in the page you linked. How to get that eq.? Thank you Mar 30 at 8:25
• @James Webb : I joined Physics SE as diracpaul in June'15 and I quit the site in Sep'15 for personal reasons. I came back as Frobenius in Mar'16. Under my answers as former diracpaul now you could see the name user82794. My 2015 answer here Two sets of coordinates each in frames O and O′ - Lorentz transformation gives the details for your question about the Lorentz transformation along an arbitrary direction. Mar 30 at 15:46
• @James Webb : It's prerequisite to know about the 1+1-Lorentz transformation along the $x-$axis. Sincerely, I didn't have any reference with the proof of this more general Lorentz transformation (called also "boost"). I have produced this many years ago by my own. Mar 30 at 15:48

I need a calculation which can show how length contraction actually helps in cancelling out FB.

The answer by @Frobenius shows the calculation using ordinary vectors. Since the question asks about length contraction, it is a question about relativity. So this answer shows the calculations using the relativistic framework of four-vectors and tensors.

The electric and magnetic fields are combined into the antisymmetric electromagnetic field tensor $$F^{\mu \nu }=\left( \begin{array}{cccc} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \\ \end{array} \right)$$ and the four-current $$J^\mu=(\rho,j_x,j_y,j_z)$$ The four-force on a charge distribution is then given by $$f_\mu = J^\nu F_{\mu\nu}$$. These equations hold in any reference frame.

Now, in your case you have a neutral wire with a steady current in the $$x$$ direction lying along the $$x$$ axis. This wire has a four-current of $$J^{\mu }=(0,i,0,0)$$ There is also a test charge of magnitude $$q$$ which is located a distance of $$r$$ away from the origin in the $$z$$ direction. This test charge has a four current of $$q^\mu=(q,0,0,0)$$

At the location of the test charge, the current produces the EM tensor $$F^{\mu \nu }=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -\frac{i}{r} \\ 0 & 0 & 0 & 0 \\ 0 & \frac{i}{r} & 0 & 0 \\ \end{array} \right)$$ So the force on the test charge is $$f_{\mu }=q^{\nu }F_{\mu \nu } =(0,0,0,0)$$ which, as expected, means that there is no force on the charge in the rest frame.

Now, when we boost everything to a primed frame moving at $$v$$, represented by primes on the indices, we get the following: $$J^{\mu '}=(\gamma i v,\gamma i,0,0)$$ where the non-zero first term indicates that the wire is no longer neutral in the primed frame, but has some net charge. $$q^{\mu '}=\left(q',j',0,0\right)=\left(\gamma q,\gamma q v,0,0\right)$$ where the non-zero second term indicates that the test charge also has some "test current" because it is moving in this frame. And $$F^{\mu ' \nu '}=\left( \begin{array}{cccc} 0 & 0 & 0 & -\frac{i v \gamma }{r} \\ 0 & 0 & 0 & -\frac{i \gamma }{r} \\ 0 & 0 & 0 & 0 \\ \frac{i v \gamma }{r} & \frac{i \gamma }{r} & 0 & 0 \\ \end{array} \right)$$ where the new terms indicate the electric field produced by the charge of the wire in the primed frame at the location of the test charge. In this frame the force is calculated as $$f_{\mu '}=q^{\nu '} F_{\mu ' \nu '}=\left( 0,0,0,\frac{\gamma i \left(j'-v q'\right)}{r}\right)=(0,0,0,0)$$ Notice that regardless of $$v$$ the term of force from the wire's electric field acting on the test charge, $$\gamma i v q'/r$$, is exactly canceled out by the term of force from the wire's magnetic field acting on the test current, $$\gamma i j'/r$$. So if the force is 0 in one frame it is 0 in all frames.

• ${\gamma i ( j' - vq') }/r$ . The value is either zero or not zero and it depends on $(j' -vq')$. Can you tell what frames these values ( $j'$, $v$, $q'$) belong to and how the expression above becomes zero. Feb 19 at 7:47
• $j'$ is current per unit area as I guess and therefore dependent on area of cross section. Area of cross-section doesn't get contracted since it is perpendicular to velocity. And therefore there is no reason i can guess to say $j' - vq'$ should be zero always. Feb 19 at 7:53
• @JamesWebb from the expression above $q^{\mu '}=\left(q',j',0,0\right)=\left(\gamma q,\gamma q v,0,0\right)$ so $q’=\gamma q$ and $j’=\gamma q v$ so $j’-vq’=0$ always
– Dale
Feb 19 at 14:38
• Actually what you wrote in answer was going out of my head i don't understand these four force representation. I understand only simple form equations like in answer of @frobenius. I can calculate contraction and other stuff separately and then put them together in force equation. That's what i understand. Feb 19 at 15:37
• @JamesWebb that is fine. Frobenius’ answer is valid too. Other people with the same question may prefer the four vector approach, so it is good to have both
– Dale
Feb 19 at 15:46

If you want to know why moving magnet doesnt affect moving charge, if their relative speed is zero:

You can think of it as if they both emit EM field. Moving magnet emitts magnetic and electric fields, mostly magnetic while speed is significantly below c. And moving charge emits electric and magnetic fields, mostly electric while speed is significantly below c.

As you take speed higher and higher as a frame of reference, but their relative speed remains zero, you can measure strong EM fields in your frame of reference, from either of them. But fields cancel out on their frame of reference, where their relative speed is zero.