SMT359 - 2008

General remarks: The solutions contained on this page have been provided by current and former SMT359 students according to their best endeavours. Users are encouraged to make amendments, comment on subjective points, make suggestions, and develop solutions for questions not yet solved.

Question 1
The magnitudes of the forces on A due to B and due to C are equal, and given by



F=\dfrac{q^2}{4\pi\epsilon_0(2a)^2} $$ The components of these forces in the

$$x$$ -direction cancel. The forces have equal components in the

$$y$$ -direction,

$$F\sin60^{\circ}$$ , so the resultant force has magnitude



2\times\dfrac{q^2}{4\pi\epsilon_04a^2}\times\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}q^2}{16\pi\epsilon_0a^2} $$, and is in the

$$y$$ -direction.

Question 2
(a) An electrostatic field is conservative and the curl of a conservative vector field is zero. We test the given field for conservativeness using the formula for curl of a field in cartesian coordinates which is given in the formula sheet. Most of the components in the expression for curl turn out to be zero because the given vector field E has no component along

$$\mathbf{e}_z$$ and the components along

$$\mathbf{e}_x$$ and

$$\mathbf{e}_y$$ depend only on the

$$x$$ and

$$y$$ coordinates.


 * $$\mathbf{E}=axy^2\mathbf{e}_x + ax^2y\mathbf{e}_y$$

\begin{align} \text{curl(}\mathbf{E}\text{)}&=\left(\dfrac{\partial{E}_y}{\partial{x}}-\dfrac{\partial{E}_x}{\partial{y}}\right)\mathbf{e}_z\\ &=\left(2axy-2axy\right)\mathbf{e}_z\\ &=0 \end{align} $$ The curl of the given vector field is zero, we conclude that

$$\mathbf{E}$$ could represent an electrostatic field. (b) The equation that relates an electrostatic field to its corresponding charge density is the differential form of Gauss' law which states that

$$\text{div}\left(\mathbf{E}\right)=\dfrac{\rho}{\epsilon_0}$$

The left hand side of this equation is solved using the expression for the divergence of a vector field in cartesian coordinates which is given in the formula sheet.

\text{div}\left(\mathbf{E}\right)=\dfrac{\partial{E_x}}{\partial{x}}+\dfrac{\partial{E_y}}{\partial{y}}+\dfrac{\partial{E_z}}{\partial{z}}=ay^2+ax^2 $$ Equating this result with the right-hand side of Gauss' law gives us

ay^2+ax^2=\dfrac{\rho}{\epsilon_0} $$ which can be rewritten as

\rho = a\epsilon_0\left(x^2+y^2\right) $$

Question 3
According to the Lorentz force law the electromagnetic force experienced by a charged particle is given by

\mathbf{F}=q\left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right) $$ The particle of interest in an electron, with charge

$$q$$ of

$$-e=-1.6\times10^{-19}\mathrm{C}$$

We are given the electric field E, the magnetic field B, and the velociy of the particle v, we can calculate the force by straightforward application of the Lorentz Force Law.

Firstly we calculate the cross-product

$$\mathbf{v}\times\mathbf{B}$$

\mathbf{v}\times\mathbf{B} = 1\times10^7\left[-3\mathbf{e}_x - 6\mathbf{e}_y\right] $$

The force on the electron becomes

\begin{align} \mathbf{F}&=-1.6\times 10^{-19}\times\left[3.0\times 10^7\mathbf{e}_x - 3.0\times 10^7\mathbf{e}_x - 6.0\times 10^7\mathbf{e}_y\right]\mathrm{ N}\\ &=-1.6\times 10^{-19}\times\left[-6.0\times 10^7\mathbf{e}_y\right]\mathrm{ N}\\ &=9.6\times 10^{-12}\mathbf{e}_y \mathrm{ N} \end{align} $$

Question 4
A current can be driven around the circuit by means of an induced EMF, the cause of the induced EMF could be one of two things. Firstly, a time-varying magnetic field, according to Faraday, will induce a current driven by mean of a non-conservative electric force. Secondly, if the magnetic field is not changing over time but the circuit is expanding (or moving), then current is driven by magnetic forces.

(a) According to Lenz' Law, an induced current will flow to appose the change that induced it. As rail DG moves the amount of flux passing through circuit CDGF into the page increases with time because the area of the circuit is expanding, this time-varying magnetic flux induces a current to flow in circuit CDGF driven by magnetic forces. The induced current will flow such that the magnetic field that it induces apposes the increasing magnetic flux through CDGF which means that the induced current will flow in a counterclockwise direction.

(b) The integral version of Faraday's Law is expressed mathematically as

\int{\mathbf{E}\cdot\mathrm{d}\mathbf{l}}=-\dfrac{d}{dt}\int{\mathbf{B}\cdot\mathrm{d}\mathbf{S}} $$ The equation says that the induced EMF around a closed circuit is equal to the negative of the time-varying magnetic flux through the open surface enclosed by the circuit. As rail DG moves to the right, the area bound by the conducting circuit increases into the page. We choose the direction of positive circulation to be clockwise with the normal to the surface bounded by the circuit pointing into the page (we could choose the opposite direction of circulation and normal, it would make no difference to the result).

This means that

$$\int{\mathbf{B}\cdot\mathrm{d}\mathbf{S}}$$ is increasing with time in the direction of

$$\mathbf{B}$$ which is into the page making

$$-\dfrac{d}{dt}\int{\mathbf{B}\cdot\mathrm{d}\mathbf{l}}$$ a negative quantity, which says that

$$\int{\mathbf{E}\cdot\mathrm{d}\mathbf{l}}$$ is negative. Our choice was that the direction of positive circulation is clockwise, so we conclude that the induced EMF drives current in the counterclockwise direction.

Question 5
This question requires us to use our knowledge of boundry conditions at the boundry of a dialectric material. It would be worth referring to the formula sheet provided in the exam as soon as you have read the question. Many of the formulas relating to this section of work look very similar and use the same constants, referring to the formula sheet may help to prevent confusion.

(a) The electric field is normal to the surface of the dialectric slab, there is no component of E parallel to the surface. We should recall the boundry condition for the component of the electric field perpendicular to the surface which is that

$$\epsilon_1 E_{1\perp} = \epsilon_2 E_{2\perp}$$ . We can calculate the magnitude of the electric field inside the dialectric which is

E_{dialectric} = E_{2} = \dfrac{\epsilon_1}{\epsilon_2} E_{1\perp} = \dfrac{1}{5}\times 1\times 10^4 \mathrm{Vm^{-1}} = 2\times10^3 \mathrm{Vm^{-1}} $$ The electric displacement vector

$$\mathbf{D}=\epsilon_0\epsilon\mathbf{E}$$ , we calculate the magnitude of this vector as

\begin{align} D&=\epsilon_0\epsilon_2\mathbf{E}\\ &= \left(8.85\times10^{-12}\right)\times 5 \times \left(2\times10^3\right)\mathrm{Cm^{-2}}\\ &= 8.9\times10^{-8}\mathrm{Cm^{-2}} \end{align} $$ The polarization vector is often expressed in terms of electrical susceptibility as

$$\mathbf{P}=\chi_E\epsilon_0\mathbf{E}$$ , this can be rewritten as

$$\mathbf{P}=\left(\epsilon - 1\right)\epsilon_0\mathbf{E}$$ .

We calculate the magnitude of the polarization to be

\begin{align} \mathrm{P}&=\left(\epsilon_2 - 1\right)\epsilon_0 E\\ \mathrm{P}&=\left(5-1\right)\times\left(8.85\times10^{-12}\right)\times\left(2\times 10^3\right)\\ &= 7.1\times 10^{-8}\mathrm{Cm^{-2}} \end{align} $$ (b) Volume charge density

$$\rho_b=-\mathrm{div}\left(\mathbf{P}\right)$$ . We calculated the magnitude of the polarization vector in part (a), the direction of which will lie along the same direction as the electric field. Let us choose this direction to be

$$\mathbf{e}_x$$ so that the polarization vector becomes

$$\mathbf{P}=7.1\times 10^{-8}\mathbf{e}_x\mathrm{Cm^{-2}}$$ . We are told that the slab is made of LIH dialectric material and that the electric field is uniform, we therefore expect that there is no net charge density anywhere within the dialectric slab. By applying the formula for

$$\rho_b$$ we find

\rho_b=-\mathrm{div}\left(\mathbf{P}\right)=0 $$

Which is consistant with our reasoning that there is no net bound charge density within the slab.

We need to calculate the bound surface charge density on both the left face and the right face of the slab. Bound surface charge density is calculated as

$$\sigma_b=\mathbf{P}\cdot\hat{\mathrm{n}}$$ where

$$\hat{\mathrm{n}}$$ is the unit normal to the face of the slab. We chose earlier that the direction of polarization was along

$$\mathbf{e}_x$$ and we will stick to this convention. At the left face we take the normal to the surface to point along

$$\mathbf{-e}_x$$ so that

$$\sigma_{left}=\mathbf{P}\cdot\hat{\mathrm{n}}= -\mathrm{P} = -7.1\times 10^{-8}\mathrm{Cm^{-2}}$$ At the right face the normal to the surface points along

$$\mathbf{e}_x$$ such that

$$\sigma_{right}=\mathbf{P}\cdot\hat{\mathrm{n}} = \mathrm{P} = 7.1\times 10^{-8}\mathrm{Cm^{-2}}$$

Question 6
(a) The boundary conditions between two magnetic materials with no free currents at the interface are

$$B_{1\perp}=B_{2\perp}$$ and

$$\dfrac{B_{1\parallel}}{\mu_1}=\dfrac{B_{2\parallel}}{\mu_2}$$ for the magnetic field

$$\mathbf{B}$$

and

$$H_{1\parallel}=H_{2\parallel}$$ and

$$\mu_1 H_{1\perp}=\mu_2 H_{2\perp}$$ for the magnetic intensity

$$\mathbf{H}$$ (b) In the first material the angle that

$$\mathbf{B_1}$$ makes with the normal to the boundry is

$$30^\circ$$ , this is given to us. It is straightforward to calculate the angle that the magnetic field

$$\mathbf{B_2}$$ makes with the normal to the boundry in material 2 by applying the formula

$$\dfrac{\tan{\theta_1}}{\tan{\theta_2}}=\dfrac{\mu_1}{\mu_2}$$ where

$$\theta_1=30^\circ$$ and the relative permeabilities of the two materials are given to us. We solve for

$$\theta_2$$

\begin{align} \tan{\theta_2}&=\dfrac{\mu_2}{\mu_1}\times \tan{\theta_1}\\ &=\dfrac{3}{2}\times\tan{30^\circ}\\ &=0.87 \end{align} $$

Therefore

\theta_2=\arctan{\left(0.87\right)} = 40.9^\circ $$

Question 7
(a) We are told that the electric current is constant in the block and that it flows radially outwards. This tells us that

$$I=\int{\mathbf{J\left(r\right)}\cdot\mathrm{d}\mathbf{S}}$$ is constant, such that for any spherical surface with radius

$$r > a$$ we have

$$I=J\left(r\right)\times 4\pi r^2$$

We know that the current points radially outwards, therefore current density must point radially outwards such that

\mathbf{J\left(r\right)}=\dfrac{I}{4\pi r^2}\mathbf{e}_r $$ (b) The current density and the electric field in the block are related by

$$\mathbf{J}=\sigma\mathbf{E}$$ and the current flows as a result of the electric field so both point in the same direction. We find the electric field is

\begin{align} \mathbf{E\left(r\right)}&=\dfrac{\mathbf{J\left(r\right)}}{\sigma}\mathbf{e}_r\\ &=\dfrac{I}{4\pi r^2\sigma}\mathbf{e}_r \end{align} $$ (c) We find the resistance through

$$R=\dfrac{V}{I}$$ . First we need to find

$$V$$ by integrating the electric field using a lower bound of

$$a$$ which is the radius of the metal sphere, and an upper bound of

$$\infty$$



\begin{align} V&=\int_a^\infty{\mathbf{E}\cdot\mathrm{d}\mathbf{r}}\\ &=\dfrac{I}{4\pi\sigma}\int_a^\infty{\dfrac{1}{r^2}\mathrm{dr}}\\ &=\dfrac{I}{4\pi\sigma a} \end{align} $$ Now we can divide this expression for

$$V$$ by

$$I$$ to find the resistance, this gives us

R=\dfrac{V}{I}=\dfrac{1}{4\pi\sigma a} $$

Question 9
(a) The speed of the electromagnetic radiation inside the dialectric is

$$v=\dfrac{1}{\sqrt{\epsilon\epsilon_0\mu\mu_0}}$$ and we know that

$$c=\dfrac{1}{\sqrt{\epsilon\epsilon_0}}$$ so that

$$v=\dfrac{c}{\sqrt{\epsilon\mu}}$$ We are told that

$$\mu=1$$ so that

$$v=\dfrac{c}{\sqrt{\epsilon}}$$ . The refractive index of the dialectric is the ratio of the speed of light in a vacuum to the speed of light (of the same frequency) in the dialectric so rearranging the last equation we have

$$n=\dfrac{c}{v}=\sqrt{\epsilon}=\sqrt{2.5}=1.6$$ Note that we could have jumped straight to the formulas

$$n=\sqrt{\epsilon}$$ although I have taken the approach of deriving the formula for the refractive index here, which is useful for understanding the solution to part (b).

(b) The speed of the radiation in the dialectric

$$v=\dfrac{c}{n}=\dfrac{3\times10^8\mathrm{ms^{-1}}}{1.6}=1.9\times10^8 \mathrm{ms^{-1}}$$ (c) The frequency of the radiation does not change when it enters the dialetric. The angular frequency of the radiation is given to be

$$\omega=2.0\times10^9\mathrm{s^{-1}}$$ meaning that its frequency is

$$\dfrac{\omega}{2\pi}=3.2\times10^8\mathrm{s^{-1}}$$

Note that it is unclear whether the question is simply asking for the angular frequency of the radiation, which was given to us in the question, although strictly speaking we are asked for frequency rather than angular frequency.

(d) The wavenumber of the electromagnetic radiation in a vacuum is

$$k=\dfrac{\omega}{c}$$ . When the radiation enters the dialectric it slows down by a factor of

$$\dfrac{1}{n}$$ although its frequency is unchanged, this means that its wavelength becomes proportionately smaller in the dialectric. The wavenumber is inversely proportional to the wavelength of the radiation so as the wavelength becomes smaller the wavenumber becomes larger. We find that

$$k_1=\dfrac{n_1\omega}{c}$$ where

$$n_1$$ is the refractive index of the dialectric.

k_1=\dfrac{n_1\omega}{c}=\dfrac{1.6\omega}{c}=10.7\mathrm{m^{-1}} $$ Note that is it useful to remember that the wavenumber

$$k=\dfrac{2\pi}{\lambda}$$ which shows the inverse relationship between wavenumber and wavelength.

Question 10
(a) The incident light is randomly polarized, the electric field of the incident light will have components in the scattering plane and normal to the scattering plane. At the Brewster angle any light polarized in the scattering plane will be completely transmitted, all reflected light will therefore be polarized normal to the scattering plane.

(b) For light incident on a dialectric surface at an angle of incidence

$$\theta_i$$ , the Brewster angle

$$\theta_b$$ is calculated as

\begin{align} \theta_b&=\arctan{\left(\dfrac{n_2}{n_1}\right)}\\ &=\arctan{\left(1.3\right)}\\ &=52.4^\circ \end{align} $$

(c) The scattering plane for sunlight reflected from a horizontal surface is vertical, reflected light will be mostly polarized normal to the scattering plane. Polaroid sunglassed have filters that are orientated to allow light polarized in the vertical direction to pass, which will block most of the reflected light.

Question 13
(a) Capacitance is the ability of a body to store charge for a given potential difference. The SI unit of capacitance is the farad (F).

(b) We assume that the charge of the solid sphere is uniformly distributed such that the electric field permeates radially outwards. Due to the symmetry of the charge distribution we know that the electric field has spherical symmetry.

Gauss' Law in integral form is

$$\int_S{\mathbf{E}\cdot\mathrm{d}\mathbf{S}}=\dfrac{Q_{enc}}{\epsilon_0}$$ and we choose our Guassian surface S to be a spherical surface of radius

$$r$$ with

$$a \leq r < b$$ and

$$Q_{enc}$$ is that charge contained within our Guassian surface, which is

$$Q$$ because surface S contains all of the solid sphere. The electric field points along

$$\mathbf{e}_r$$ which is in the same direction as the normal to surface S at all points on the Gaussian surface. We therefore have

\int_S{\mathbf{E}\cdot\mathrm{d}\mathbf{S}}=E\times 4\pi r^2=\dfrac{Q}{\epsilon_0} $$ We solve for

$$E$$

E=\dfrac{Q}{4\pi \epsilon_0 r^2} $$ and we have established that the electric field points radially outwards so we have, for the electric field

\mathbf{E\left(r\right)}=\dfrac{Q}{4\pi\epsilon_0 r^2}\mathbf{e}_r $$ (c) We calculate the capacitance of the system by the formula

$$C=\dfrac{Q}{\delta V}$$ where

$$\delta V$$ is the potential difference between the two plates of the capacitor (in this case between the solid sphere and the spherical shell). The spherical shell at radius

$$b$$ is earthed, we take the spherical shell to be our surface of zero potential. First we find

$$\delta V$$ by integrating the electric field found in part (b)

\begin{align} \delta V&=-\int_b^a{\mathbf{E}\cdot\mathrm{d}\mathbf{r}}\\ &=-\dfrac{Q}{4\pi\epsilon_0}\int_b^a{\dfrac{1}{r^2}dr}\\ &=\left[\dfrac{Q}{4\pi\epsilon_0 r}\right]_b^a\\ &=\dfrac{Q}{4\pi\epsilon_0}\left[\dfrac{1}{a}-\dfrac{1}{b}\right]\\ &=\dfrac{Q}{4\pi\epsilon_0}\left[\dfrac{b-a}{ab}\right] \end{align} $$ The solid sphere carries charge

$$Q$$ and the capacitance is calculated as

C=\dfrac{Q}{\delta V}=\dfrac{4\pi\epsilon_0ab}{b-a} $$ We can calculate the energy stored in the capacitor by applying the formula

$$Energy = \dfrac{1}{2}QV = \dfrac{1}{2}\dfrac{Q^2}{4\pi\epsilon_0}\left(\dfrac{b-a}{ab}\right)$$

Note that the long way around is to find the energy density of the field and then integrate over the volume between the solid sphere and the shell. Doing this would take much longer but we would arrive at exactly the same answer.

(d) In part (a) we calculated the electric field in the capacitor, the magnitude of which is

$$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$$ where

$$a \leq r < b$$ . We can see from this expression that the electric field takes on its maximum value when

$$r$$ is minimized which is when

$$r=a=0.01\mathrm{m}$$ such that

$$E=\dfrac{Q_{max}}{4\pi\epsilon_0\left(0.01^2\right)}=3\times 10^6\mathrm{Vm^{-1}}$$ . We solve for

$$Q_{max}$$ to find the maximum charge that can be placed on the solid sphere.

\begin{align} Q_{max}&=3\times 10^6\mathrm{Vm^{-1}}\times 4\pi\epsilon_0\times\left(0.01^2\right)\\ &=3.3\times 10^{-8}\mathrm{C} \end{align} $$ (e) In part (d) we found that the maximum charge possible on the solid sphere to avoid electrical breakdown was

$$Q_{max}$$ . In order to keep the electrical field strength below

$$3\times 10^6 \mathrm{Vm^{-1}}$$ whilst maintaining the charge at

$$Q_{max}$$ then from the expression

$$E=\dfrac{Q_{max}}{4\pi\epsilon_0 a^2}$$ we would require

$$a$$ to become larger. We see therefore that in order to increase the maximum charge possible on the solid sphere we should make the solid sphere larger.

Note that this conclusion makes sense if we consider why we see corona discharge at sharp points of conductors when an object is charged, the electric field strength is greatest for greater curvature, which for a sphere is where the radius is smallest (larger surface gradient). The reverse argument is true so for the answer in part (e) we see that if we increase the radius of the solid sphere we reduce the electric field strength for a given amount of charge, because the curvature is reduced at the larger radius.

Question 14
(a) The given current distribution possesses axial and translational symmetry. Using cylindrical coordinates,the resulting magnetic field will only have a component along the

$$\phi$$ direction and only depend on the distance from the central axis

$$r$$ . We therefore have

$$\mathbf{B\left(r\right)}=B_{\phi}\left(r\right)\mathbf{e}_\phi$$ . Ampere's Law relates the current distribution to the resulting magnetic field by

$$\int_C{\mathbf{B}\cdot\mathrm{d}\mathbf{l}}=\mu_0I_{pen}$$ where I_{pen} is the current that penetrates the open surface bound by the loop C. Inside the cylinder where

$$r < R$$ we have

$$I_{pen}=I\times\dfrac{r^2}{R^2}$$ because the current density is uniform in the cylinder. The line integral of

$$B$$ around loop C is e

$$B\times2\pi r$$ so we have

B\times 2\pi r = \mu_0 I\dfrac{r^2}{R^2} $$ and therefore

$$B=\dfrac{\mu_0 Ir}{2\pi R^2}$$ for

$$r < R$$ Outside of the cylinder where

$$r > R$$ we have

$$I_{pen}=I$$ because any open surface bound by a closed loop around the cylinder contains all of the cross-sectional area of the cylinder. We now have

B\times 2\pi r = \mu_0 I $$ and therefore

$$B=\dfrac{\mu_0 I}{2\pi r}$$ for

$$r > R$$ We see that when

$$r=R$$ at the surface of the cylinder both of these equations become

$$B=\dfrac{\mu_0I}{2\pi R}$$ which must be the case to avoid discontinuity at the surface of the cylinder.

(b) Inside the cylinder:



\begin{align} \text{curl}\left(\mathbf{B}\right)&=\dfrac{1}{r}\dfrac{\partial}{\partial r}\left(rB_{\phi}\right)\mathbf{e}_z\\ &=\dfrac{1}{r}\dfrac{\partial}{\partial r}\left(\dfrac{\mu_0 Ir^2}{2\pi R^2}\right)\mathbf{e}_z\\ &=\dfrac{\mu_0 I}{\pi R^2}\mathbf{e}_z \end{align} $$ According to Ampere's Law

$$\text{curl}\left(\mathbf{B}\right)=\mu_0 \mathbf{J}$$ so from the result above we see that

$$\mathbf{J}=\dfrac{I}{\pi R^2}\mathbf{e}_z$$ which is the current per unit of cross-sectional area flowing through the cylinder. The expression does not depend on

$$r$$ so the current density uniform within the cylinder, consistent with the information given in the question.

Outside the cylinder

\begin{align} \text{curl}\left(\mathbf{B}\right)&=\dfrac{1}{r}\dfrac{\partial}{\partial r}\left(rB_{\phi}\right)\mathbf{e}_z\\ &=\dfrac{1}{r}\dfrac{\partial}{\partial r}\left(\dfrac{\mu_0 I}{2\pi r}\right)\mathbf{e}_z\\ &=0 \end{align} $$ In this case we see from Ampere's Law that

$$\mu_0 I=0$$ outside of the cylinder which means that the current density is zero. (c) We need to find the magnitude of the force experienced by the section of the long thin wire of length

$$\mathrm{d}\mathbf{l}$$ . Imagine the long thin wire is above the cylinder at a distance

$$2R$$ from the central axis of the cylinder. From our result in part (b) we know that the magnetic field at a point on this wire will point out of the page in the

$$\mathbf{e}_\phi$$ direction. To make the situation easier to visualize let us consider the direction out of the page to be

$$\mathbf{e}_x$$ in a cartesian coordinate system with the current flowing along

$$\mathbf{e}_z$$ . If we point the fingers of our right hand along the direction of current flow with our fingers curling towards us (along

$$\mathbf{e}_x$$ ) then our thumb will point downwards towards the cylinder in the direction

$$\mathbf{e}_y$$ , which tells us the direction of the force experienced by a section of the long thin wire when the currents flow in the same direction along both conductors. This is consistent with the general principle that two parallel wires carrying current in the same direction will attract each other. To calculate the magnitude of the current per unit length we apply

$$F=I\mathrm{dl}B$$ which gives us

F=Idl\times B=Idl\times\dfrac{\mu_0I}{2\pi\times 2R}=\dfrac{\mu_0 I^2dl}{4\pi R} $$ The force per unit length is the same at all points on the thin wire so we divide by

$$\mathrm{dl}$$ to find that

$$\dfrac{F}{L}=\dfrac{\mu_0 I^2}{4\pi R}$$ and the force points along

$$\mathbf{e}_y$$ towards the cylinder so we have

\dfrac{\mathbf{F}}{L}=\dfrac{\mu_0 I^2}{4\pi R}\mathbf{e}_y $$

If the current in the wire was in the opposite direction then the force per unit length would be the same in magnitude but point in the opposite direction such that

\dfrac{\mathbf{F}}{L}=-\dfrac{\mu_0 I^2}{4\pi R}\mathbf{e}_y $$