A long, straight wire carrying a 2.00-A current lies along the \(x\) -axis. A particle with charge \(q=-3.00 \mu \mathrm{C}\) passes parallel to the \(z\) -axis through the point \((x, y, z)=(0,2,0)\). Where in the \(x y\) -plane should another long, straight wire be placed so that there is no magnetic force on the particle at the point where it crosses the plane?

Applying Biot-Savart Law, we can find the magnetic field due to the current carrying wire along the \(x\)-axis. The equation for the magnetic field at a point \((x,y,z)\) due to a current carrying wire along the \(x\)-axis is given by: $$ \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi r_1} \hat{k} $$ where \(\mu_0\) is the permeability of free space, \(I_1\) is the current in the wire, and \(r_1\) is the distance from the point to the wire (which is equal to \(y\) in this case). The direction of the magnetic field is in the \(z\)-axis direction (\(\hat{k}\)). Since the particle's position is given by \((x, y, z) = (0, 2, 0)\), the magnetic field at the charged particle due to the first wire is: $$ \vec{B}_1 = \frac{\mu_0 (2.00)}{2 \pi (2)} \hat{k} $$

Now let's calculate the magnetic force acting on the charged particle. The magnetic force on a charged particle is given by: $$ \vec{F} = q \,(\vec{v} \times \vec{B}) $$ where \(q\) is the charge of the particle, \(\vec{v}\) is the velocity vector, and \(\vec{B}\) is the magnetic field. Since the magnetic field is along the \(z\)-axis and the charged particle moves parallel to the \(z\)-axis, the velocity vector of the charged particle is given by: $$ \vec{v} = v \hat{k} $$ Substituting this into the equation for the magnetic force, we get: $$ \vec{F} = q (\vec{v} \times \vec{B}_1) = 0 $$ This means that there is no magnetic force acting on the particle in the \(z\)-direction due to the first wire.

Now we need to find the position of the additional wire in the \(xy\)-plane so that there is no net magnetic force on the charged particle. To cancel out the magnetic force due to the first wire, the additional wire must create a magnetic field of the same strength, but in the opposite direction. Therefore, let the position of the additional wire be \((x_2, y_2)\). The magnetic field due to the second wire at the charged particle is given by: $$ \vec{B}_2 = -\frac{\mu_0 I_2}{2 \pi r_2} \hat{k} $$ where \(I_2\) is the current in the second wire and \(r_2\) is the distance from the point to the second wire (which is equal to \(y_2\) in this case). The direction of the magnetic field is in the negative \(z\)-axis direction (\(-\hat{k}\)). For the net magnetic force on the particle to be zero, the magnitudes of the magnetic fields created by both wires must be equal. Therefore, we have: $$ \frac{\mu_0 I_1}{2 \pi r_1} = \frac{\mu_0 I_2}{2 \pi r_2} $$ Since the currents in both wires are the same (\(I_1 = I_2 = 2.00\) A), we can simplify the equation as follows: $$ r_1 = r_2 $$ This means that the distance from the charged particle to both wires must be equal. Since the distance from the charged particle to the first wire is \(y = 2\), the distance from the charged particle to the second wire in the \(xy\)-plane must also be \(2\) units. Therefore, the additional wire should be placed at either \((0, 0)\) or \((0, 4)\) in the \(xy\)-plane so that there is no magnetic force on the charged particle at the point \((0, 2, 0)\).