Question

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?

Short answer

Assume that there is a long, straight wire carrying a current of 2.00 A along the x-axis. Answer: The additional wire should be placed at either (0, 0) or (0, 4) in the xy-plane to cancel out the magnetic force acting on the charged particle at the point (0, 2, 0).

Step-by-step solution

Biot-Savart Law for a Wire along the \(x\)-axis

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: $$ \vec{B}_1 = \frac{\mu_0 (2.00)}{2 \pi (2)} \hat{k} $$

Magnetic Force on The Charged Particle

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.

Determine the Position of the Additional 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 \((0, 2, 0)\).

Key concepts

Biot-Savart Law

The Biot-Savart Law helps us understand the magnetic field produced by a current-carrying wire. This fundamental law indicates that the magnetic field, denoted as \( \vec{B} \), around a current-carrying conductor is directly proportional to the current, \( I \) and inversely proportional to the distance \( r \) from the point to the current element.

The law's mathematical expression is \[ \vec{B} = \frac{\mu_0}{4\pi} \int \frac{I \, d\vec{l} \times \vec{r}}{r^3} \[\frac{T\cdot m}{A}\]\], where \( \mu_0 \) is the magnetic constant, \( d\vec{l} \) is a small segment of the wire carrying the current \( I \) and \( \vec{r} \) is the position vector from the current element to the point of interest.

In layman's terms, the law describes how current creates the magnetic field around it, with the field's strength decreasing as the distance from the wire increases. This is crucial when predicting how magnetic fields interact with charged particles nearby, prompting forces that could even create motion. For a straight, long wire, the Biot-Savart Law simplifies, and we can calculate the magnetic field strength at a particular point with ease.

Magnetic Field of a Current-Carrying Wire

The magnetic field produced by a current-carrying wire is not just a theoretical concept, but a real force field having a definite shape and direction. Using the Biot-Savart Law, for an infinitely long, straight wire, the magnetic field \(\vec{B}\) can be simplified as \[ \vec{B} = \frac{\mu_0 I}{2 \pi r} \hat{n} \[\frac{T}{A}\]\], where \(\mu_0\) is the permeability of free space, \(I\) the current flowing through the wire, and \(r\) the shortest distance to the point of measurement, with \(\hat{n}\) representing the direction of the field using the right-hand rule.

An important characteristic to note is that the magnetic field lines form concentric circles around the wire, with their direction determined by the direction of the current—fingers curled in the direction of current and thumb pointing along the field lines. The 'right-hand rule' is a mnemonic that helps us visualize the direction of the magnetic field with relation to the current's flow. This knowledge is vital for understanding how forces act on charged particles near current-carrying wires, and how we can manipulate these fields in various technological applications.

Cross Product in Magnetic Force

Magnetic force on a moving charge in a magnetic field is a classic example of the cross product in action. The force \( \vec{F} \) on a charged particle with charge \( q \) moving with velocity \( \vec{v} \) in a magnetic field \( \vec{B} \) is given by the equation \[ \vec{F} = q (\vec{v} \times \vec{B}) \[\frac{N}{C}\]\]. The cross product, represented by \( \times \), determines both the magnitude and direction of the force.

The cross product results in a vector that is orthogonal (at a right angle) to the plane formed by the two vectors being multiplied. This means the direction of the force is perpendicular to both the velocity of the particle and the magnetic field. The magnitude of this force is proportional to the charge, the speed of the particle, the strength of the magnetic field, and the sine of the angle between the velocity and the field. If the charged particle moves parallel or antiparallel to the magnetic field, the angle between \( \vec{v} \) and \( \vec{B} \) is 0° or 180° respectively, making the sine of the angle zero – as such, the particle would experience no magnetic force. This principle is applied when ensuring that our charged particle in the exercise moves in such a way as to experience no magnetic force due to the presence of the currents in the wires.