Question

A long, straight wire is located along the \(x\) -axis \((y=0\) and \(z=0)\). The wire carries a current of \(7.00 \mathrm{~A}\) in the positive \(x\) -direction. What is the magnitude and the direction of the force on a particle with a charge of 9.00 C located at \((+1.00 \mathrm{~m},+2.00 \mathrm{~m}, 0),\) when it has a velocity of \(3000 . \mathrm{m} / \mathrm{s}\) in each of the following directions? a) the positive \(x\) -direction c) the negative \(z\) -direction b) the positive \(y\) -direction

Short answer

Answer: The resulting magnitudes and directions of the forces on the charged particle are: a) 0 N b) 37.8 N in the positive z direction c) 37.8 N in the negative y direction.

Step-by-step solution

Determine the magnetic field produced by the wire

We can determine the magnetic field produced by the wire using the formula: $$\vec{B} = \frac{\mu_0 I}{2 \pi r}\hat{θ}$$ In this case, the current \(I = 7\,\mathrm{A}\), \(r = 2\,\mathrm{m}\) and \(\mu_0 = 4\pi\times10^{-7} \,\mathrm{T\cdot m/A}\) (permeability of free space).

Calculate the magnetic field due to the wire

Plug the values from Step 1 into the formula to find the magnetic field: $$\vec{B} = \frac{4\pi\times10^{-7} \,\mathrm{T\cdot m/A}\cdot 7\,\mathrm{A}}{2\pi\cdot2\,\mathrm{m}}\hat{θ} = 1.4\times10^{-6} \,\mathrm{T}\cdot\hat{θ}$$ The magnetic field at the position of the charged particle has a magnitude of \(1.4\times10^{-6}\,\mathrm{T}\) and points in the \(θ\) direction, which is in the positive \(z\) direction in the given coordinate system.

Determine the force for each velocity direction

In each case, use the formula \(\vec{F} = q\vec{v} \times \vec{B}\), with \(q = 9.00\,\mathrm{C}\) and the magnetic field from Step 2 to find the force on the particle due to its velocity in each direction. a) $$\vec{F_{x}} = q\vec{v_{x}} \times \vec{B} = (9.00\,\mathrm{C})(3000\,\mathrm{m/s})\hat{i} \times (1.4\times10^{-6}\,\mathrm{T}\hat{θ}) = 0 \,\mathrm{N}$$ Since the velocity is parallel to the magnetic field in this case, the force is zero. b) $$\vec{F_{y}} = q\vec{v_{y}} \times \vec{B} = (9.00\,\mathrm{C})(3000\,\mathrm{m/s})\hat{j} \times (1.4\times10^{-6}\,\mathrm{T}\hat{θ}) = 37.8\,\mathrm{N}$$ The force is in the \(\hat{k}\) direction (positive \(z\) direction) with a magnitude of \(37.8\,\mathrm{N}\). c) $$\vec{F_{z}} = q\vec{v_{z}} \times \vec{B} = (9.00\,\mathrm{C})(-3000\,\mathrm{m/s})\hat{k} \times (1.4\times10^{-6}\,\mathrm{T}\hat{θ}) = -37.8\,\mathrm{N}$$ The force is in the \(\hat{j}\) direction (negative \(y\) direction) with a magnitude of \(37.8\,\mathrm{N}\). The resulting magnitudes and directions of the forces on the charged particle for each given velocity direction are: a) 0 N b) 37.8 N in the positive \(z\) direction c) 37.8 N in the negative \(y\) direction.

Key concepts

Biot-Savart Law

The Biot-Savart Law is a fundamental principle in electromagnetism that describes how current flowing through a conductor creates a magnetic field around it. According to this law, the magnetic field \textbf{dB} at a point in space due to an infinitesimal segment of current-carrying wire \textbf{dl} carrying a current I is directly proportional to the current and the length of the wire segment, and inversely proportional to the square of the distance r from the segment to the point in space.

This relationship is mathematically represented as: \[\begin{equation}\vec{dB} = \frac{\mu_0}{4\pi} \frac{I \vec{dl} \times \vec{r}}{r^3}\end{equation}\]where \(\mu_0\) is the permeability of free space, \(\vec{dl}\) is the direction of the current, and \(\vec{r}\) is the position vector from the current element to the point of interest. Because the magnetic field contribution from each segment of wire is a vector quantity, the total magnetic field at a point is the vector sum of the fields due to each segment.

For practical calculations involving long, straight wires, the Biot-Savart Law simplifies to:\[\vec{B} = \frac{\mu_0 I}{2\pi r}\hat{\theta}\]The simplified formula provides a straightforward way to calculate the magnetic field produced by a straight wire, as was done in the initial step of the exercise in question.

Lorentz Force

The Lorentz force describes the force experienced by a charged particle moving through a magnetic field. This force is central to our understanding of the behavior of charged particles in magnetic fields, a concept that has widespread applications, from the paths of electrons in a magnetic field to the operation of electrical motors.

The Lorentz force equation is given by:\[\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\]In this formula, \(\vec{F}\) is the force on the charged particle, \(q\) is the charge of the particle, \(\vec{E}\) is the electric field, \(\vec{v}\) is the velocity of the particle, and \(\vec{B}\) is the magnetic field. In situations where there is no electric field, such as in the provided textbook exercise, the equation reduces to:\[\vec{F} = q(\vec{v} \times \vec{B})\]Here, the velocity of the particle and the magnetic field interact to produce a force that is perpendicular to both the velocity vector and the magnetic field vector. This force's direction and magnitude can be determined using the right-hand rule and the magnitude of the vectors involved.

Magnetic Field of a Current-Carrying Wire

The magnetic field of a current-carrying wire is fundamental in explaining the magnetic effects of electric currents. This magnetic field is the result of the moving charges in the wire, and it encircles the wire in a direction that can be determined by the right-hand grip rule: if you grip the wire with your right hand so that your thumb points in the direction of current flow, your fingers will curl in the direction of the magnetic field.

For a long, straight wire carrying a current \(I\), the magnitude of the magnetic field at any point distance \(r\) from the wire is given by:\[\vec{B} = \frac{\mu_0 I}{2\pi r}\]The magnetic field decreases with increasing distance from the wire and has no component along the length of the wire; it only has a circular component around the wire. As associated with the problem under discussion, knowing the magnetic field behavior around a current-carrying wire enables us to predict the interaction of this magnetic field with a charged particle near the wire.

Cross Product in Magnetic Force

The concept of the cross product is critical when dealing with the magnetic force on a moving charged particle. The cross product is a vector operation that takes two vectors and returns a third vector that is perpendicular to the plane containing the first two.

Mathematically, for two vectors \(\vec{A}\) and \(\vec{B}\), their cross product \(\vec{C}\) is given by:\[\vec{C} = \vec{A} \times \vec{B}\]In the context of magnetic force, when a charged particle with charge \(q\) moves with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\), the cross product is used to calculate the force \(\vec{F}\) acting on the particle. This force is given by the equation:\[\vec{F} = q(\vec{v} \times \vec{B})\]The direction of \(\vec{F}\) is orthogonal to the plane formed by \(\vec{v}\) and \(\vec{B}\), and its magnitude is \(|\vec{F}| = qvB\sin(\theta)\), where \(\theta\) is the angle between the velocity and the magnetic field vectors. If the velocity vector is parallel or antiparallel to the magnetic field, as in part (a) of the exercise, there is no force because \(\sin(0\degree)\) or \(\sin(180\degree)\) is zero. This principle is essential to understanding magnetic interactions and is elegantly demonstrated in the problem solved above.