A long, straight wire lies along the x -axis and carries...

A long, straight wire lies along the $x$ -axis and carries current $I=60.0 \mathrm{~A}$ in the $+x$ -direction. A small particle with mass $3.00 \times 10^{-6} \mathrm{~kg}$ and charge $8.00 \times 10^{-3} \mathrm{C}$ is traveling in the vicinity of the wire. At an instant when the particle is on the $y$ -axis at $y=8.00 \mathrm{~cm},$ its acceleration has components $a_{x}=-5.00 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}$ and $a_{y}=+9.00 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2} .$ At that instant what are the $x$ - and $y$ -components of the velocity of the particle?

A long, straight wire lies along the $x$ -axis and carries current $I=60.0 \mathrm{~A}$ in the $+x$ -direction. A small particle with mass $3.00 \times 10^{-6} \mathrm{~kg}$ and charge $8.00 \times 10^{-3} \mathrm{C}$ is traveling in the vicinity of the wire. At an instant when the particle is on the $y$ -axis at $y=8.00 \mathrm{~cm},$ its acceleration has components $a_{x}=-5.00 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}$
and $a_{y}=+9.00 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2} .$ At that instant what are the $x$ - and $y$ -components of the velocity of the particle?

Key Concepts

Newton's Second Law in the Context of Charged Particle Motion

Newton's second law, F = ma, links the net force acting on a particle to its acceleration. In problems involving magnetic forces, equating the magnetic force components (obtained from the cross product) to m times the corresponding acceleration components allows for the determination of unknown quantities, such as the velocity components of the charged particle.

Vector Cross Product and Component Decomposition

The use of the cross product in F = q(v × B) necessitates breaking down both the velocity and magnetic field vectors into their components. This process allows us to analyze the effects of the magnetic force in each coordinate direction separately, providing a means to relate given acceleration components to the unknown velocity components.

Lorentz Force on a Charged Particle

A moving charged particle in a magnetic field experiences the Lorentz force, given by F = q(v × B), where q is the charge, v is the velocity vector, and B is the magnetic field vector. This force is always perpendicular to both the velocity of the particle and the magnetic field, affecting only its direction of motion and not its speed.

Magnetic Field due to a Long Straight Conductor

A long, straight current?carrying wire produces a magnetic field whose magnitude is given by ??I/(2?r), where ?? is the permeability of free space, I is the current, and r is the radial distance from the wire. The direction of this field is determined by the right-hand rule, meaning that if you point your thumb along the direction of conventional current, your fingers curl in the direction of the magnetic field lines.

Transcript

00:01 Okay, in this question we're talking about the magnetic force exerted on a particle, which is near a current.

00:10 So what we're given is the distance that the particle is from the wire carrying the current and the acceleration that the particle is experiencing.

00:21 And what we want to find is the velocity of the particle at that time.

00:27 So again, what we're given is the acceleration and the acceleration x direction is minus 5 kilometers per second, and the acceleration in the y direction is 9 kilometers per second.

00:44 All right, we're given the distance that the particle is from the current, r is 0 .08 meters.

00:52 And then we're also given the characteristics of a particle.

00:56 Its mass is 3 times 10 to minus 6 kilograms.

01:03 And its charge is eight times 10 to the negative three grew lumps.

01:11 Okay, so we have to come up with a strategy for solving this problem.

01:17 So what we have is acceleration and what we need is velocity.

01:21 So if you remember the formula for the force created by a magnetic field is equal to qv cross b.

01:31 And that's useful because it has velocity.

01:35 And so we can solve for v if we know what the force is, if we know the charges, know what we know what the magnetic field is.

01:42 We can figure out what the magnetic field is using this formula, or b is equal to nu not i over 2 pi times the direction, or 2 pi r times the direction.

02:02 We are given the charge, so we just have to set this force equal to another force so that we can have all three of these values be known in solve for velocity.

02:14 So we'll remember newton's second law, force is equal to mass time acceleration.

02:22 And thankfully we were given the acceleration.

02:24 So the force, the particle is experiencing, is equal to m times minus five kilometers per second, nine kilometers per second, and then zero in the c direction.

02:40 So that vector is going to be something like this angle...

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