An electron has an initial velocity of (12.0 ĵ+15.0 k̂) km / s and a constant acceleration of (2

An electron has an initial velocity of (12.0 ĵ+15.0 k̂) km /...

An electron has an initial velocity of $(12.0 \hat{\mathrm{j}}+15.0 \hat{\mathrm{k}}) \mathrm{km} / \mathrm{s}$ and a constant acceleration of $\left(2.00 \times 10^{12} \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{in}}$ a region in which uniform electric and magnetic fields are present. If $B=(400 \mu \mathrm{T}) \hat{\mathrm{i}}$ , find the electric field $\vec{E} .$

An electron has an initial velocity of $(12.0 \hat{\mathrm{j}}+15.0 \hat{\mathrm{k}}) \mathrm{km} / \mathrm{s}$ and a constant acceleration of $\left(2.00 \times 10^{12} \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{in}}$ a region in which
uniform electric and magnetic fields are present. If $B=(400 \mu \mathrm{T}) \hat{\mathrm{i}}$ ,
find the electric field $\vec{E} .$

Key Concepts

Electric Field Influence on Charged Particles

The electric field exerts a force on a charged particle directly proportional to the charge and the field strength, independent of the particle’s velocity. This force is given by qE and can accelerate the particle along the direction of the electric field. When a charged particle moves in both electric and magnetic fields, the net behavior is determined by the combined effects of both types of forces.

Magnetic Force and the Cross Product

The magnetic component of the Lorentz force, given by q(v × B), involves the cross product of the velocity and the magnetic field vectors. This cross product yields a vector that is perpendicular to both v and B, meaning that the direction and magnitude of the magnetic force depend on the orientation of the particle's motion relative to the magnetic field.

Lorentz Force

The Lorentz force is the net force experienced by a charged particle moving in electric and magnetic fields. It is given by the vector sum q(E + v × B), where q is the charge, E is the electric field, v is the particle's velocity, and B is the magnetic field. This concept is fundamental for understanding how charged particles are accelerated and deflected in electromagnetic fields.

Newton’s Second Law for Charged Particles

Newton’s second law, F = m*a, connects the force acting on a particle to its acceleration. When applied to charged particles in electromagnetic fields, it allows one to relate the net Lorentz force to the particle’s mass and acceleration, thereby facilitating the determination of unknown quantities such as the electric field when the acceleration is known.

Transcript

00:01 Okay, folks, so in this problem we have an electron having an initial velocity of blah blah blah and a constant acceleration of blah blah blah in a region in which uniform electric field and magnetic fields are present.

00:16 And if b equals 400 mu tesla in the ihat direction, find the electric field e.

00:23 So this is a relatively straightforward problem conceptually speaking.

00:28 All we need to know is the following learned force.

00:31 Equation, which is that force is equal to q v cross b plus e.

00:40 So this is a formula that tells us how much force is acting on a particle, having charged q, moving in a velocity, moving with a velocity in a background of a constant magnetic field and an e field.

00:56 It doesn't have to be constant.

00:58 Okay, the b field and e field could be, could could both be varying with space or with time, but whatever.

01:05 This is the formula, and according to newton's second law, this force is always equal to mass of the particle times the acceleration of the particle.

01:17 And what we're given is we're given the initial velocity, and we're given a constant acceleration, and we're also given the b.

01:33 Field.

01:35 Okay.

01:35 And what we're looking for is the electric field e.

01:42 So this is a relatively straightforward problem to solve because all we're really doing is we're solving this equation.

01:50 Okay.

01:51 So let's do it.

01:52 So we have, let's simplify this a little bit.

01:56 V cross b plus e equals m over q times a.

02:04 That means e equals m over q times a minus b cross.

02:17 This in principle is our answer to this problem.

02:21 If you plug in all of the numbers, all of the known numbers, for example, velocity, b field, acceleration, you will get a vector, which is the e field, which is what we're looking for.

02:41 Since i want to be a little bit more thorough than that, i don't want to leave you stranded.

02:47 I'm going to actually do the calculation for you...

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