A proton traveling at 23.0^∘ with respect to the direction of a magnetic field of strength 2.60

step 1 lyrics. So this video covers 28.1 problem one. Um, and basically, what we have here is some type

A proton traveling at 23.0^∘ with respect to the direction...

A proton traveling at $23.0^{\circ}$ with respect to the direction of a magnetic field of strength $2.60 \mathrm{mT}$ experiences a magnetic force of $6.50 \times 10^{-17} \mathrm{~N}$. Calculate (a) the proton's speed and
(b) its kinetic energy in electron-volts.

Key Concepts

Unit Conversion (Joules to Electron-Volts)

This concept involves converting energy values from Joules to electron-volts (eV), a common practice in atomic and particle physics. One electron-volt is defined as the amount of kinetic energy gained or lost by an electron when it moves through an electric potential difference of one volt, and knowing how to convert between these units is key in many physics applications.

Kinetic Energy of a Particle

Kinetic energy is the energy that a particle possesses due to its motion; it is calculated using the formula KE = 0.5mv², where m is the mass and v is the velocity of the particle. Understanding this concept is essential when evaluating the energy state of particles after interactions such as those involving magnetic forces.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving in a magnetic field. It is given by the equation F = qvBsin(?), where q is the charge, v is the speed, B is the magnetic field strength, and ? is the angle between the velocity and the magnetic field. This concept is crucial for understanding how magnetic fields influence the motion of charged particles.

Trigonometric Component in Magnetic Force

In problems involving the magnetic force on a charged particle, the angle between the particle’s velocity and the magnetic field must be accounted for by including the sine of that angle. This adjustment ensures that only the component of the velocity perpendicular to the magnetic field is considered when calculating the force.

Transcript

00:03 Lyrics.

00:04 So this video covers 28.1 problem one.

00:09 Um, and basically, what we have here is some type of magnetic field which will we will be and a particle, um, in this case, a proton traveling at a velocity b with an angle data, um, in between the magnetic field and the velocity.

00:31 So from the problem, we can just go ahead and write down everything we know.

00:37 We know that on the angles data is we know that the angle theta iss 23 degrees.

00:53 Um, the magnitude of the magnetic field is 2.6 times 10 to the negative three test, huh? on the magnetic force is given as 6.5 times 10 to the negative 17 newtons.

01:14 And we know that the charge of a proton is positive 1.6 times 10 to the negative 19 coghlan's.

01:26 And of course, it's the mass of the proton is 1.67 times 10 to the negative 27 kilograms.

01:36 Now, q and m weren't given explosively in the problem.

01:38 But since you know that it's a proton, those air values you could look up on google.

01:43 Um, so, first of all, we are trying to find ah, proton speed on dhe.

01:51 It's important to note that speed is a magnitude, so it's the magnitude of velocity, so our lives are a little bit easier because of that.

02:00 We know that the magnetic force f equals q.

02:06 Times v cross be when it's just equals.

02:12 Q.

02:13 The b the magnitudes sign data so we can solve her v...

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