Proton in a magnetic field. A proton (e=4.80 ×.10^-10 esu⟩is projected with a velocity vector 𝐯=

Proton in a magnetic field. A proton (e=4.80 ×.10^-10 esu⟩is...

Proton in a magnetic field. A proton $(e=4.80 \times$ $\left.10^{-10} \mathrm{esu}\right\rangle$ is projected with a velocity vector $\mathbf{v}=2 \times 10^{8} \hat{\mathbf{x}} \mathrm{cm} / \mathrm{s}$ into a region where a uniform magnetic field exists described by $\mathbf{B}=1000 \hat{\mathbf{z}} \mathrm{G}$ (a) Evaluate the force (in magnitude and direction) acting on the proton immediately after its projection. (b) What is the radius of curvature of its subsequent path? (c) Locate the position of the center of its circular path if the projection point is the origin.

Proton in a magnetic field. A proton $(e=4.80 \times$ $\left.10^{-10} \mathrm{esu}\right\rangle$ is projected with a velocity vector $\mathbf{v}=2 \times 10^{8} \hat{\mathbf{x}} \mathrm{cm} / \mathrm{s}$
into a region where a uniform magnetic field exists described by $\mathbf{B}=1000 \hat{\mathbf{z}} \mathrm{G}$
(a) Evaluate the force (in magnitude and direction) acting on the proton immediately after its projection.
(b) What is the radius of curvature of its subsequent path?
(c) Locate the position of the center of its circular path if the projection point is the origin.

Key Concepts

Determining the Center of Circular Motion

Once the particle starts moving in a circle due to the magnetic force, the center of the circle can be found by using the geometry of circular motion. The center lies in a direction perpendicular to the velocity at the starting point, at a distance equal to the radius of the circular path, which is determined by the balance between the magnetic and centripetal forces.

Circular Motion of Charged Particles

When the velocity of a charged particle is perpendicular to a uniform magnetic field, the resulting magnetic force acts as a centripetal force, causing the particle to move in a circular path. By equating the magnetic force (qvB) to the centripetal force (mv²/r), one can derive the radius of the circular path (r = mv/(qB)).

Magnetic Force on a Moving Charge

This concept explains that a charged particle in a magnetic field experiences a force given by the Lorentz force law, F = q(v × B). The magnitude of this force depends on the charge, the speed of the particle, the strength of the magnetic field, and the sine of the angle between the velocity and the magnetic field vectors.

Right-Hand Rule

The right-hand rule is a method used to determine the direction of the force experienced by a moving charge in a magnetic field. With your right hand, you point your fingers in the direction of the velocity, curl them toward the magnetic field direction, and your thumb then points in the direction of the magnetic force.

Transcript

00:01 Hello everyone.

00:02 So today we are going to solve this problem which is a photon in a magnetic field.

00:09 So first we need to learn the concept when a particle thrown in a magnetic field with charge q.

00:19 Let's say this is our magnetic field v and a particle is thrown perpendicular to bit like this as well as being in this direction.

00:29 If this is perpendicular to the magnetic field, like this is our partition coordinate system, and if our magnetic field will be in this direction, and if speed of particle in this direction, then this will perform circular motion in this plane, perpendicular, side cap, and the cap plane with radius r.

01:03 So here first we will convert all these values in ss units which is standard units.

01:15 So charge of proton q plus will be 1 .6 multiplied and is to minus 90 meters velocity will be 2 multiplied and is 2 of 6 meter per second and magnetic will be 0 .1 tesla because 1 cost is equal to 10x2 over minus 4 tesla which is in pzc direction and this is in 5 cap direction and this is the charge in cycle.

01:57 Now first we need to evaluate the force so when a charged particle is thrown in the magnetic field the force acting on it is app as opposed to q into v cross b, which is v cross b.

02:23 So we write two times b vector which is 2 multiplied times 6, icap, plus magnetic field which is 0 times 1 in the direction of c cap.

02:41 So we know that from cartesian coordinate system this is over x mile and this is our z cap then i cross zp means i cross z will be a direction of negative y cap so we get 0 .2 q predicate and is 2 over 6 in the direction of minus of j cap this is our force so if we want to negative then you get 0 .2 .2 multiply 1 .6 will decline 10 to 0 .90 and addition level 10 is over 6 in your minus interaction or jackets since we want to the magnetite so we don't need the jquery directions because we want to the magnitude so the magnitude will come up to be as usual street three point two multiply 10 is so at minus 14 meter so we just another fourth now odd bit meaning the radius of the circular path actobly so the radius of the peripheral path will be given as huge right the radius of the circle path is given as m b divided by two times b that m is the mass of photo so the mass of proton is one quite c 7 magnetic light and so minus 27 kases times speed speed is through myetic light and rest 2 over 6 in a charge and myelotube so we do our calculation where it comes up to be 0 .2 .0000 so this is the release of this path in which it will be thrown is moved.

05:18 Now what's the third path? third part says, mitigate the position of center of its cellular path if the projection point is original...

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