In Fig. 28-39, a charged particle moves into a region of...

In Fig. $28-39$, a charged particle moves into a region of uniform magnetic field $\vec{B}$, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends $130 \mathrm{~ns}$ in the region. (a) What is the magnitude of $\vec{B} ?$ (b) If the particle is sent back through the magnetic field (along the same initial path) but with $2.00$ times its previous kinetic energy, how much time does it spend in the field during this trip?

In Fig. $28-39$, a charged particle moves into a region of uniform magnetic field $\vec{B}$, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends $130 \mathrm{~ns}$ in the region.
(a) What is the magnitude of $\vec{B} ?$ (b) If the particle is sent back through the magnetic field (along the same initial path) but with $2.00$ times its previous kinetic energy, how much time does it spend in the field during this trip?

Key Concepts

Charged Particle Motion in Magnetic Fields

When a charged particle enters a uniform magnetic field, it experiences a force that is perpendicular to both its velocity and the magnetic field direction. This causes the particle to follow a curved, often circular, trajectory without changing its kinetic energy, provided there are no other forces acting on it.

Lorentz Force

The magnetic component of the Lorentz force, given by the cross product of the particle's velocity and the magnetic field vector (F = q(v × B)), is responsible for the particle’s curved motion in the magnetic field. The force’s magnitude depends on the charge, the speed of the particle, and the strength of the magnetic field.

Cyclotron Motion and Frequency

Charged particles moving in a uniform magnetic field undergo cyclotron motion, where they follow circular paths with a radius determined by the balance between the centripetal force and the magnetic Lorentz force. The cyclotron frequency, independent of the particle’s speed, is defined by the ratio of the charge to the mass multiplied by the magnetic field strength and governs the time period of the circular motion.

Relationship Between Kinetic Energy and Speed

The kinetic energy of a particle is related to its speed by the equation KE = 1/2 mv². Changes in kinetic energy directly affect the speed, which in turn influences the radius of the particle’s circular path and the time it takes to traverse a given arc in the magnetic field.

Time-of-Flight Through a Magnetic Field

The time a charged particle spends in a magnetic field while traversing a circular arc depends on both the length of the arc and its speed. Since the cyclotron motion is characterized by a constant angular frequency, any changes in speed or kinetic energy will alter the time-of-flight along a fixed path within the magnetic field.

Transcript

00:03 So the question says that there is a charge particle which has just entered this magnetic field here right now the charge particle enters the magnetic field that is below this line here then it makes a half circle and then wait a minute this arrow is up was and then it and then the particle goes out from here let me change the arrow upwards now what happens is we don't know whether this particle is a proton or an electron that we must decide it says that the time that the particle spins inside this magnetic grid is 130 nanoseconds right well this is a semicircle half circle so and the time it takes is 13 nanosecond so if it was a full circle then the time it would take is 260 seconds so nanoseconds so what we can say is the time period of the particle that is the time taken by the particle for one complete rotation is equal to two times 1 30 nanoseconds so this is equal to 260 intert 10 to 4 minus 9 seconds right so this is the time period of the particle so what we have to find out is the magnitude of b now equate let's do begin that equate the centipural force f with with the force to magnetic field that is mv squared by r is equal to b so this is an electron or proton that way i don't know yet so let's just write qb qv okay so basically we will need to find out v so let us first find out whether it is a proton or an electron so if you see in this figure here and consider a point here that i noted as in red the particle this point at this point the particle has velocity downwards the force is towards the center because you know cycle motion the force is towards the center so it is across the right right now the magnetic field at this point is out of the screen right because b is ours right so if you use the right here then and the force is on the left then what we can say that the electron is positive which are sorry the particle is positively charged right so what we can say the particle is positive charged particle is positive from right -hand rule so if this is a proton so for a proton mass of proton is 1 .67 this we may require sometimes into 20 minus 27 kg and the charge of a proton q is equal to is electronic charge e that is equal to plus 1 .6 into 10 power minus 19 all right okay so let us carry on here so let us cancel out…

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