A proton is initially at rest and moves through three...

A proton is initially at rest and moves through three different regions as shown in the figure. In region $1,$ the proton accelerates across a potential difference of $3330 \mathrm{V} .$ In region $2,$ there is a magnetic field of $1.20 \mathrm{T}$ pointing out of the page and an electric field (not shown) pointing perpendicular to the magnetic field and perpendicular to the proton's velocity. Finally, in region $3,$ there is no electric field, but just a $1.20 \mathrm{T}$ magnetic field pointing out of the page. (a) What is the speed of the proton as it leaves region 1 and enters region $2 ?$ (b) If the proton travels in a straight line through region 2 , what is the magnitude and direction of the electric field? (c) In region $3,$ does the proton follow path 1 or $2 ?$ (d) What is the radius of the circular path in region $3 ?$

A proton is initially at rest and moves through three different regions as shown in the figure. In region $1,$ the proton accelerates across a potential difference of $3330 \mathrm{V} .$ In region $2,$ there is a magnetic field of $1.20 \mathrm{T}$ pointing out of the page and an electric field (not shown) pointing perpendicular to the magnetic field and perpendicular to the proton's velocity. Finally, in region $3,$ there is no electric field, but just a $1.20 \mathrm{T}$ magnetic field pointing out of the page. (a) What is the speed of the proton as it leaves region 1 and enters region $2 ?$ (b) If the proton travels in a straight line through region 2 , what is the magnitude and direction of the electric field?
(c) In region $3,$ does the proton follow path 1 or $2 ?$
(d) What is the radius of the circular path in region $3 ?$

Key Concepts

Circular Motion in a Magnetic Field

When a charged particle moves in a region with only a magnetic field, it experiences a magnetic force perpendicular to its velocity. This force acts as a centripetal force, causing the particle to follow a circular path. By equating the magnetic force (qvB) to the centripetal force (mv²/r), one can determine the radius of the circular path.

Energy Conversion from Electric Potential Difference

This concept involves converting electric potential energy into kinetic energy. By applying the work-energy theorem, a charged particle that is accelerated across a potential difference gains kinetic energy equal to the product of its charge and the voltage. This energy conversion is essential for calculating the speed of the particle after acceleration.

Lorentz Force and Force Equilibrium

The Lorentz force law describes the net force on a charged particle moving in electric and magnetic fields, expressed as F = q(E + v × B). When a particle travels in a straight line in a region with both fields, the electric force and the magnetic force must cancel each other. This balance, often written as qE = qvB, is key to finding the required magnitude and direction of the electric field.

Right-Hand Rule for Direction Determination

The right-hand rule is used to determine the direction of forces resulting from the cross product in the Lorentz force equation. For a positively charged particle, pointing the fingers in the direction of velocity and curling them towards the magnetic field, the thumb gives the direction of the magnetic force. This rule is critical for understanding the geometry and orientation of particle trajectories in both the presence of magnetic fields alone and combined electric and magnetic fields.

Transcript

00:01 High in the given problem potential difference which is accelerating the proton is given as v is equal to 3 ,330 volt magnetic field in the region 2 is given as b is equal to 1 .20 tesla now in the first part of the problem here we have to find the speed of the proton when it enters from region 1 to region 2.

01:06 That will be given by using work energy theorem.

01:14 According to which the kinetic energy gained by the proton will be equal to work done on it by this potential which will be given.

01:25 By the product of charge over the proton with the potential difference through which it is being accelerated.

01:33 So using that we obtain an expression for the velocity for the speed achieved by this proton which comes out to be 2 times of e v divided by mass having a square root.

01:47 So finally plugging in all known values here this is 2 times of charge over an electron means 1 .6 into 10 dash per minus 19 multiplied by the potential 3 ,330 volt divided by mass of the proton, which is 1 .67 into 10 dash power minus 27 kilogram.

02:12 So here it comes out to be 6 ,380 .8 into 10 .8 into 10 dash power 8.

02:24 Meter per second it should be and finally it comes out to be approximately 8 .0 into 10 dash 5 meter per second which becomes the answer for the first part of the problem velocity achieved by the proton when it enters from region 1 to region 2 now in the second part of the problem we have to find the necessary electric field in magnitude and direction both in order to keep the proton undeflected in region 2.

03:02 So as the proton moves undeflected in region 2, it means there is no net force acting on it, means electrostatic force acting on it should be equal to magnetic chlorants force.

03:46 Acting on it now as the electron as the proton is moving perpendicular to the magnetic field so expression for electrostatic force this is just a product of charge over the proton with the electric field and for magnetic lawrence force it will be e into v cross b the cross product of velocity vector with the magnetic field vector so finally it will be e v b sine theta and for theta this is 90 degree and velocity vector and magnet field vector are perpendicular to each other...

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