Moving Charged Particle

Electricity and Magnetism Level 2

The above diagram shows a charged particle starting from position a , moving through two separate uniform electric fields in a uniform magnetic field, and eventually arriving at e . The direction of the magnetic field is upward, vertical to the plane. The path that the charged particle moves along is a $\to$ b $\to$ c $\to$ d $\to$ e . Which of the following explanations is correct?

a) The charged particle is positively charged.
b) The speed of the charged particle from position b to position c is constant.
c) The speed of the charged particle at position d is larger than that at position b .

c) only a) only a), b) and c) b) only

9 solutions

Shravan Jain
Mar 7, 2014

A) Option A is TRUE.

Explanation: Since the direction of motion of charged particle and that of Electric field is same, it implies that it is a positively charged particle.

B) Option B is TRUE.

Explanation: F=q (V x B). Since F is always perpendicular to Velocity , it can only change its direction and not its magnitude. Thus the magnitude of the velocity of the charged particle from position b to c does not change.

C)Option C is TRUE.

Explanation: F=qE. Since a force acts on the charged particle in the direction of motion, the particle gets accelerated and thus the velocity increases

I disagree from option C.because magnetic and electric field are uniform and how is this possible the magnitude of velocity is different at b and d?

Muhammad Arslan - 7 years, 3 months ago

Hey Muhammad, it is true that the magnetic field can't speed the particle up. However, the electric field does increase the kinetic energy of charged particles.

Since the particle passes through an electric field between points c and d , it should end up with a greater speed than it has at b or c .

Josh Silverman Staff - 7 years, 3 months ago

yup... that's right because the charge starts next half cycle with a greater kinetic energy like jerk.

Adnan Afzal - 7 years, 3 months ago

as we know, F=qE ie, ma=qE a=qE/m. since electric field produces acceleration in the direction of the motion of particle, the velocity of the particle would increase.

Shravan Jain - 7 years, 3 months ago

i think it will be solved in right hand thumb rule

Pratyush Kumar Rout - 7 years, 3 months ago

no need for using it as the resultant of cross product (say C ) of 2 vectors (say A and B) is always perpendicular to both the vectors (ie, C is always perpendicular to both A and B )

Shravan Jain - 7 years, 3 months ago

Can you get the direction along the mutually perpendicular direction using the mutually perpendicular argument alone? How would you determine the particle's charge?

Josh Silverman Staff - 7 years, 3 months ago

@Josh Silverman – we can determine the charge of the particle by observing that it is going in the direction of electric field which implies that it is a +ve charge (as F=qE. if q=+ve, F is in the direction of electric field and if q=-ve, f is in the opposite direction of electric field)

Shravan Jain - 7 years, 3 months ago

but a positive particle move in a direction of electic field

sristi guleria - 7 years, 3 months ago

that's what i have explained

Shravan Jain - 7 years, 3 months ago

i think all are true

zamir alam - 7 years, 3 months ago

i think Option C is true..

Adnan Afzal - 7 years, 3 months ago

all options are true

Shravan Jain - 7 years, 3 months ago

i have a question, you guys says that particle accelerate because of increase in its kinetic energy, but why we ignore a magnetic field although it is uniform , its direction is outward, so how the speed increases

salman Ahmed - 7 years, 3 months ago

Hey Salman, are you asking if the magnetic field increases the speed of the particle?

Josh Silverman Staff - 7 years, 3 months ago

no i just say that why the particle accelerate in uniform magnetic field

salman Ahmed - 7 years, 3 months ago

@Salman Ahmed – because the velocity changes its direction due to uniform magnetic field and change in velocity (either in direction or magnitude or both) is called acceleration. thus uniform magnetic field accelerates a moving charged particle

Shravan Jain - 7 years, 3 months ago

i does not agree with the option c in the presence of magnetic field

salman Ahmed - 7 years, 3 months ago

@Salman Ahmed – both magnetic field and electric field produces acceleration

but acceleration due to magnetic field changes the direction of velocity of the particle.

and acceleration due to electric field changes the magnitude of velocity of the particle

Shravan Jain - 7 years, 3 months ago

i have found the answer of my own question that because magnetic field produce deflection only, not any other effect, is it true

salman Ahmed - 7 years, 3 months ago

@Salman Ahmed – Yes, exactly. When the magnetic field is constant (stays the same over time), the only effect it can have on the particle is to change its direction.

Josh Silverman Staff - 7 years, 3 months ago

i think all options is true but i left a question in previous post for sharavan

salman Ahmed - 7 years, 3 months ago

how option b and c can be true at the same time

ranjeet kumar - 7 years, 3 months ago

because the magnitude of velocity does not change between path b to c but changes when the particle travels form c to d

Shravan Jain - 7 years, 3 months ago

its confusing

ahmad raza - 7 years, 2 months ago

Valentin Bogatu
Feb 27, 2014

The Force on a charge moving an a combined electric and magnetic field is

F = q*(E + v x B)

where x is the cross product and * in the dot product.

Hence between a and b and c and d because there is only electric parallel to the speed of particle field the particle will accelerate. F =q*E.

Between b and c and d and e there is only magnetic field perpendicular to the speed of particle. F =q(v x B). Hence the speed will only change direction. Also using the rule of right screw (for the cross product direction) we find the particle as being positively charged.

Sitesh Pattanaik
Mar 6, 2014

plz refer class 12 chapter physics electricity and magnetism for solution

Everybody does not have a CBSE class XII Physics book.

This setup is similar to that of Cyclotron.

Shravan Jain - 7 years, 3 months ago

Arijit Banerjee
Mar 6, 2014

bulleted First of all from " Fleming's Left Hand Rule" we get the direction of the current which shows that the charged particle is positive.
bulleted Next in a magnetic field kinetic energy of a moving charged particle does not change so velocity at' b 'and 'c' points is same.
bulleted Now when it again enters into the field it gains acceleration and hence velocity increases .

thanks

vaishnav garg - 7 years, 3 months ago

Senthil Kumar
Mar 12, 2014

The Time Period is constant and hence the charged particle moves with lower speed for lesser radius and moves with higher speed for larger radius

Yogesh Ghadge
Mar 12, 2014

is can be explained by the flemings left hand rule where it is said that the magnitude of velocity depends upon the magnetic field .and electric charge flows from positive to negative.

Chetan Yadav
Mar 11, 2014

a) particle is positively charged ,as per flemings left hand rule the particle shoud follow clockwise direction

b) the velocity from b to c dosen't change because it experience centrifugal force and so work done by the particle is zero( velocity perpendicular to displacement) and so momentum of the particle is also conserved

c) from c to d particle experience a force due to electric field in the same direction of motion and hence it's velocity should increse

Rushyanth Reddy
Mar 7, 2014

F=q(VxB).this force always acts towards the centre ..by using vectors we find the direction.so it has to be a positively charged particle.. from r=mv/qb..r is directly proportional to velocity..

Andre Yudhistika
Feb 26, 2014

in this situation Florentz=Fsentrifugar therefore

Bqv=mv^2/R hence

R=(mv)/Bq

the conclusion is the greater the radius, the greater the velocity

Moving Charged Particle

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