Home
Class 12
PHYSICS
A charged particle (electron or proton) ...

A charged particle (electron or proton) is introduced at the origin `(x = 0, y = 0, z = 0`) with a given initial velocity `vecv`. A uniform electric field `vecE` and a uniform magnetic field `vecB` exist everywhere. The velocity `vecv`, electric field `vecE` and magnetic field `vecB` are given in columns 1,2 and 3 respectively. The quantities `E_(0), B_(0)` are positive in magnitude.
`{:("Column-1","Column-2","Column-3"),("Electron with" vecv=2(E_(0))/(B_(0))hatx, (i)vecE=E_(0)vecz,(P) vecB=-B_(0)hatx),((II)"Electron with" vecv=(E_(0))/(B_(0))haty,(ii)vecE=-E_(0)hatx,(Q)vecB=B_(0)hatx),((VI)"Proton with" vecv=0,(iii)vecE=-E_(0)hatx,(R)vecB=B_(0)haty),((VI)"Proton with" vecv=2(E_(0))/(B_(0))hatx,(iv)vecE=E_(0)hatx,(S)vecB=B_(0)hat2):}`
In which case will the particle describe a helical path with axis along the positive z-direction ?

A

(IV) (i) (S)

B

(II) (ii) (R)

C

(III) (iii) (P)

D

(IV) (ii) (R)

Text Solution

AI Generated Solution

The correct Answer is:
To determine in which case the charged particle will describe a helical path with the axis along the positive z-direction, we need to analyze the given scenarios based on the relationship between the velocity vector \(\vec{v}\), electric field \(\vec{E}\), and magnetic field \(\vec{B}\). ### Step 1: Understand the conditions for a helical path A charged particle moves in a helical path when: 1. There is a component of velocity along the direction of the magnetic field. 2. The electric and magnetic forces must be perpendicular to each other. 3. The magnetic force provides the centripetal force necessary for circular motion, while the electric force provides the linear acceleration along the direction of the magnetic field. ### Step 2: Analyze each case Let's analyze each case provided in the columns: 1. **Case (i)**: - \(\vec{v} = \frac{2E_0}{B_0} \hat{x}\) - \(\vec{E} = E_0 \hat{z}\) - \(\vec{B} = -B_0 \hat{x}\) Here, the velocity is in the x-direction, the electric field is in the z-direction, and the magnetic field is also in the x-direction. The magnetic force will not contribute to motion in the z-direction, so this case cannot describe a helical path along the z-axis. 2. **Case (ii)**: - \(\vec{v} = \frac{E_0}{B_0} \hat{y}\) - \(\vec{E} = -E_0 \hat{x}\) - \(\vec{B} = B_0 \hat{x}\) The velocity is in the y-direction, the electric field is in the negative x-direction, and the magnetic field is in the x-direction. The electric force acts in the negative x-direction while the magnetic force acts in the z-direction (due to the cross product). This case does not have a component of velocity in the z-direction, so it cannot describe a helical path along the z-axis. 3. **Case (iii)**: - \(\vec{v} = 0\) - \(\vec{E} = -E_0 \hat{x}\) - \(\vec{B} = B_0 \hat{y}\) Here, the particle is initially at rest. Since the velocity is zero, it cannot describe any path, let alone a helical path. 4. **Case (iv)**: - \(\vec{v} = \frac{2E_0}{B_0} \hat{x}\) - \(\vec{E} = E_0 \hat{x}\) - \(\vec{B} = B_0 \hat{z}\) In this case, the velocity is in the x-direction, the electric field is also in the x-direction, and the magnetic field is in the z-direction. The electric force will act in the x-direction, while the magnetic force will act in the y-direction (due to the cross product of velocity and magnetic field). This case does not have a component of velocity in the z-direction, so it cannot describe a helical path along the z-axis. ### Conclusion None of the cases provided in the columns lead to a helical path along the positive z-direction. However, if we were to modify the conditions so that the velocity had a component in the z-direction while maintaining the perpendicularity of the forces, we could achieve a helical path.
Promotional Banner

Topper's Solved these Questions

  • ELECTROMAGNETIC INDUCTION

    MOTION|Exercise EXERCISE-4 (LEVEL-I)|11 Videos
  • ELECTRO MAGNETIC WAVES

    MOTION|Exercise EXERCISE - 3 (SECTION - B)|8 Videos
  • ELECTROMAGNETISM

    MOTION|Exercise Exercise|136 Videos

Similar Questions

Explore conceptually related problems

A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0 ) with a given initial velocity vecv . A uniform electric field vecE and a uniform magnetic field vecB exist everywhere. The velocity vecv , electric field vecE and magnetic field vecB are given in columns 1,2 and 3 respectively. The quantities E_(0), B_(0) are positive in magnitude. {:("Column-1","Column-2","Column-3"),("Electron with" vecv=2(E_(0))/(B_(0))hatx, (i)vecE=E_(0)vecz,(P) vecB=-B_(0)hatx),((II)"Electron with" vecv=(E_(0))/(B_(0))haty,(ii)vecE=-E_(0)hatx,(Q)vecB=B_(0)hatx),((VI)"Proton with" vecv=0,(iii)vecE=-E_(0)hatx,(R)vecB=B_(0)haty),((VI)"Proton with" vecv=2(E_(0))/(B_(0))hatx,(iv)vecE=E_(0)hatx,(S)vecB=B_(0)hat2):} In which case will the particle move in a straight line with constant velocity ?

A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0 ) with a given initial velocity vecv . A uniform electric field vecE and a uniform magnetic field vecB exist everywhere. The velocity vecv , electric field vecE and magnetic field vecB are given in columns 1,2 and 3 respectively. The quantities E_(0), B_(0) are positive in magnitude. {:("Column-1","Column-2","Column-3"),("Electron with" vecv=2(E_(0))/(B_(0))hatx, (i)vecE=E_(0)vecz,(P) vecB=-B_(0)hatx),((II)"Electron with" vecv=(E_(0))/(B_(0))haty,(ii)vecE=-E_(0)hatx,(Q)vecB=B_(0)hatx),((VI)"Proton with" vecv=0,(iii)vecE=-E_(0)hatx,(R)vecB=B_(0)haty),((VI)"Proton with" vecv=2(E_(0))/(B_(0))hatx,(iv)vecE=E_(0)hatx,(S)vecB=B_(0)hat2):} In which case would the particle move in a straight line along the negative direction of y-axis (i.e., move along haty)

If an electron is going in the direction of magnetic field vecB with the velocity of vecv then the force on electron is

The dimension of the quantity vecExxvecB , where vecE represents the electric field and vecB the magnetic field may be given as

In electromagnetic wave, if vecE,vecB and vecv are the electric field vector, magnetic field vector and velocity vector of electromagnetic wave, then

The dimensions of the quantity vecE xx vecB , where vecE represents the electric field and vecB the magnetic field may be given as :-

A homogenous electric field E and a uniform magnetic field vecB are pointing in the same direction. A proton is projected with its velocity parallel to vecE . It will

In an electromagnetic wave the electric field vector and magnetic field vector are given as vecE=E_(0)hati and vecB=B_(0)hatk respectively. The direction of propagation of electromagnetic wave is along: