Tuesday, July 27, 2021

Motion of a charged particle in a uniform electric and magnetic field | Class 12 Physics Notes

 What are Motion of a charged particle in a uniform electric and magnetic field  ? it is most important topic of NCERT   in Class 12.  Questions are frequently asked in the CBSE  board ,   ICSE Board and other competitive exam ( IIT JEE, NEET,  AIIMS, State Engineering exam  from  Motion of a charged.

Motion of a charged particle class 12 ncert notes “ will be very beneficial for the students who are engaged in the preparation of  upcoming board exam and competitive exam.

In this  topic, the following terms will be illustrated.

Contents :

* Motion of a charged particle in a uniform electric field

* Motion of charged particle in uniform magnetic field

* Lorentz force

Let Us Start

Motion of a charged particle in a uniform electric and magnetic field | Class 12 Physics Notes


Motion of a charged particle in a uniform electric field 


* Consider that a charged particle of charge +q  and mass m enters at right angles to a uniform electric field of strength E with velocity v along  OX –axis.

* The electric field is along OY-axis and acts over a horizontal distance x.

* Since the electric field is along OY-axis, so only vertical  force will act along OY .

* No horizontal force acts on the charged particle entering the field, Therefore, the horizontal velocity v of the charged particle remains the same throughout the journey.

* Vertical Force on the charged particle,  Fy = qE

* Acceleration of the charged particle, along OY  ,  a = qE/m

* Time taken to traverse the field,   t = x / v

 In same time t , vertical distance covered by particle,


Where ,  

* This is the equation of a parabola. Therefore, inside the electric field, the charged particle follows a parabolic path OA.

* As the charged particle leaves the electric field at A, it follows a straight line path AB tangent to path OA at A.

Important point :

(i) When a charged particle moves perpendicular to a uniform electric field, it follows a parabolic path in the electric field.

(ii) When a charged particle of charge +q is at rest or moving in the direction of electric field, it will experience a force qE in the direction of E. As a result, charged particle is accelerated in the direction of E.

(iii) If the charged particle of charge +q is moving opposite to the direction of E, it will experience a retarding force -q E and hence the motion of the particle is retarded.


Motion of charged particle in uniform magnetic field

When a charged particle of charge +q and mass m moving with velocity v enters a uniform magnetic field B, it experiences a magnetic force F,  given by

      F = q  (v x B)

* The magnetic force F acts at right angles to the plane containing v and B.

* The magnitude of this force is F = qvB sinθ ,   where  θ  is angle between v and B .

 

Case 1-  When the initial velocity of the particle is parallel to the magnetic field

*  In this case,  θ = 00   

So ,   F = qvB sin0   = 0

* Thus in this case, the magnetic field does not exert any force on the charged particle. Therefore , the charged particle will continue to move parallel to the field.

 

Case 2 - When the initial velocity is anti parallel to the magnetic field

In this case ,    θ= 1800

So,   F = qvB sin1800  =  0

Therefore , the particle will continue to move in the original direction.

 

Case 3-  When the initial velocity of the particle is perpendicular to the magnetic field.

In this case ,    θ= 900

So,   F = qvB sin900  =   qvB

* In this condition, force acting on charge will be maximum.

So,      Fm  =  qvB

* The direction of force is perpendicular to direction of velocity.

Therefore, Force cannot change the speed of the charged particle; it only changes the direction of motion of the charged particle.

* As a result, the charged particle moves in a circle of radius r perpendicular to the field.

 


(a) To find radius of circular path

* Centripetal force required to maintain circular motion = qvB

Thus,      qvB =  mv2 / r

where ,    r=  radius of circular path

Thus,   r = mv2 /  qvB

-So,          r =  mv / qB

*  for given mass, charge and magnetic field ,      r  α  v

* it means, fast particle moves in large circle and slow particle moves in small circle.

 

(b)  To find time period of circular motion

Time period ,  T=  2πr / v

                       T =  ( 2π /v )  x   ( mv / qB )

     So ,           T=  2πm / qB

 

( c )  Frequency of revolution,

       f =  1/T   =  1 /  2πm / qB

so,     f =   qB / 2πm

 

(d)  Angular frequency, 

        ω =  2π / T

 So,   ω =  qB / m

* For constant uniform magnetic field,   Time period , frequency and angular frequency   of a charge do  not depend on speed (v )  and  radius ( r). 

* All charge particles  with same q/m value and moving in a uniform magnetic field will have same value of   T ,  f  and  ω  .

 

Case 4-   When the initial velocity of the charged particle makes an angle θ to the direction of magnetic field.

* Suppose the charged particle moving with velocity v enters in  a uniform magnetic field B making an angle θ to the direction of the magnetic field .


* The velocity v can be resolved into two perpendicular components

(a)  v1 = v cosθ  ,   acting in the direction of the field.

(b) v2 = v sinθ  ,    acting perpendicular to the direction of the field.

* The  component ( v2 = v sinθ  ) moves the charged particle in a circular path perpendicular to the field.

* The component ( v1 = v cosθ  )  moves charge in the direction of the magnetic field.

* Thus,  the charged particle covers circular path as well as linear path simultaneously   .

* In other words , circular path advance in direction of magnetic field with speed  v1 = v cosθ .  i.e   the charged particle follows a helical path .

 

*   Parameters of Helical motion,  

(a)  Radius of helical path

     r =  mvsinθ / qB

(b)  Time period of circular motion,

    T =  2πm / qB

(c)  Frequency ,     

     f =  qB / 2m

(d)  Angular Frequency,

     ω =  qB / m

 

Pitch of helix:


It is the linear distance covered by the charged particle when it completes one circular revolution i.e. it is the linear distance covered by the charged particle during time T.

So ,  Pitch of helix, 

  d =  velocity component along magnetic field  x   Time period

ie.     d =  v1  x  T

or,     d =   v cosθ. ( 2πm / qB)

 

Lorentz force 

The Net force experienced by a moving charged particle in a region of  electric and magnetic fields is called Lorentz force.

* Suppose a charge +q moves with velocity v in region of both electric field E and magnetic field B.

* Electric field exert a force on charged particle. This force acts in the direction of E and does not depend on the velocity of the charged particle.

Electric force,     Fe = qE

* Magnetic  field exert a force on moving charged particle . This force acts perpendicular to the plane containing v and B and depends on the velocity of the charged particle.

Magnetic force,    Fm =  q ( vx B )

* The total force experienced by the charged particle due to both electric and magnetic fields

F = Fe + Fm    (  Vector sum of electric and magnetic force )

Thus,      F = qE +q  ( V x B )

Or ,        F =  q  [ E + ( V x B ) ]

 


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