Tag: Magnetic effect of current

CYCLOTRON

In this topic we will, discuss about working of cyclotron , its principle, construction and its limitations.  (** this topic is removed from the syllabus of class 12th, board examination).To know the complete syllabus click on the given link-

Before to know about the cyclotron students must know  Force Acting on a Charge Particle Moving in a Uniform Magnetic Field; To get the notes on this topic click here-

Cyclotron-

It is a device used to accelerate positive charged particles like proton, deutron , alpha particles etc . with desired velocity.

Principle– Its working based on the principle that a positive charge particle can be accelerated to a sufficient high energy with the help of oscillating electric field with the help of perpendicular strong magnetic field .

Construction– It consist two D-shaped chamber ( Dees)  D₁ and D₂ , which are placed slightly separated with each other between two poles of the magnet N- pole and S–pole ( as shown in figure. ) Dees are connected with an high frequency oscillator ( cause of variable electric field).

Working and theory –  suppose initially the D₁ is negative and D₂ is positively charged , then the positive charged particle at point P starts to move towards D₁ , but due to perpendicular magnetic field it starts moving on a circular path. If r is the radius of the circular path then ,

mv²/r  = qvB

So r = mv/qB ;

Time taken by the charged particle to complete half cycle  t= ∏r/v = ∏m/qB ;

When particle reaches to D₁ at the same time D₁ becomes positively charge and D₂ negatively then again charge particle starts to move towards D₂ with the circular path with greater radius . similarly charge particle remain describing the circular path with greater radius and greater speed and at an instant it ejected through the window with high kinetic energy .

Maximum energy of charged particle–  suppose v₀ , r₀ is the  is the maximum velocity and maximum  radius of the circular path followed by the charged particle ,

Then ,m v₀ ²/ r₀ = B q v₀  or ,  v₀ = Bq r₀/m ,

Therefore maximum kinetic energy  k = ½ mv₀² = B² q² r₀² / 2m ;

Cyclotron frequency –

Angular velocity  ω=  Bqr/mr = Bq/m ……………..(3)

here frequency  f=ω/2∏ = Bq/2∏m  ( here frequency is independent of velocity)………….(4)

limitations of cyclotron-

1. When a positive ion is accelerated by the cyclotron , it moves with greater speed when it becomes closer to speed of light or comparable with speed of light , then mass of the particle increases according to the given relation m= m₀ /√(1-v²/c²) where m₀ is the mass of the charged particle , v is the speed of charged particle , c is the speed of light . But cyclotron doesn’t discuss about variable masses .
2. Cyclotron is suitable for accelerating heavy particles only . i.e electron can not be accelerated through cyclotron .
3. The uncharged particle can not be accelerated by the cyclotron.

To watch the video on motion of a charged particle in magnetic field  click here-

 

 

 

 

 

 

 

State and explain Biot-Savart’s law

In this topic we will discuss about Biot-savart’s law and magnetic field at a point due to finite current carrying wire or conductor.

Biot- Savart law – Biot-Savart’s law explain the magnetic field due to small current carrying element

As we know , whenever the current passes through a conductor magnetic field setup around it  . To find the magnetic field at any point first given by Jean Biot and Felix savart . according to Biot- Savart , magnetic field intensity at any point due to small current carrying element is given  as dB=µ₀idl sinθ/r² .

EXPLANATION- Let us consider a small element of length  ‘dl’ carrying a current I through it  . If dB is the magnetic field at a point r distance away from the element , then according to this law

• dB α I …………………….1
•  dB α dl …………………2
•   dB α sin θ …………….3
•   dB α 1/r² ………………4

after combining these equations ,we get

dB  α   I dl sinθ/r²
so ,   dB= k I dl sinθ/r²

(Where K= µ₀/4Π = 10^-7 Wb/ (A m ))

Magnetic field due to straight current carrying wire of finite length –****( derivation is not in syllabus for 12^th board examination, but we must know the result )

Suppose  current ‘I’ is flowing through  a wire XY , and we have to find the magnetic field at point ‘P’.as shown in figure .

 

 B =  (µ₀I/4Πa)(sinɸ₁ + sinɸ₂) ,

Special cases – Case-(1) —  When the wire is of infinite length and point P is near the wire in that case                        ɸ₁=90⁰ , ɸ₂=90⁰ ;

So magnetic field at that point P ,   B =  (µ₀I/4Πa)(sin90 + sin90)  =  µ₀I/2Πa

Case-(2)–  When the wire is of infinite length and point P is near the  one end of the wire in that case                        ɸ₁=90⁰ , ɸ₂=0⁰ ;

So magnetic field at the point P ,  B =  (µ₀I/4Πa)(sin90 + sin0)  = µ₀I/2Πa ;

Case-(3)—If length of the conductor is finite (L) and point P lies on right bisector of conductor , then ɸ₁=ɸ₂= ɸ ; and sinɸ=L/( 4a²+L²)^1/2 .

Then magnetic field B= (µ₀I/4Πa)(sinɸ + sinɸ)= (µ₀I/4Πa)2sinɸ=(µ₀I/4Πa)[ L/( 4a²+L²)^1/2];

Direction of magnetic field – The direction of magnetic field through a current carrying wire can be found using right hand thumb rule .

Right hand thumb rule – According to this rule we held the current carrying straight wire in our hand grip of our right hand so that thumbs directed towards the direction of flow of current then the curled finger gives the direction of magnetic field around the conducting wire . As shown in figure.

to see the video on Biot-Savart’s law click on the link given below-