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# Explain the term self-induction of a coil

Self – Induction

It is the property of an electric circuit or coil by virtue of which it opposes any change of flux or current in it by inducing a current in itself is called self- induction. In another words we can say “self-induction is the inertia of electricity “. Due to self -induction an electric circuit resist any change of current through it .

As we know magnetic flux linked with a coil is directly proportional to the current flowing through the circuit

i.e.    ɸ α I ( ɸ= flux linked with the coil, and I is the current in the circuit )

ɸ = LI ( L is the constant called coefficient of self- induction or self-inductance  or only inductance . )

If current I =  1 ampere, then flux ɸ= L  . so we can define self-induction is the flux linked with the coil when unit current passes through the coil.

as we know,

emf  e= – dɸ/dt

e= -d (LI)/ dt

e= -L ( dI/dt)

L= –  e/(dI/dt)

So we can define self-inductance of a circuit is numerically equal to the induced emf setup in it when the rate of change of current through it is unity.

Unit of self-inductance is Henry (H) .  We can define self-inductance of a coil is one Henry if 1 volt emf induced in the coil when rate of change of current through the coil is unity.

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## What is electric current

What is electric current

What is electric current ? what is its unit ? define Ampere.

Electric current-When electric charges is in motion , they constitute what is called an electric current . electric current is defined as the rate of flow of charge through an area of the conductor . if  Q is the charge flows through a conductor in time T then  electric current I = Q/T

Or  I= dQ /dt . unit of current is Ampere .

We can define 1 Ampere  is the current flowing through a conductor when 1 Coulomb  of charge flows through a conductors in unit time.

In the terms of drift velocity current I = n e A Vd  where n is the charge density , A  is the area of cross-section of the conductor , e is the charge on electron and Vd is the drift velocity.

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In terms of Ohm’s law – current  I = V/R  V is the potential difference across the conductor and R is the resistance of the conductor .

So we can say 1 Ampere current flows through a conductor when unit potential difference setup across a conductor of resistance 1 Ohm .

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# Faraday’s law of electromagnetic induction

Electromagnetic induction – As we know when electric current passes through a conductor then magnetic field developed across the conductor , first discovered by Oersted . But Michael Faraday found its converse is also possible i.e. Whenever magnetic field changes around a coil ( and hence magnetic flux linked  across the coil changes  ) electric current induced in the coil , such phenomena is called electromagnetic induction (E.M.I.) . And the current induced in the coil  is known as induced current and emf developed is known as induced emf .

Faraday’s laws of electromagnetic induction – On the basis of experiments Faraday’s gave the two laws –

(i) Whenever electric flux linked with the coil changes then emf induced in the coil . The emf lasts only for the time for which flux is changing .

(ii)  The magnitude of induced emf is directly proportional to rate of change of magnetic flux linked with the coil.   I.e.  emf   e  α ɸ2 – ɸ1 / t (time taken).

Or  e =  – k dɸ/ dt . ( Here negative sign shows the  induced emf oppose any change in the magnetic flux through the coil.

After combining these two equations Faraday’s law states “ An induced emf is produced in any closed circuit if there is a varying magnetic flux . the magnitude of induced emf  is equal to the negative of the time rate change of magnetic flux through the circuit .”

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# USES OF POTENTIOMETER

Uses of potentiometer

• Measurement of emf of cell-

Let ,

E is the  emf of  driver cell  and ,  E1  is the emf of cell to be measured

Here potential gradient K= [ {E / (R+r)} r]/l .

Let l1  is the length of the wire from zero end where galvanometer shows no deflection

Then E1 = K l1

E1 =   [ {E / (R+r)} r]/l X l1

• Comparison of EMFs of two cells

• To find the emf of two cells we connect the two cells with the potentiometer  as shown in figure  having emfs E1  and E2  .

When 1 and 2 is connected together let null point is at length l1  from zero end,

Then E1 = K l1 …….(i)

Again when 2 and 3 are connected , null point  is at length  l2 from zero end ,

Then  E2 = Kl2 ……..(ii)

By dividing (i) and (ii) we get E1 /E2 = l1 / l2 ………..(req. equation)

To find the internal resistance of a cell –

Let,

E – is the emf of the cell of which we have to find the internal resistance

R – is the external resistance connected with  the cell

R –  internal resistance of the cell

Let initially  switch K’ is off and l1 is the length of the null point from the zero end .

then E = K l1 ……….(i)

After that switch K’ is switched on then the null point is at l2  from zero end

then ( potential across R )       V = K l2 ……… (ii)

E/V = l1 /l2 …………..(iii)

But,  internal resistance r = [(E/V)-1] x R ……………..(iv)

So, from equation (iii) and (iv)

r =  [(l1 / l2 )-1] x R …….(req. equation)

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# POTENTIOMETER

## THIS TOPIC EXPLAIN ABOUT THE POTENTIOMETER

Potentiometer – An instrument used to measure accurately the EMF or potential difference is called potentiometer.

Principle– Its working based on the principle that the potential drop across a conductor is directly proportional to the length of the conductor , if constant current passes through the conductor having uniform area of cross-section .

Construction –

As we know ,

potential V=IR

V = (ρl/A) I

Since current is constant and area of cross-section is uniform  so V α l

Or V = K l

K = V/l ( K is a constant called potential gradient)

[ calculation of potential gradient ( students don’t have to derive it in case of potentiometer it is useful for the numerical questions) – if Rh = r and resistance of the wire is r and E is the emf of  driver cell  then ,  current through the circuit is , i = E/ (R+r)

There fore potential difference at the end  the wire is v= i r

V= {E/ (R+r) } r . so potential gradient  K = V/l = {E/ (R+r) } r/l. ]

Potantiometer is preferred over the galvenometer to measure  potential difference across a conductor

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## 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-

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## 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 12th 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 ɸ12= 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-

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## What is Ampere circuital law

Ampere circuital law – After the discovery of Oersted and Biot-Savart, Ampere found a useful relation between magnetic field and electric current.     [ In the case of all basic laws of physics , Ampere’s law can’t be derived , its validity is based on the correctness of the result.] . According to Ampere circuital law that the linear integral of magnetic field (B) along a imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium µ0 . for a  long straight current carrying conductor  , magnetic field

B= µ0 I/2ΠR

So,

∫B.dl =   ( µ0 I/2Πr) ∫dl    ( integrate with limit 0 to 2Πr )                                                      = ( µ0 I/2ΠR) × 2Πr = µ0I

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