Conversion of galvanometer into voltmeter

In this topic we will discuss about the ‘ Conversion of galvanometer into voltmeter ‘ . What is ammeter? and how can we convert a Galvanometer into voltmeter?

Before to read this topic ‘ Conversion of galvanometer into voltmeter ‘  students must know about the galvanometer and shunt .

To get the notes on galvanometer click here-

To get the notes on shunt click here-

 

This topic is very important topic  for class 12th board examination and for other competitive examination ( for medical and engineering entrance examinations )

Voltmeter-

It is high resistance galvanometer used to measure potential difference between two points of a conductor or circuit.

A galvanometer can be converted into voltmeter by connecting high value resistance in series with the galvanometer( as shown in figure).

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Let, G is the resistance of the galvanometer,

I_G  is the current passes through the galvanometer,

R is the high value resistance  in series with the galvanometer.

Let,

V is the potential difference of the galvanometer.

So we can write V = I_G ( R + G)

Hence  R =( V/I_G)- G ;

Resistance of the voltmeter R_s = R + G ;

So we can say ,for ideal voltmeter the resistance of the voltmeter should be infinity.

This topic is very important for all state  class 12th board examination and for other entrance examination , therefore it is important to solve the assignment of this topic .

In the assignment , we provide all kind of questions like , objective , short answer type, long answer type , numericals and hots . After solving the assignment of this chapter students will be able to improve there knowledge on this chapter.

To get the complete assignment of magnetic effect of current click here-

But students must solve the questions base on its previous topic.

We are providing the link , so that students should clear all topic of this chapter- click here for ammeter

Conversion of galvanometer into ammeter

In this topic we will discuss about the ‘ Conversion of galvanometer into ammeter ‘ . What is ammeter? and how can we convert a voltmeter into ammeter?

Before to read this topic ‘ Conversion of galvanometer into ammeter ‘  students must know about the galvanometer and shunt .

To get the notes on galvanometer click here-

To get the notes on shunt click here-

This topic is very important topic  for class 12th board examination and for other competitive examination ( for medical and engineering entrance examinations )

 Ammeter –

It is a low resistance galvanometer used to measure the electric current in a circuit.

Let, I and I_G are the total current passes through the circuit and the galvanometer respectively. G , and S are the resistances of the galvanometer and shunt respectively.

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As shown in figure,  potential across S is V_S = (I- I_G ) S

Potential across  G is  V_G = I_G G

Since both are in parallel combination , hence  is  V_S = V_G

So, ( I-I_G) S =  I_G G

S=  I_G G/( I-I_G) ……………………….( i)

Using this  equation we can find the value of shunt to get the ammeter of our desired value.

Resistance of the Ammeter ;  Let R is the resistance of the ammeter.

Then we can write ;  1/g  + 1/s = 1/ R

R = S+G/SG ;

For ideal ammeter value of R should be  zero.

After studding this topic students can visit the next topic which is conversion of galvanometer into voltmeter. To get the next topic click on the link given below.

NEXT TOPIC -CONVERSION OF GALVANOMETER INTO VOLTMETER. CLICK HERE-

To get the notes on the complete assignment of moving coil galvanometer click here-

To get the notes on objective questions of moving coil galvanometer click here-

Shunt

In this topic we will discuss about, what is shunt? what are the uses of shunt? and how can we convert a galvanometer into ammeter?

Before to go with this topic students must know about the moving coil galvanometer, to get the notes of moving coil galvanometer click here-

Shunt-

Shunt is a low resistance connected in parallel with the galvanometer or ammeter to protect it from high current. It is used to convert the galvanometer into an ammeter, by connecting it in parallel.

When high value current passes through the galvanometer or a low range ammeter then it can damage the instrument . To overcome this difficulties a low value resistance(shunt) is connected in parallel , so that maximum current can pass through the shunt ,and instrument will remain  safe.

Let, I ,I_S and I_G are the total current passes through the circuit , galvanometer and circuit respectively. G , and S are the resistances of the galvanometer and shunt respectively.

As shown in figure,  potential across S is V_S = I_S S

Potential across  G is  V_G = I_G G

Since both are in parallel combination , hence  is  V_S = V_G

So,  I_S S =  I_G G

But I_s = I – I_G

So we can write , ( I-I_G ) S =  I_G G ;and hence  I_G = I (S /S+G)

Similarly , I_S = I (G /S+G) ;

This result shows that shunt can be used as the (i) convert the galvanometer into ammeter and (ii) To increase the range of galvanometer.

NEXT TOPIC CONVERSION OF GALVANOMETER INTO AMMETER. CLICK HERE-

Torque on a current carrying coil placed in magnetic field .

In this topic we will find the expression for the torque on a current carrying coil placed in magnetic field. But before to derive the expression for ‘Torque on a current carrying coil placed in magnetic field’ , students must know the expression for the force experienced on the current carrying coil placed in uniform magnetic field. To get the notes on this topic click here-

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Torque on a current carrying coil placed in a magnetic field-

Suppose a rectangular coil PQRS carrying current I is placed in a uniform magnetic field B as shown in figure ( a) .

Let,  PQ = RS = l and QP=SP=b , suppose I is the current flowing through the coil.

Let, ϴ is the angle between the plane of the coil with the magnetic field.

As we know force on a current carrying coil is given as F = BIL sinϴ ,

Then we can find force on PQ is given as , F₁ = BIl sinϴ [ but ϴ=90⁰]

Therefore,  F₁ = BIl sin90⁰  out side the plane ( using Fleming’s left hand rule)

Similarly, we can find force on SR is given as , F₃ = BIl sinϴ [ but ϴ=90⁰]

Therefore,  F₃ = BIl sin90⁰   Inside the plane ( using Fleming’s left hand rule).

Again,  Force on QR is given as , F₂ = BIb sinϴ

Therefore,  F₂ = BIb sinϴ  directed downward ( using Fleming’s left hand rule),

And again, Then we can find force on SP is given as , F₄ = BIb sinϴ

Therefore,  F₄ = BIb sinϴ  directed upward ( using Fleming’s left hand rule)

It is clear that F₂ and F₄ will cancel out each other,

But due to pair of force F₁ and F₃ these force form a couple , and cause of turning effect,

So, torque Ʈ = either force x perpendicular distance

= F₁ bcosϴ = BIlb cos(90-α)=BIl sinα ;

But here lx b = area ‘A’, ; So we can write , Ʈ= BIA sinϴ,

If there are N number of turns in the coil then we can write net torque , Ʈ= BI NA sinϴ.

 

Next topic – Moving coil Galvenometer click here.

Force between two parallel current carrying linear conductors

In this topic we will find the expression for Force between two parallel current carrying linear conductors when they are separated by some distance , and current flowing through them is in same direction ( also when current flowing through them is in different directions).

Before to know about In this topic  expression for Force between two parallel current carrying linear conductors when they are separated by some distance, students must know the expression for current carrying conductors placed in uniform magnetic field. To know about this topic click here

Force between two parallel current carrying linear conductors –

Two long straight conductors are placed parallel two each other at distance ‘r’  and current flowing through them are I₁ and I₂ . Since each conductors are carrying current produces magnetic field and hence each conductor experience force on each other.

As shown in figure (a) due to current I₁ through C₁D_1­ magnetic field at a point P on the wire C₂D₂ is B₁ , then   B₁ =  μ₀I₁ /2∏r ,

Since current flowing through the wire C₂D₂ is I₂,

Then force experienced on the wire C₂D₂ is given as ,

F₂ = B₁ I₂ l

Then  , Force per unit length F₂/l = B₁ I₂ = μ₀I₁I₂ /2∏r (Towards left i.e. towards C₁ D₁) .

Similarly , As shown in figure (b) due to current I₂ through C₂D_2­ magnetic field at a point Q on the wire C₁D₁ is B₂ , then   B₂ =  μ₀I₂ /2∏r ,

Since current flowing through the wire C₁D₁ is I₁,

Then force experienced on the wire C₁D₁ is given as ,

F₁ = B₂ I₁ l

Then  , Force per unit length F₁/l = B₂ I₁ = μ₀I₁I₂ /2∏r (Towards right i.e. towards C₂ D₂) .

So we can say if the current through each wire are in same direction then there will be force of attraction between them .

If the direction of current in the both the wires are opposite then there will be force of repulsion between them, as shown in figure ( c )below, with the same force  μ₀I₁I₂ /2∏r .

Force acting on the current carrying conductor placed in magnetic field

In this topic we will discuss about Force acting on the current carrying conductor placed in magnetic field, where magnetic field is uniform and current carrying coil is laced at some angle with the magnetic field.

Before to know about this topic students must know these topics –

(i)force on a moving charged particle in uniform magnetic field and . To get the notes on this topic click here-

(ii) Motion of a charged particle in uniform magnetic field . To get the notes on this topic click here-

Force acting on the current carrying conductor placed in magnetic field –

Suppose a cylinder of length ‘l’ area of cross-section ‘A’ carrying current ‘I’ , placed in uniform magnetic field ‘B’.

Angle between them is ‘ϴ’ as shown in figure. If it is considered that ‘n’ is the number density of electron,Then total number of electrons present in the conductor is given as,

N = A l n. [ where volume V = Al ]

As we know when a charge moves in a magnetic force , force experienced on it is given by

f =  q v B sinϴ  [ ϴ is the angle between v and B]

so for each electron f=e v_d B sinϴ ( where v_d is the drift velocity of the electron) ‘,

So, net force on the electrons F = Nf = N e v_d B sinϴ =  Aln e v_d B sinϴ = n e A v_d l B sinϴ

But we know from I = neAv_d  ,

So, we can write  F = I l B sinϴ……………eq.

Special case-

(i)  if ϴ=0⁰ or 180⁰ then ,

Force F = 0 . i.e. a linear conductor carrying current placed parallel or anti parallel to the magnetic field experience no force.

(ii) If ϴ= 90⁰ then, sinϴ=sin90 = 1

Then force will be maximum given by F = BIl .

The direction of the force is given by the Fleming’s left hand rule.

Lorentz force and velocity filter

In this topic we will discuss about Lorentz force and velocity filter.In which we will know different conditions for Lorentz force and complete explanation of velocity filter .

Lorentz force-

The force experienced by a charged particle due to both electric and magnetic field at a space called Lorentz’s force .

Suppose a charge particle of mass ‘m’ and charge ’q’ is moving in the both electric and magnetic field of Field intensity ’E’ and ‘B’ respectively.

Then force due to electric field is given as F_e= q E

And force due to magnetic field will be F_b=  q v B sinϴ ( where ϴ is the angle between v and B).

Then, Lorentz force F = F_E+F_B  = q( E + v B sinϴ)………………Eq.

Special cases –

Case 1- When v , E and B all are collinear –

In this case F_e = q E ,

And F_b= o ( since ϴ= 0 ; sin 0 = 0 )

acceleration a = F_e/m =  qE/m ,

case 2 – When v , E and B are mutually perpendicular to each other –

then net force F = F_e +F_b = 0

therefore acceleration  a= 0 . It means particle will move without any deflection with the same velocity .

Velocity filter –

It is an arrangement of cross electric and magnetic field in a space which help to select from a beam , charged particles of given velocity irrespective of their charge and mass.

It consist two slits S₁ and S₂ placed parallel , with common axis at some distance . Uniform electric field ( E ) and magnetic field(B) applied  which are perpendicular to each other , which is shown in figure. When a beam of charged particles of different charges and masses after passing through S₁ enters into the region where crossed electric and magnetic field are present . The particles which is moving with velocity ‘v’ the electrostatic force and magnetic force are equal and opposite , then    q E = q v B   ,

Or, v= E/B

Such kind of particles go without changes its path and filtered out the region through the slits S₂. Therefore the particles emerging from S₂ will have the same velocity even though their charges and masses may be different.

Velocity filter is used in mass spectrograph which help to find the mass and specific charge of the charged particle.

 

 

Motion of a charged particle in uniform magnetic field

In this topic Motion of a charged particle in uniform magnetic field we will discuss about how a charged particle moves in uniform magnetic field, when it is projected with some velocity at an angle with magnetic field.

Before to know about this topic students must know about the force acting on a charge particle when it moves in a uniform magnetic field. To know about this topic force on a charge particle in uniform magnetic field click here.

Motion of a charged particle in uniform magnetic field –

Suppose a charged particle of mass ‘m’ and charge ‘q’ is projected  with velocity ‘v’ at angle ‘ϴ’ with the magnetic field  ‘B’ as shown in figure (a) . Here v has two components  (v₁= vcosϴ ) along x -axis and ( v₂= v sinϴ) along y-axis.

Due to v₁ it moves along x-axis and due to v₂ a force F acts which is given as F = qv₂ B = qvBsinϴ. Since this force F acts perpendicular to the velocity and magnitude of v₂ not changes hence it moves on a circular path.  As shown in fig(b).

let r is the radius of the circular path , due to these two components of v₁ and v₂ particle follows a helical path as shown in fig(a) .

As we know when body moves on a circular path the centripetal forc F = mv₂²/r = mv²sin²ϴ/r ;

Which is provided by magnetic force F= qvBsinϴ

So we can write  mv²sin²ϴ/r = qvBsinϴ;

So, r = mvsinϴ/qB ……………….(1)

or vsinϴ=qBr/m …………………….(2)

angular velocity  ω= vsinϴ/r = Bqr/mr = Bq/m ……………..(3)

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

and time period T = 1/f = 2∏m/Bq ……………………………(5).

The pitch of the helix = vcosϴ x T = v cosϴ 2∏m/Bq .

Special cases –

Case (i) -If ϴ=0⁰ i.e. v₁=v and v₂=0

So there will be no force and particle will move in the direction of B .

Case (ii) – if ϴ=90⁰ then v₁=0 and v₂=v  i.e. body will move only in the circular path

And radius of the circular path  r= mv/qB  or  r= v/(q/m)B .

 

Solenoid and Toroid

Before to know Solenoid and Toroid  , students must know Ampere’s circuital law , proof of Ampere’s circuital law and its application.        To know all these thing click here-

In this topic we will discuss about  one of the application of Ampere’s circuital law . We will define solenoid and Toroid , we will find the magnetic field due to Solenoid and Toroid .

Solenoid –

It is the closely wound coil in the form of helix . its length is very large as compared to its diameter.

 

Magnetic field due to a solenoid –

Let current I is flowing through the coil , each turn of solenoid regarded as a circular loop carrying current which produces a magnetic field . Total magnetic field is vector sum of magnetic field due to current through all the turns in the coil .

Let n be the number of turns per unit length of the solenoid . Consider a rectangular loop PQRS  near the middle of the solenoid as shown in figure.

PQ=L . hence total numbers of turn in length L = nL .

The line integral of magnetic field over the closed path PQRS is ,

At a point near the end of the solenoid magnetic field B = μ₀nI/2 .

If the solenoid is filled by material of permeability μ in side then magnetic field B = μnI = μ NI/L.

If we draw a plot magnetic field B vs r (distance) from the centre of the solenoid we get the following curve.

 

Toroid –

Toroid is the endless solenoid in the form of ring . or we can define ‘The toroid is the hollow circular ring on which a large number of insulated turns of a metallic wire are closely wound’. As shown in figure below.

Magnetic field due to current in a toroid–  Let n be the number of turns per unit length of the toroid , I be the current flowing through the toroid . When current passes through the solenoid magnetic field of constant magnitude setup in side the turn of toroid in the form of circular magnetic field . We draw three circle having radii r₁,r₂ and r₃ as shown in fig (b). Let B₁ is the magnetic field along loop 1  then using Ampere’s law –

The magnetic field at any point inside the empty space surrounded by toroid or outside the toroid magnetic field is zero .