In this topic we will discuss about Ampere’s circuital law and proof of Ampere’s circuital law ( using Biot- Savart’s law).We will also discuss the applications of Ampere’s circuital law (Magnetic field due to infinite long straight wire carrying current,magnetic field due to current through very long circular cylinder or thick wire, solenoid and Toroid)

Ampere’s circuital law –

According to this law the line integral of the magnetic field around any closed path in free space is equal to μ_{0} times the total current passing through the surface enclosed by the closed path .

In this topic we will discuss about Magnetic field at a point on the axis of a circular coil carrying current , and using its derivation we can find the magnetic field and we will also discuss about magnetic moment due current carrying coil.

Magnetic field at a point on the axis of a circular coil carrying current –

Suppose a circular coil of radius ‘a’ with center ‘O’ . Let current I is flowing through the coil we have to find the magnetic field at point ‘P’ , which is x distance away from the center .

Suppose two small element ‘dl’ of the coil C and D which is diametrically opposite points as shown in figure.

Here PC =PD = √(a^{2}+x^{2}), and we consider <COP = ɸ = <DPO .

As shown in figure dBcosɸ is cancelled by each other , then the net magnetic field dB sinɸ will be in the same side .

Here magnetic field due to small current carrying element dB = (µ_{0}/4Π) I dl sinθ/r^{2} ; here r=√(a^{2}+x^{2}),

So we can write , dB=(µ_{0}/4Π) Idl sinθ/(a^{2}+x^{2}) ,

So magnetic field at point p due to the circular loop

Special case 1- when point P lies at the center of the circular coil then , x = 0

Then B= (µ_{0}/4Π) 2∏nI/a = µ_{0}nI/2a ,

Case 2 – When point P is far away from the center then a^{2}+x^{2}=x^{2}

Then B= (µ_{0}/4Π) 2nIA/x^{3} [ since ∏a^{2} = A (area)]

Here nIA= M (magnetic moment)

So we can write , B= (µ_{0}/4Π) 2M/x^{3}

So we can define the magnetic moment due to current carrying coil is given as the product of ampere turns and area of current loop . SI unit of magnetic moment is A-m^{2} .

To watch the video related to this topic, Magnetic field at a point on the axis of a circular coil carrying current ( By Nayan jha sir) go to the link given below-

The polarity of magnetic dipole due to the current loop is decided as , if the current from one side is clock wise direction it gives south pole and on another face direction of current is anti-clock wise it gives north pole , as shown in figure-

Case 3- The variation of magnetic field induction with distance of a point on the axis of coil carrying current is given as –

class 12th physics syllabus removed . How it is beneficial for the students , see the video given below-

Magnetic field at the center of a circular current carrying coil is the one of the application of Biot-Savart’s law . here we will derive the expression for Magnetic field at the center of a circular current carrying coil .

Magnetic field at the center of a circular current carrying coil – Consider a circular coil of radius ‘r’ having center ‘O’. suppose I be the current flowing through the coil , and we have to find the magnetic field at the center .

Suppose a small element ‘dl’ which is the part of coil create a magnetic field dB at the center.

According to Biot-savart’s law dB = (µ_{0}/4Π) I dl sinθ/r^{2} . but ϴ=90^{0},

So we can write dB = (µ_{0}/4Π) I dl sin90^{0}/r^{2} = dB = (µ_{0}/4Π) I dl /r^{2} .

Then magnetic field at the center due to complete coil

B=∫ (µ_{0}/4Π) I dl sinθ/r^{2} ( Taking limit 0 to 2∏)

We get B= dB = (µ_{0}/4Π) I 2∏r/r^{2} = µ_{0} I /2r

For an arc which is making angle ϴ at the center will be given as

B= (µ_{0} I /4∏r)(Angle at the center )

Or , B= (µ_{0} Iϴ /4∏r) ;

The direction of magnetic field due to current carrying coil may be give by right hand rule , according to it if curled finger shows the direction of current then stretched thumb gives the direction of magnetic field .

In this topic Force on a moving charge in a Magnetic field , define the magnetic field and units and dimension of magnetic field .We will also know about Fleming’s left hand rule.

Suppose a positive charge ‘q’ is moving with velocity ‘v’ at an angle ‘ϴ’ with magnetic field ‘B’. Then experimentally it is found that force ‘F’ experienced depends on

F α q ………(i)

F α B……………….(ii)

F α v sinϴ …………(iii)

On combining these three equations we get ,

F α q B v sinϴ

Or, F =k q B v sinϴ ; where k is constant of proportionality k=1

Then we can write F = q B v sinϴ or, F = q( x )

Here the direction of force is given by Fleming’s left hand rule or right hand screw rule .

Fleming’s left hand rule –

According to this rule when we stretch our left hand’s fore finger, middle finger and thumb such that they are perpendicular to each other , if fore finger shows the direction of field, middle finger shows the direction of current( +ve charge) then thumbs gives the direction of force .

Definition of B (magnetic field intensity) –

As we have seen in the equation F = q B v sinϴ ;

If q= 1C , v=1m/s ϴ=90^{0} i.e. sinϴ = 1 then B= F ;

So we can define magnetic field intensity at a point is equal to the force experienced by a unit charge moving with a unit velocity perpendicular to the direction of magnetic field at that point .

Unit of magnetic field ‘B’ –

From the equation F = q B v sinϴ ,

B = F/qv sinϴ ; then unit of B is NA^{-1}m^{-1} = Tesla (T) ,

Magnetic effect of current, Oersted experiment and Amperes swimming rule –

In this topic we will discuss about magnetic effect of current,Oersted experiment and Ampere’s swimming rule.

Magnetic effect of current–

When electric current passes through a conductor (conducting wire) then magnetic field developed around the conductor.

The intimate relationship between electricity and magnetism was discovered 200 years ago , Oersted discovered in the year 1820 that a straight wire carrying current cause a deflection in a nearby magnetic compass needle.

Oersted’s experiment and Ampere’s swimming rule-

According to Oersted’s experiment , we take a needle NS which is free to rotate . We place the needle above a current carrying wire AB . If current flows through the wire in the direction A to B and from B to A then deflection in the needle shown in the figure . Since magnetic needle can deflect only in the interaction of the another magnetic field . so it is confirmed that due to flow of current in a conductor magnetic field setup across around the wire.

The direction of deflection in the magnetic needle due to current in the wire is given by Ampere swimming rule . According to this rule ‘ If a man swimming along the wire in the direction of current with his face turned always towards the needle , so that the current enters through his feet and leaves in his had , then north pole of the magnetic needle will be deflected towards his left hand’ .

This assignment contains all type objective questions of magnetism and matter. it will be helpful for the students preparing for class 12th board examination or other competitive examinations like JEE/ NEET .

Magnetism and matter

With the tangent galvanometer it is desirable to have a deflection near 45°; then the percentage error:

(a)is less in the reading of deflection

(b)is negligible in the reading of deflection

(c)is less in the measurement of current

(d)is large in the measurement of current

The sensitivity of a moving coil galvanometer depends on:

(a)the angle of deflection

(b)the earth’s magnetic field

(c)torsional constant of the spring

(d)the moment of inertia of the coil

An ammeter can be converted into a voltmeter by connecting:

(a)a high resistance in series

(b)a low resistance in parallel

(c)a low resistance in series

(d)a high resistance in parallel

In a moving coil galvanometer the deflection of the coil θ is related to the electric current I by the relation:

(a)I ꭀ tan θ (b) I ꭀ θ (c) I ꭀ θ^{2} (d) I ꭀ √θ

A voltmeter of range 2V and resistance 300 Ω cannot be converted into ammeter of range:

(a)1 A (b) 1 mA (c) 100 mA (d) 10 mA

The effect due to uniform magnetic field on a freely suspended magnetic needle is as follows:

(a)both torque and net force are present

(b)torque is present but no net force

(c)both torque and net force are absent

(d)net force is present but not torque

A voltmeter has a resistance of G ohm and range V volt. The value of resistance used in series to convert it into voltmeter of range nV volt is:

(a)nG (b) (n – 1)G (c) G/n (d) G/(n – 1)

An ammeter has a resistance of G ohm and a range of I The value of resistance used in parallel to convert it into an ammeter of range nI amp is:

(a)nG (b) (n – 1)G (c) G/n (d) G/(n – 1)

To reduce the range of a voltmeter, its resistance need to be reduced. Which of the following resistance when connected in parallel will convert it into a voltmeter of range (V/n)?

(a)nR_{0 }(b) (n + 1)R_{0} (c) (n – 1)R_{0} (d) None of these

this assignment contains all important objective type questions of magnetic effect of current . Students preparing for 12th board examination or other competitive examination like JEE/NEET.

The force acting on a charge q moving with a velocity ʋ in a magnetic field of induction B is given by:

(a) q/(ʋ ₓ B) (b) (ʋ ₓ B)/q (c) q/(ʋ ₓ B) (d) (ʋ . B)q

Two free parallel wire carrying currents in the opposite directions:

(a)attract each other

(b)repel each other

(c)do not affect each other

(d)get rotated to be perpendicular to each other

Two parallel wires carrying currents in the same direction attract each other because of:

(a)potential difference between them

(b)mutual inductance between them

(c)electric forces between them

(d)magnetic force between them

A free charged particle moves through a magnetic field. The particle may undergo a change in:

(a)speed (b)energy (c) direction of motion (d) none these

An electron and a proton travel with equal speeds and in the same direction, at 90° to a uniform magnetic field. They experience forces which are initially:

(a)in opposite direction and differing by a factor of about 1840

(b)in the same direction and differing by a factor of about 1840

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-

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) D_{1} and D_{2} , 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 D_{1} is negative and D_{2} is positively charged , then the positive charged particle at point P starts to move towards D_{1} , but due to perpendicular magnetic field it starts moving on a circular path. If r is the radius of the circular path then ,

mv^{2}/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 D_{1} at the same time D_{1} becomes positively charge and D_{2} negatively then again charge particle starts to move towards D_{2} 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 v_{0} , r_{0} is the is the maximum velocity and maximum radius of the circular path followed by the charged particle ,

Then ,m v_{0}^{2}/ r_{0} = B q v_{0} or , v_{0} = Bq r_{0}/m ,

Therefore maximum kinetic energy k = ½ mv_{0}^{2} = B^{2} q^{2} r_{0}^{2} / 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-

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= m_{0} /√(1-v^{2}/c^{2}) where m_{0} 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 .

Cyclotron is suitable for accelerating heavy particles only . i.e electron can not be accelerated through cyclotron .

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-

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=µ_{0}idl sinθ/r^{2} .

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/r^{2} ………………4

after combining these equations ,we get

dB α I dl sinθ/r^{2 } so , dB= k I dl sinθ/r^{2}

(Where K= µ_{0}/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 = (µ_{0}I/4Πa)(sinɸ_{1} + sinɸ_{2}) ,

Special cases – Case-(1) — When the wire is of infinite length and point P is near the wire in that case ɸ_{1}=90^{0} , ɸ_{2}=90^{0} ;

So magnetic field at that point P , B = (µ_{0}I/4Πa)(sin90 + sin90) = µ_{0}I/2Πa

Case-(2)– When the wire is of infinite length and point P is near the one end of the wire in that case ɸ_{1}=90^{0} , ɸ_{2}=0^{0} ;

So magnetic field at the point P , B = (µ_{0}I/4Πa)(sin90 + sin0) = µ_{0}I/2Πa ;

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

Then magnetic field B= (µ_{0}I/4Πa)(sinɸ + sinɸ)= (µ_{0}I/4Πa)2sinɸ=(µ_{0}I/4Πa)[ L/( 4a^{2}+L^{2})^{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-