Magnetic Field XII

aMagnetic Field Magnetic Field is the region around a magnetic material or a moving electric charge within which the of magnetism acts. It is a vector quantity. It is represented by 'B' or 'H' and its unit is tesla (T). Magnetic Lines of ForceMagnetic lines of force (field lines) are imaginary lines around the magnet which shows the path traced by a unit north pole when it is allowed to move freely. A tangent drawn at magnetic lines of force gives the direction of magnetic field.Properties of magnetic lines of force1. Magnetic field lines never cross each other2. The density of the field lines indicates the strength of the field3. Magnetic field lines always make closed loops4. Magnetic field lines always emerge or start from the north pole and terminate at the south pole.

Figure 1:Magnetic Lines of force around a bar magnet

Figure 2:Magnetic lines of force around straight conductor

Figure 3:Magnetic lines of force in a solenoid

Magnetic FluxMagnetic flux ɸ through any surface area in a magnetic field is defined as the number of magnetic lines of force crossing through the surface.

The magnetic flux passing through a surface area A in a magnetic field B is given by𝜙=B⋅A=BACos𝜃If 𝜃=0∘ then, 𝜙=BAWhere:a

𝜙	= magnetic flux (Wb)
B	= magnetic flux density (T)
A	= cross-sectional area (m2)

This means the magnetic flux is:Maximum =BA , when cos(𝜃)=1 therefore 𝜃=0∘. The magnetic field lines are perpendicular to the plane of the areaMinimum =0 when cos(𝜃)=0 therefore 𝜃=90∘. The magnetic fields lines are parallel to the plane of the area Magnetic Flux Density We Know, Magnetic flux density𝜙=BAor B=𝜙

AThus, Magnetic field intensity (B) is defined as the magnetic flux (normal) per unit area.Oersted's DiscoveryDuring a lecture demonstration, while setting up his apparatus, Oersted noticed that when he turned on an electric current by connecting the wire to both ends of the battery, a compass needle held nearby deflected away from magnetic north. This shows that there is magnetic field associated with the current. It gave a great milestone in the physics which established a connection between electricity and magnetism. Limitations1. Oersted experiment is only for DC electric circuit.2. It is applicable in the circuit with resistor only, without capacitor and inductors. Force on a moving Charge in a magnetic Field

When an electric charge moves through a magnetic field, the field exerts a force on the charge. Let us consider a charge +q moves with velocity v in XZ plane with magnetic field B along x-axis.Experimentally it is found that magnitude of force isi) directly proportional to the magnitude of the chargeF∝qii) directly proportional to the velocity of the chargeF∝viii) directly proportional to the strength of the magnetic fieldF∝Biv) directly proportional to the sine of angle between B and vF∝sin𝜃Combining allF∝Bqvsin𝜃or, F=kBqvsin𝜃where k is a proportionality constant whose value is I in SI units.

F=Bqvsin𝜃

In Vector formF=q(v×B)Direction of the ForceThe force is perpendicular to the plane containing v and B, ie it is along y-axis. It is determined by the Fleming right hand rule.Special Case:F=Bqvsin𝜃i) when 𝜃=0∘ or 180∘,sin𝜃=0a

	F=Bqv(0)
	=0

Hence, F=0, if the charge particle is moving parallel or antiparallel to the magnetic field.ii) when 𝜃=90∘F=Bq, which is max. valueHence, Force is maximum if the charge particle is moving perpendicular to the magnetic field.iii) When v=0, F=0Hence, No force is experienced by an stationary charge.iv) When, q=0F=0Hence, Force is experienced by charged particle only. Force on a current carrying conductor

Let us consider a straight segment of a conductor of length l and cross-sectional area A carrying a steady current I from bottom to top as shown in figure. Let it be placed in a region of uniform magnetic field B which is parallel to the plane of paper and 𝜃 be its inclination with the direction of field. We have assumed the conventional direction of current and hence the moving charges are positive.Let vd be the drift velocity of these charges which also must have an inclination of 𝜃 with the magnetic field. The magnetic Lorentz force experienced by each charge of magnitude q is given by,a

F	=q(vd×B)
F	=Bqvdsin𝜃 ..............(i)

The direction of this force is into the plane of paper as defined by Fleming's left hand rule.If n be the number of charge per unit volume of the conductor, the total number of charge in the conductor is,N=nV=nAl................(ii)So, the total magnetic force experience by all the charges in the conductor is,F=NBqvdsin𝜃……(iii)From equations (ii) and (iii), we get,5a

F	=(nAl)Bqvdsin𝜃
∴F	=(nqAvd)(Blsin𝜃) .............(iv)

We know, current in the form of drift velocity of the charge particle is given by,I=vdqnA....................(v)So, from equations (iv) and (v), we get,a

	F=IBlsin𝜃
	F=I(l×B)

Thus, total force experienced by the conductor placed in uniform magnetic field is perpendicular to both length and field (perpendicularly inward into the plane of paper which contains both l and B in this case)Case I. If the conductor is parallel or antiparallel to the field, thena

𝜃	=0∘ or 180∘
∴F	=IBlsin0∘ or IBlsin180∘
F	=0

Thus, a conductor placed parallel to uniform magnetic field experiences no force.Case II When the conductor is placed perpendicular to magnetic fielda

𝜃	=90∘
∴F	=IBlsin90∘=BIl

This is the maximum force experienced by the conductor.Direction of ForceAs F=I(l×B) The direction of force is perpendicular to the plane containing B and l. It is determined using Fleming's left hand rule.Torque on a Rectangular Coil in a Uniform Field

Let us consider a rectangular coil PQRS of length ' l ' and breadth ' b ' suspended in uniform horizontal magnetic field ' B '. Let the plane of the coil make an angle ' 𝜃 ' with the direction of magnetic field.When current ' I ' is passed through the coil, the magnetic force produced on the various arms of the coil are:i) Force on arm PQa

	a F1 =BIlsin𝜃 =BIl sin90∘ =BIl

∴Fl is perpendicular to both l and B and is directed outwards.ii) Force on arm RS, a

F3	=BIlsin𝜃
	=BIlsin90∘
	=BIl

∴F3 is perpendicular to both land B and is directed inwards.

The forces F2 and F4 are equal and opposite and they pass through same line of action, so they cancel each other resulting to no torque or force on the coil. The forces F1 and F3 are also equal and opposite but they pass through different lines of action, so they constitute couple or torque which is given by:a

𝜏	=(F1 or F3)× Arm of couple
	=BIl×bcos𝜃
	=BIAcos𝜃[∵A=l×b]

If there are ' N ' number of turns in the coil, then the total torque is:𝜏=N×BAcos𝜃∴𝜏=BINAcos𝜃Special Cases:i) If 𝜃=0∘, 𝜏= BINA (maximum value)When the plane of the coil is kept parallel to the magnetic field, torque on the coil is maximum.ii) If 𝜃=90∘, 𝜏=0 (minimum value)When the plane of the coil is kept perpendicular to the magnetic field, torque on the coil is minimum.Moving Coil Galvanometer

A moving coil galvanometer is a device which is used to detect and measure small amount of electric current. Principle: It is based on the principle that when current carrying loop is placed in uniform magnetic field, the coil experiences a torque.Construction: It consists of a rectangular coil having large number of turns wound over non-magnetic metallic frame. The coil is suspended between two poles of permanent magnet which are concave towards the coil and current is passed to the coil by connecting strip and spring with terminal as shown in the figure. The core of the coil consists soft iron rod in cylindrical shape.Working: Let N, A be the number of turns of the coil and area of the plane of the coil respectively, and B be the uniform magnetic flux density between the poles of the magnet. As we pass current through the coil, deflecting torque (𝜏d) is produced and is given by𝜏d= BINA sin𝛼Here 𝛼 is the angle between B and normal to the plane of the coil. Due to the curved surface of the magnet, the field becomes radial i.e the lines of the force becomes always parallel to the plane of the coil.a

	i. e 𝛼=90∘
	∴𝜏d= BINA.

Due to the deflecting torque, restoring torque is developed in the strip and is given by𝜏c=k𝜃Here, k= spring constant and 𝜃= angular displacement of the coil.In equilibrium,𝜏x=𝜏dor, k𝜃=BINAor, 𝜃=BNA

kIHere, B,N and A are constant∴𝜃∝Ii.e more deflection in the galvanometer represents more current.This is the working principle of the scale of a moving coil galvanometer. Thus, linear scale is designed in a moving coil galvanometer.Sensitivity of the galvanometerA galvanometer is said to be sensitivity if small amount of current flowing though the coil of galvanometer produces large deflection in it. A galvanometer can be converted into ammeter or voltmeter so it has two types of sensitivity.Current SensitivityThe current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer per unit current flowing through it. Current sensitivity =𝜃

I=BINA/k

I=BNA

KCurrent sensitivity of a galvanometer can be increased either by1. Increasing the magnetic field B2. Increasing the number of tums N.3. Increasing the area of the coil A. But it will make the galvanometer bulky and ultimately less sensitive.4. Decreasing the value of restoring force constant k by using a flat strip of phosphor - bronze instead of circular wire of phosphor - bronze.Voltage SensitivityVoltage sensitivity of a galvanometer is defined as the deflection produced in the galvanometer per unit voltage applied to it. Volatage sensitivity =𝜃

V=BINA/k

IR=BNA

KRVoltage sensitivity can be increased by1. Increasing number of turns of the coil (N)2. Increasing magnetic field intensity (B)3. Increasing area of the coil (A)4. Decreasing restoring torque per unit twist of the suspension (k)5. Decreasing resistance (R)Advantage of Moving Coil Galvanometer1. The sensitivity of the galvanometer can be increased by increasing N,B and A while decreasing the value of k.2. The instrument has a linear scale.3. Since the instrument uses high value of B, the deflection is undisturbed by the earth's magnetic field.4. As the coil is wound on a nonmagnetic metallic frame, damping is produced by eddy currents. As a result the coil quickly assumes the final position.Hall EffectWhen a magnetic field is applied to a current carrying conductor, a voltage is developed across the specimen in the direction perpendicular to both the current and the magnetic field. This effect is called Hall effect.

The figure above shows a copper strip of width d and carrying current I. As the charge carrier are electrons, the direction of drift velocity is opposite to the direction of current (fig I). Magnetic field is pointing inward to the plane of paper. So that the moving electron is deflected right. After sometime, electrons age gathered at the right edge and positive charge on the right. ( fig. II). This causes an electric field E towards right.Now the electrons experiences another force (electrostatic) due to the electric field. Deposition of the electron in right edge continues till the magnetic force in right direction equals the electrostatic force in left direction. After that electron moves straight through the conductor.The potential difference developed isV=Ed. At the equilibrium condition,a

	FE=FB
	eE=B e vd.

E=BvdAs we know, I=vdenAVd=I

neASo eqn (ii) becomesa

	E=BI neA
	V d=IB neA
	n=Bld VeA

n=BI

Vet.where t=A

d is the thickness of the strip and A is cross section area.Now, the p. d called Hall voltage is given by=BvddV=Ed=BI

neAdV=BT

netWe found that, the hall voltage is greater for the material with less value of n. (e.g. n-type semiconductor)Hall CoefficientThe Hall coefficient is defined as the ratio of the induced electric field to the product of the current density and the applied magnetic field. i.e RH=E

JB=1

vd𝜃n×vdRH=1

neApplication of Hall EffectThe Hall effect has many applications. It is used to accurate measurement of magnetic field, Hall mobility etc. It is also used to determine the nature of materials. It is also used to determine whether the specimen is metal, semiconductor or insulator. Hall ProbeA Hall probe can be used to measure the magnetic flux density between two magnets based on the Hall effect.To measure the magnetic flux density between two magnets, the flat surface of the probe must be directed between the magnets so the magnetic field lines pass completely perpendicular to this surface.

Hall probe is used for:• Non contact signal transmitter.• Magnetic field camera• To determine thickness• Door locking , gear shift sensor etc in automobiles.• For detecting wheel speed Magnetic Field of a Moving ChargeIf a charge is moving in a space, it produces magnetic field around it. Let a charge q moves in free space with velocity v, Then magnetic field B due to the moving charge at point P , at a distance of r is given by, B=𝜇0

4𝜋 qv⨯ r

r2 Biot and Savart Law

Consider a conductor XY through which current I is flowing. dl be the length element of conductor. P be any point at a distance r from dl making angle 𝜃. Then according to Biot and savart law, the magnetic field at point P due to current I in length element dl is 1) directly propoetional to the current dB∝ I2) directly proportional to length of the length element dB∝ dl3) directly proportional to sine of the angle between the conductor and the line joining r dB∝ sin𝜃4) inversely proportional to square of the distance between the element and the point P dB∝ 1

r2Combining above equations, we geta

dB	∝Idlsin𝜃 r2
dB	=kIdlsin𝜃 r2

where k is proportionality constant. In SI-units, k=𝜇0

4𝜋 The constant 𝜇0 is the permeability of free space and has value 4𝜋×10-7 henry per meter, So, we can writedB=𝜇0

4𝜋Idsin𝜃

r2Biot and Savart Law in Vector FormIf we consider length of the element as dl, distance of P as displacement vector r and unit vector along CP us r thena

dB=𝜇0 4𝜋Idl×r r2

dB=𝜇0 4𝜋I(dl×r) r3

where unit vector, r=r

|r|=r

r. The element vector dl has direction in the direction of the current.Application of Biot Savart Law1. Magnetic Field at hte centre fo the current carrying circular coil

Consider a circular coil of radius r. Let N be the number of turns of the coil and I be the current flowing throught the coil, then according to Biot -Savart law magnetic field at the dentre of coil due to small elemnt dl is given by, dB = 𝜇o

4𝜋 Idlsin𝜃

r2since, the radius r is perpendicular to length element dl𝜃=900so dB = 𝜇o

4𝜋 Idl

r2 ..........(i)Now the total magnetic field 'B', due to a whole circular coil is obtained by integrating eqn. (i) from 0 to 2𝜋rB=∫dBB = 2𝜋r∫0𝜇o

4𝜋 Idl

r2 =𝜇oI

4𝜋r2 2𝜋r∫0 dl=𝜇oI

4𝜋r2 [dl]2𝜋r0=𝜇oI

4𝜋r2 [ 2𝜋r - 0]=𝜇oI2𝜋r

4𝜋r2 =𝜇oI

2rThe magnetic field due to total, N number of turns of coil is,

B = 𝜇0NI

2r

Direction of BThe magnetic field produce is perpendicular to the plane of coil. At the centre of coil it is pointed inward to the plane of paper. 2. Magnetic Field on the axis of Current carrying circular coil

Let there be a circular coil of radius 'a' and carrying current 'I'. Let P be any point on the axis of a coil at a distance 'x' from the center and which we have to find the field. To calculate the field consider a current element dl at the top of the coil pointing perpendicularly outward from plane of paper. Then according to Biot Savart Law, the magnetic field due to length element dl is given bydB = 𝜇o

4𝜋 Idlsin𝛼

r2.........................(i)Here, angle 𝛼 between dl and 'r' can be taken as 𝜋

2. So the equation (i) becomes,dB = 𝜇o

4𝜋 Idl

r2This, magnetic field dB, is perpendicular to dl and r both as shown in figure. dB can be resolved in to components, dBcos𝜃 and dBsin𝜃. The vertical components dBcos𝜃 cancle each other being equal and opposite. So only the horizontal component dBsin𝜃 contibute to the total magnetic field.so, total magnetic field due to whole coil is given by integratind dB from 0 to 2𝜋a B = ∫dB =2𝜋a∫0𝜇o

4𝜋 Idl

r2sin𝜃 =𝜇o

4𝜋 I

r2sin𝜃2𝜋a∫0dl = 𝜇o

4𝜋 I

r2sin𝜃 [l]2𝜋a0 = 𝜇o

4𝜋 I

r2sin𝜃 [2𝜋a - 0] = 𝜇oIa

2r2sin𝜃 From the figure, sin𝜃 = a

rso, B = 𝜇oIa

2r2.a

r or, B = 𝜇oIa2

2r3 As, r=(a2+x2)so, B = 𝜇oIa2

2(a2+x2)3/2 If the coil has N number of turns

B = 𝜇oNIa2

2(a2+x2)3/2

Special Case,i) when point P is at the centre of the coil x = 0 so B = 𝜇oNIa2

2(a2)3/2 = 𝜇oNI

2a Which is the expression for the magnetic field at the centre of a circular coil.ii) When point P is far away from the centre of the coil, in such case, x≫ a so that B = 𝜇oNIa2

2(x2)3/2 B = 𝜇oNIa2

2x3Direction of BAs the current is in anticlockwise direction, the magnetic field at the axis is the coil is pointing right towards positive x- axis. ( By applying right hand thumb rule). 3. Magnetic Field due to straight current carrying conductor

Let us take a straight conductor, XY carrying current (I) in upward direction. We have to calculate magnetic field at point (P) at a distance (a) from the straight conductor i.e. OP = a. Let us take small element of length (dl) at point (C) i.e. OC = l Let us join C and P i.e. CP = r In figure, ∠OPC = 𝜃and ∠OCP= 𝜙 According to Biot and Savart law, the magnetic field at point P due to small element the (dl) at point (C) is dB=𝜇0

4𝜋Idl sin 𝜙

r2 ................(i)Then in Δ POCsin𝜙=a

r=cos𝜃or r=a

cos𝜃Also, tan𝜃=l

aor l=atan𝜃Differentiating both sides with respect to 𝜃, we get or dl=asec2𝜃d𝜃Putting these values in eqn. (i)dB=𝜇0

4𝜋I(asec2𝜃d𝜃)

(a2

cos2𝜃)cos𝜃a

=𝜇0 4𝜋Iasec2𝜃d𝜃 a2cos2𝜃cos𝜃

dB=𝜇0 4𝜋Icos𝜃d𝜃 a ........................(ii)

Magnetic field due to whole conductor AB can be calculated by integrating Eq. (ii) from -𝜃1 to 𝜃2. (By convention 𝜃1, being anticlock-wise is taken as negative).a

B=𝜃2∫-𝜃1dB

=𝜇0I 4𝜋a𝜃2∫-𝜃1cos𝜃d𝜃

=mal4𝜋[sin𝜃]𝜃2-𝜃1 4𝜋

=𝜇0I 4𝜋⋅a[sin𝜃2-sin(-𝜃1)]

B=𝜇0I 4𝜋a(sin𝜃2+sin𝜃1)

B=𝜇0I 4𝜋a(sin𝜃1+sin𝜃2) ............(iii)

This gives the value of B at point P due to a conductor of finite length.If the conductor of infinite length is taken,𝜃1=𝜃2=𝜋/2.a

B=𝜇0I 4𝜋a(sin𝜋/2+sin𝜋/2)=𝜇0I 4𝜋a(1+1)

B=𝜇0I 2𝜋a

Direction of BThe direction of magnetic field B is obtained by using right hand rule which is directed inward to the plane of paper.4. Magnetic Field due to a solenoid

Let us consider a solenoid of radius 'a' having 'n' number of turns per unit length. Current 'I' flows through the solenoid as shown in the diagram. Let 'P' be any any point on its axis. XY be length element of length dl which makes angle d𝜃 at point P. From the figure, XP = r, XD =a , and ∠XPD = 𝜃. Since magnetic field produced by a single coil is B=𝜇0Ia sin 𝜃

2r2 so a small magnetic field produce by the element dl having ndl turns at point P is given bydB=𝜇0Iasin𝜃

2r2ndl ..............(i)Here, for small length dl, XP≈YP =rIn 𝛥YCP, sin d𝜃= YC

YP=YC

rwe have for small angle d𝜃,, sind𝜃=d𝜃.so, d𝜃= YC

ror, YC= rd𝜃In 𝛥XYC, sin𝜃=YC

XY=YC

dlor YC=dlsin𝜃Then, we have,a

rd𝜃	=dlsin𝜃
or dl	=rd𝜃 sin𝜃.

Using value of dl in Eq. (i), we geta

dB	=𝜇0Iasin𝜃 2r2nrd𝜃 sin𝜃
or dB	=𝜇0Ian 2rd𝜃

Again from △XDP,a

sin𝜃=a r

r=a sin𝜃.

Putting value of r in (i), we geta

dB	=𝜇0I a nd𝜃 2sin𝜃 a
or dB	=𝜇0nIsin𝜃 2d𝜃.............(ii)

If the radius voctor r makes angles 𝜙1 and 𝜙2 at P from two ends of coil, the total magnetic field at point P can be obtained by integrating Eq. (ii) from 𝜙1 to 𝜙2. Soa

B	=𝜙2∫𝜙1𝜇0nIsin𝜃 2d𝜃
	=𝜇0nI 2𝜙2∫𝜙1sin𝜃d𝜃
	=𝜇0nI 2[-cos𝜃]𝜙2𝜙1
B	=𝜇0nI 2(cos𝜙1-cos𝜙2)..........(iii)

If the solenoid is very long then 𝜙1=0∘ and 𝜙2=𝜋. So Eq. (iii) becomes,a

B=𝜇0nI 2(cos0∘-cos𝜋)=𝜇0nI 2(1+1)

∴B=𝜇0nI

Direction of B: Direction of B can be determined by right hand first rule. Hence direction of the field is along the axis DP of the solenoid. Ampere's LawAmpere's law is an alternative to Biot and Savart law. However, it is useful for calculating magnetic field only in situations with symmetric condition. The law states that the line integral of the magnetic field around any closed path in free space is equal to 𝜇0 times the total current enclosed by the path,∮B⋅dl=𝜇0Iwhere 𝜇0 is permeability of free space, B is magnetic field, dl is the small element and I is current. The closed path is called Amperian loop.Proof: Consider a straight conductor carrying current I as shown in figure. The conductor produces magnetic field in which the magnetic lines of force are concentric circles with centre at the conductor.

Magnitude of magnetic field at P at a distance r from the conductor is given byB=𝜇0I

2𝜋rLet the closed loop is a circle of radius r. XY be a small element of length dl. Since B is tangent at every point on the loop, so dl and B are in same direction. Therefore, the line integral of B over this closed loop is given by∮B⋅dl=∮Bdlcos0∘=∮BdlSubstituting value of B in above equation,a

	∮B⋅dl=∮𝜇0I 2𝜋rdl=𝜇0I 2𝜋r∮dl=𝜇0I 2𝜋r2𝜋r
∴	∮B⋅dl=𝜇0I

This proves Ampere's law. In second figure, current I2 is passing in opposite to I1, then the currents enclosed by the loop is (I1-I2). Ampere's law is then given by,∮B⋅dl=𝜇0(I1-I2)Note: In order to use this law, it is necessary to choose a closed path for which it is possible to determine the line integral of B. For this reason, this law has limited use. This law is not a universal law and cannot be used to find out magnetic field at the centre of the current carrying loop. Applications of Ampere's Law 1. Magnetic field due to a straight current carrying conductor:

Consider a long straight conductor carrying current I as shown in figure. It is desired to find out the magnetic field at a point P at a perpendicular distance r from the conductor. The magnetic lines of force are concentric circles centred at the conductor. If we choose a circle of radius r as the closed path, the magnitude of B at every point on the path is same. The field B is tangent to dl and at every point on the circle. Thus applying Ampere's law to this closed path, we get∮B⋅dl=𝜇0Ior ∮B⋅dlcos0∘=𝜇0Ior B∮dl=𝜇0Ior B2𝜋r=𝜇0IB=𝜇0I

2𝜋rThis is the same expression as obtained by using Biot and Savart law.2. Magnetic Field due to current carrying solenoid

Consider a very long solenoid having 'n' number of turns per unit length. Let current I be current flowing through the solenoid, so that a magenetic field 'B' is produced. Inside the solenoid, the magnetic field (B) is uniform, strong and directed along the axis of the solenoid. The magnetic field outside the solenoid is very weak and can be neglected.In order to use Ampere's law to determine the magnetic field inside the solenoid, let us draw a dosed path PQRS as shown in the figure, where PQ=l. The line integral of B over the closed path PQRS is given by,∮B⋅dl=Q∫PB .dl+R∫QB.dl+S∫RB.dl+P∫SB.dl ..............(i)Now, Q∫PB⋅dl=Q∫PBdlcos0∘=BlAnd, R∫QB⋅dl=P∫SB.dl=0 ( since QR is perpendicular to B)Also S∫RB⋅dl=0 ( Magnetic field outside the solenoid is considered negligible.)∮B⋅dl=Bl ..............(ii)According to Ampere's law, we have∮B⋅dl=𝜇0× net current enclosed by the loop PQRSP=𝜇0× number of turns in PQRSP ×I∮B⋅dl=𝜇0nlI .......................(iii)From Eqs.(ii) and (iii), we getBl=𝜇0nlIB=𝜇0nI3. Magnetic field due to a current carrying Toroid

A toroid is just a circular solenoid (doughnuts shaped) whose both ends are joined together. Generally, a toroid is a circular hollow ring on which several copper wires are wrapped closely with no gaps between the turns.Let us consider a toroid of radius 'r' with current 'I' . . Let us consider a closed path inside the toroid (Ampere's loop). The magnetic field B is same everywhere on the closed loop. The magnetic field is tangent to the loop and thus angle between B and dl is zero. Thus line integral over the closed loop is ∮B.dl= ∮B dl cos 00 = B ⨯2𝜋r According to Ampere's law, ∮B.dl = 𝜇0 ⨯ net current enclosed by the closed loopIf toroid has total N number of turns, then total current =N Iso, ∮B.dl = 𝜇0 ⨯ N Ior, B⨯2𝜋r = 𝜇0 ⨯ N Ior, B=𝜇0 I N

2𝜋ror,

B = 𝜇0 n I

where, n=N

2𝜋ris the no of turns per unit length. Force between two parallel current carrying conductorsi) When current pass in same direction

Let us consider two very long parallel conductors X and Y carrying current I1 and I2 respectivelly in the same direction and are seperated by a distance 'r'.The magnetic field at point P (on Y) due to current I1 flowing in X isBx = 𝜇0I1

2𝜋r.............(i)The direction of magnetic field is perpendicular to the plane of paper and is directed outward.Here, Current I2 is flowing in conductor Y, placed in the magnetic field due to current in conductor X.The force experienced by Y due to current in length element dl of Y isFY=BX I2dlor, FY=𝜇0I1

2𝜋r I2dlThe force per unit lenght on Y isFY

dl=𝜇0

2𝜋rI1 I2or,

F=𝜇0

2𝜋r I1I2

This is a force on Y due to conductor x and is directed towards X.Similarly, The force on X due to current flowing in conductor Y is

F=𝜇0

2𝜋r I1I2

This force is ditected towards conductor Y.Thus, the force in the conductors is attractive, when current flows in same direction.ii). When current pass in opposite direction

If the current in the conductor flow in opposite direction then it is found that,FX=-FYIt means that the force produced in the conductor have same magnitude but in opposite direction. Thus there is repulsive force between two conductors with current in opposite direction. (Antiparallel)Define One Ampere CurrentWe, force between two parallel current carrying conductor is given byF=𝜇0

2𝜋r I1I2If, I1=I2=1 A, and r =1mthen F= 𝜇0

2𝜋where, 𝜇0 is the permeability of the medium, whose value is 4𝜋⨯10-7so, F= 4𝜋⨯10-7

2𝜋= 2⨯10-7NThus, One ampere current is that much current which when passed throught two parallel conductors seperated by distance of one metre produces force of 2⨯10-7N between them.