Magnetic Field of Moving Charge

aMagnetic 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

ocks>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.