The magnetic field of steady current; The second kind of field which enters into steady of electricity and magnetism is, of course, the magnetic field. Such field have been known since ancient time, when the effect of the naturally occurring permanent magnet magnetite (Fe 3 O 4 ) were first observed. Magnetism was a little used and still less understood phenomenon until the early nineteenth century, when Oersted discovered that electric current produced a magnetic field. This work, together with the later work of Gauss, Henry, Faraday and other, has brought the magnetic field in to prominence as a partner to the electric field.. 1 شريك شهره

2 Magnetic induction: The electric field was defined as the ratio of the force on a test charge to the value of test charge, (1) The assumption that charge is at rest. For purpose defining the magnetic induction it is convenient to define F m, the magnetic force (Lorentz force), as the part of force exerted on a moving charge which is neither electrostatic nor mechanical. The magnetic induction B, is then defined as the vector which satisfies, F m = qV×B (2) For all velocities If two measurement of F m are made for two mutually perpendicular velocities V 1 and V 2 B=1/q (F 1 ×V 1 )/(v 1 2 )+K 1 V 1 (3) B=1/q (F 2 ×V 2 )/(v 2 2 )+K 2 V 2 (4) K :scalar arbitrary

Taking the scalar product of each of these with V 1 and remembering that V 1 and V 2 are perpendicular, we obtain K 1 v 1 2 =1/q F 2 ХV 2.V 1 /v 2 2 (5) using eq.(3) B =1/q F 1 ХV 1 /v /q ( F 1 ХV 2.V 1 /v 1 2 v 2 2 )V 1 (6) 3

4 Forces on current-carrying conductor: From the definition of B, an expression for the force on an element dF of a current carrying conductor can be found. If dl is an element of conductor with its sense taking in the direction of the current I which it carries, then dl is parallel to the velocity V of the charge carriers in the conductor, if there are N charge carriers per unit volume in the conductor, the force on the element dl is dF= NA|dℓ|qV×B (7) Where A is the cross-sectional area of the conductor and q is the charge per charge carrier. Since V and dl are parallel, an alternative form of eq.(7) dF = Nq|V|Adℓ×B (8) However, Nq|V|A is the current for single species of carriers, therefor the expression dF=Idℓ×B (9) If the circuit in question is represented by the contour C, then (10)

5 The law of Biot and Savart: In 1820, just a few weeks after Oersted announced his discovery that currents produce magnetic effects, Ampere presented the results of a series of experiments which may be generalized and expressed in modern mathematical language as, (11) The force F 2 is the force exerted on circuit 2 due to the influence of circuit 1, the dℓ and r are explained by the figure. The number μ 0 ⁄4π plays the same role here as 1⁄(4πϵ 0 )played in the electrostatics, μ 0 /4π=10 (-7) n/Amp 2 (є 0 =8.85x coul/N.m 2 )

6 The magnetic flux at the point due to the closed circuit is (12) This equation is the generalization of the Biot and Savart law, which name used both for eq. (12) and differential form (13) Because i=J.nda, therefore eq.(13) can be written as (14) (15) It is an experimental observation that all magnetic induction field can be described in term of a current distribution this implies that there are no isolated magnetic poles and that Div B=0 (16) Eq. (16) is true for any B of the form (14)&(15).

7 Elementary application of the Biot and Savart law: Ex.1: magnetic field due to a current carrying infinite straight wire: Let us consider an infinite wire AB along the x- direction carrying a current i. let p be a point at distance a from this wire as shown in the diagram. Let dx be an elementary length of this wire then according to Biot – Savart law we get 0 Or

8 6-2 Force on a current element in a magnetic flux B: Let us consider an elementary length dl of a conductor placed in magnetic flux as shown in the diagram. Let the current through the conductor be i. According to Lorentz force we know that if charge q moves with a velocity V in a magnetic field of intensity B, then the force on the charge is =qV ⃗ × B ⃗. Let the area of cross -section of the elementary dℓ be A. The volume of the element =dℓ.A If there are N charge carriers /unit volume and charge on each carriers is q, then the charge in the element is Q=NqdℓA (19) ∴ The force on the element is (dF) ⃗ =Q ⃗ V ⃗ ×B Where V is the drift velocity. Or (dF) ⃗ =NqdℓAV ⃗ ×B ⃗ (20)

9 In eq.(20) (dℓ ⃗ )has the same direction as V ⃗ therefore we can write (dF) ⃗ =NqVA(dℓ ⃗ )×B ⃗ (21) In eq.(11) NqVA=i ∴ (dF) ⃗ =i(dℓ) ⃗ ×B ⃗ (22) Equation (22) can be integrated to give the force on a complete (or closed) circuit. If the circuit in question is represented by the contour C, then the total force F ⃗ = ∮ C i (dℓ ⃗ )×B ⃗ (23) If B is uniform, then F ⃗ = ∮ C i (dℓ) ⃗ ×B ⃗ =iB ∮ C dℓ=0 (24) Another interesting quantity is the torque on complete circuit. Since torque is moment of force, the infinitesimal torque dτ is given by dτ=r×dF= ı r×(dℓ×B) The torque on a complete circuit is τ=I ∮ C r×(dℓ×B)

Ex1: force between two finite current carrying parallel wires: Let us consider that AB &CD are two parallel wires carrying current i 1, and i 2 as shown in the diagram. Let the distance between the two wires be (a) let p the any point on the wire CD. The intensity of the magnetic flux at the point p due to the element dl of the wire AB is Now we know that force on element of wire CD is given by Combining eq.(26)and (27) we get θ=90 Therefore Force /unit length on wire CD due to AB,force of attraction. 10

Amperes law: States that the line integral of magnetic flux B over a closed circuit is equal to μ 0 time the current passing through the closed circuit. Mathematically; this law is written as Where B is the magnetic flux density and i is the current through the closed circuit. Ex1: magnetic flux due to an infinite current wire: Let PQ be a wire carrying a current i, let R be a point at a distance (a) from the wire in the diagram(consider a closed circle through R, of radius a. then applying Amperes law we can write

12 Ex2: magnetic field on the axis of circular current loop. Q: calculate the magnetic field at an axial point p a distance x from center of the loop. Consider a circular wire loop of the radius R in the yz plane and carrying a steady current I as shown in figure. The length element perpendicular to the vector r ̂ of the location of the element thus The direction of dB is perpendicular to plane formed by r ̂ and

13 (The summed over all elements around the loop is zero) (The circumference of the loop). At the center of field x=0