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EECS 117 Lecture 15: Biot-Savart Law, Ampere’s Law and Magnetic Dipole

University of California, Berkeley Biot Savart Law and Ampere’s Law In the last lecture, we have shown that the magnetic force exerted on a small segment of wire flowing a current I with length dl is equal to r dFm = Idl × B where B is the density, and B = µH. The H is called the and µ is the permeability of the medium, which is the medium properties just like ε.

The intensity H induced by a current carrying wire segment has been shown to be r r I(r')dl × (rr − rr') J (r')× (rr − rr') d H (r) = = dv 4π rr − rr' 3 4π rr − rr' 3 Biot Savart Law and Ampere’s Law (cont.) rr is a vector pointing towards the point of interest from the origin, and r r 'is a vector towards the wire segment. The second expression is for a non-uniform current distribution in the magnetic field generating system (like a wire).

Ampere’s circuital law states that the magnetic field intensity integrated along a closed path is equal to the current flowing through the surface enclosed by the path. r r ∫ H ⋅ dl = I Off-axis Magnetic Field of a Straight Wire z (r,φ,z’) Consider a segment of a straight wire of r r length 2a. Using Biot-Savart Law, the ar − r' α magnetic flux density due to a small segment r dl is given by r µ Idzr × (rr − rr') d B(r,φ, z') = φˆ 0 y 4π rr − rr' 3 I µ I sinαdz -a = φˆ 0 4π (r 2 + (z − z')2 ) The B vector is pointing into the paper. a µ I sinαdz a µ Irdz B(r,φ, z') = φˆ 0 = φˆ 0 ∫ 2 2 ∫ 2 2 3/ 2 −a 4π (r + (z − z') ) −a 4π (r + (z − z') ) a µ Ir(z − z') µ I  z'+a z'−a  = 0 = 0  −  4πr 2 (r 2 + (z − z')2 )1/ 2 4πr 2 2 2 2 −a  r + (z'+a) r + (z'−a)  On-axis Magnetic Field of a Straight Wire

If the point of interest is at the midpoint of the wire, z’ = 0, and µ Ia B = φˆ 0 2πr r 2 + a2 On-axis Magnetic Field of a Circular Loop

Consider now the magnetic z field along the axis of a circular (0,0,z’) loop wire. r The green segment induces a R magnetic field in green. Because of circular symmetry, the segment (red) on the a y opposite side create a field (red) x I of equal magnitude. The resultant field is along the z- axis. On-axis Magnetic Field of a Circular Loop r µ Idl × Rˆ d B(0,0, z') = 0 r 2 4π R µ I adl µ I a2dφ dB = 0 = 0 z 4π (a2 + z 2 )3/ 2 4π (a2 + z 2 )3/ 2 r µ I a2 2π µ I a2 B = kˆ 0 dφ = kˆ 0 2 2 3/ 2 ∫ 2 2 3/ 2 4π (a + z ) 0 2 (a + z ) r µ I At the center of the loop, B = 0 kˆ 2a On-axis Magnetic Field of a Solenoid

z

-l l Using the result above, we can calculate the on-axis magnetic field inside the solenoid

Consider a solenoid with n number of windings per meter. The length of the solenoid is 2l. µ nI a2dz dB = 0 z 2 (a2 + z 2 )3/ 2 l µ nIa2 l dz µ nIa2 z µ NI B = 0 = 0 = 0 z 2 ∫ (a2 + z 2 )3/ 2 2 2 2 2 2 2 −l a a + z −l a + l On-axis Magnetic Field of a Solenoid (cont.)

where N = 2nl, the total number of windings in the solenoid.

For an infinitely long solenoid, Bz = µ0nI Ampere’s Circuital Law Version of a Solenoid D’ C’

A B The cross r B D C section of the l solenoid.

We can obtain the same information, and even learn more by using the Ampere’s circuital law.

The line integral of the B can be broken into the segment parts r B ⋅ dl = + + + ∫∫AB ∫BC ∫CD ∫DA Solenoid (Ampere’s Circuital Law) r Based on the result we obtained in the previous calculation, B inside the solenoid is only in the z direction. The geometry implies that this would be true also for the magnetic field outside the solenoid. So, = = 0 ∫BC ∫DA If we consider the path ABC’D’, the path integrals along BC’ and D’A are zero because of the above reason. Because there is not current flowing through the area enclosed by the closed path, = − ∫AB ∫C'D' This result holds if the hypothetical box ABC’D’ is drawn such that the segment C’D’ is at infinity. This implies that the magnetic field is uniform everywhere outside the solenoid. This contradicts our intuition of a field falling down to zero at infinity and also experimental observation. Solenoid (Ampere’s Circuital Law) The only solution to this contradiction is that the magnetic field outside of solenoid is zero. This coincides with experimental results. r r r r The result is: B ⋅dl = B ⋅dl = µ0nlI ∫ ∫CD If the segment CD is far away from the ends of the solenoid, B is uniform along the segment. Thus, r r B ⋅dl = Bl = µ0nlI ⇒ B = µ0nI ∫CD This result is the same as the one for the infinite solenoid.

If we move the segment CD up and down, since the current enclosed remains the same, the path integral along CD is the same. So the magnetic field is uniform inside the solenoid. For an infinite solenoid or the region where end effects are negligible, the field everywhere inside the solenoid is equal to the above result. Vector Potential r r Since there is no , can be v ∇ ⋅ B = 0 B expressed as ∇× A , the curl of vector potential.

For the two different expressions of B, we have r r µ I(r')dl ×(rr − rr') µ J (r')×(rr − rr') B(r) = = dv 4π ∫ rr − rr' 3 4π ∫ rr − rr' 3 r r µ I(r')dl µ J (r') A(r) = = dv 4π ∫ rr − rr' 2 4π ∫ rr − rr' 2 Magnetic Dipole z (r,0,θ)

x r R Top view θ ψ a φ’ x

For a current loop, the magnetic field off axis can be calculated by finding vector potential. Magnetic Dipole The vector potential due to the green segment has a direction of the green arrow. Because of symmetry, the potential due to the red segment almost cancel the green vector potential and the resultant potential due to these two segments has a direction normal to the xz plane, or in general in φ ˆ direction. Without loss of generality, we place the point of interest in the xz plane. The φ component of A due to a segment is equal to µIdl'cosφ' dA = φ 4πR The total is found as the integral around the loop µIdl'cosφ' µIa 2π cosφ'dφ' A = = φ ∫ ∫ 4πR 4π 0 R Magnetic Dipole where R 2 = r 2 + a 2 − 2ra cosψ = r 2 + a 2 − 2ra sin θ cosφ'

For r >> a a 1 1  a  R ≈ r 1− 2 sinθ cosφ' or ≈ 1+ sinθ cosφ' r R r  r  The total vector potential is µIa 2π  a  µ(Iπa2 )sinθ A = cosφ'+ sinθ cos2 φ' dφ'= φ ∫   2 4π 0  r  4πr r B are then equal to r rˆ ∂ θˆ ∂ B = ∇ × A φˆ = ()A sinθ − ()rA φ r sinθ ∂θ φ r ∂r φ µ()Iπa2 = []rˆcosθ +θˆsinθ 2πr 3 Magnetic Dipole The magnitude of the magnetic dipole moment, the counterpart of electric dipole, is defined as m = Iπa2

The dipole direction is pointing up from the loop for the counter- clockwise I direction (as in this case), and vice versa.

The vector potential can then be expressed as r µ  1  A = − 0 mr ×∇  4π  r 