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Chapter 30 Sources of the Magnetic Field A whole picture helps Charge q as source Current I as source Gauss’s Law Ampere’s Law Faraday’s Law Electric field E FE qE Magnetic field B Ampere-Maxwell Law Force on q in the field FB q v B Force on q v or I in the filed Summarized in Maxwell equations Lorentz force F qE q v B Math -- review Vector cross product: C AB C AB Magnitude of the vector C: θ B C C AB sin θ Vector cross product: B A θ D B A C Determine the direction. If D AB A FB v B The right-hand rule: 1. Four fingers follow the first vector. 2. Bend towards the second vector. 3. Thumb points to the resultant vector. Sources of Electric field, magnetic field From Coulomb's Law, a point charge dq generates electric field distance r away from the source: dq 1 dq dE rˆ 2 4 0 r r dE From Biot-Savart's Law, a point current Ids generates magnetic field distance r away from the source: dB 0 Ids dB rˆ 2 4 r Difference: 1. A current segment Ids , not a point charge dq, hence a vector. 2. Cross product of two vectors, dB is determined by the right-hand rule. P Ids Total Magnetic Field dB is the field created by the current in the length segment ds , a vector that takes the direction of the current. To find the total field, sum up the contributions from all the current elements Ids μo Ids ˆr B 2 4π r The integral is over the entire current distribution o = 4 x 10-7 T. m / A , is the constant o is called the permeability of free space Example: B from a Long, Straight Conductor with current I The thin, straight wire is carrying a constant current I Constructing the coordinate system and place the wire along the x-axis, and point P in the X-Y plane. ˆ dx cos θ k ˆ ds ˆr dx sin π2 θ k Integrating over all the current elements gives μo I θ2 B cos θ dθ θ 1 4πa μ I ˆ o sin θ1 sin θ2 , or B Bk 4πa The math part ˆ dx cos θ k ˆ ds ˆr dx sin π2 θ k μo Ids ˆ μoI cos 3 θ ˆ μo I ˆ dB r dx k cos θdθ k 4π r 2 4π a 2 4πa a adθ cos θ , x a tan θ , dx r cos 2 θ μo I θ2 B cos θ dθ θ 1 4πa μo I sin θ1 sin θ2 4πa ˆ B Bk Now make the wire infinitely long The statement “the conductor is an infinitely long, straight wire” is translated into: q1 = /2 and q2 = -/2 Then the magnitude of the field becomes μo I B 2πa The direction of the field is determined by the right-hand rule. Also see the next slide. For a long, straight conductor the magnetic field B goes in circles The magnetic field lines are circles concentric with the wire The field lines lie in planes perpendicular to to wire The magnitude of the field is constant on any circle of radius a A different and more convenient right-hand rule for determining the direction of the field is shown Example: B at the center of a circular loop of wire with current I From Biot-Savart Law, the field at O from Ids is 0 Ids 0 Ids ˆ dB rˆ k 2 2 4 r 4 r O μo I μo I μo I B ds 2πr 2 2 4πr full circle 4πr 2r Y This is the field at the center of the loop μo I ˆ B or B Bk Off center points 2r are not so easy to calculate. X r Ids How about a stack of loops? Along the axis, yes, the formula is μo I BN 2r N is the number of turns. When the loop is sufficiently long, what can we say about the field inside the body of this electric magnet? We need Gauss’s Law Oops, typo. We need Ampere’s Law. Sorry, Mr. Ampere. Ampere’s Law Connects B with I Ampere’s law states that the line integral of B ds around any closed path equals oI where I is the total steady current passing through any surface bounded by the closed path: B ds μ I o This is a line integral over a vector field B from a long, straight conductor re-calculated using Ampere’s Law Choose the Gauss’s Surface, oops, not again! Choose the Ampere’s loop, as a circle with radius a. Ampere’s Law says B ds μ I o B is parallel with ds , so B ds B2π a μoI μoI B 2π a When the wire has a size: with radius R Outside of the wire, r > R B ds B( 2πr ) μ o I μo I B 2πr Inside the wire, we need I’, the current inside the ampere’s circle r2 B ds B( 2πr ) μo I ' I ' R 2 I μo I B r 2 2πR Plot the results The field is proportional to r inside the wire The field varies as 1/r outside the wire Both equations are equal at r = R Magnetic Field of a Toroid The toroid has N turns of wire Find the field at a point at distance r from the center of the toroid (loop 1) B ds B( 2πr ) μ N I o μo N I B 2πr There is no field outside the coil (see loop 2) Magnetic Field of a Solenoid A solenoid is a long wire wound in the form of a helix A reasonably uniform magnetic field can be produced in the space surrounded by the turns of the wire The field lines in the interior are nearly parallel to each other uniformly distributed The interior of the solenoid field people use. Magnetic Field of a Tightly Wound Solenoid The field distribution is similar to that of a bar magnet As the length of the solenoid increases the interior field becomes more uniform the exterior field becomes weaker Ideal (infinitely long) Solenoid An ideal solenoid is approached when: the turns are closely spaced the length is much greater than the radius of the turns Apply Ampere’s Law to loop 2: B ds B ds B B ds B o NI path 1 B μo N ds B path 1 I μo n I n = N / ℓ is the number of turns per unit length Magnetic Force Between Two Parallel Conductors Two parallel wires each carry steady currents The field B2 due to the current in wire 2 exerts a force on wire 1 of F1 = I1ℓ B2 Substituting the equation for gives μo I1 I 2 F1 2π a Check with right-hand rule: same direction currents attract each other opposite directions currents repel each other The force per unit length on the wire is FB μo I1 I 2 2π a And this formula defines the current unit Ampere. PLAY ACTIVE FIGURE Definition of the Ampere and the Coulomb The force between two parallel wires is used to define the ampere When the magnitude of the force per unit length between two long, parallel wires that carry identical currents and are separated by 1 m is 2 x 10-7 N/m, the current in each wire is defined to be 1 A The SI unit of charge, the coulomb, is defined in terms of the ampere When a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross section of the conductor in 1 second is 1 C Magnetic Flux, defined the same way as any other vector flux, like the electric flux, the water (velocity) flux The magnetic flux over a surface area associated with a magnetic field is defined as B B dA unit of magnetic flux is T.m2 = Wb (weber) The Magnetic Flux Through a Plane A special case is when a plane of area A, its direction dA makes an angle q with B The magnetic flux is B = BA cos q When the field is perpendicular to the plane, = 0 (figure a) When the field is parallel to the plane, = BA (figure b) PLAY ACTIVE FIGURE Gauss’ Law in Magnetism Yes, that is Gauss’s Law, but in Magnetism. Magnetic fields do not begin or end at any point The number of lines entering a surface equals the number of lines leaving the surface Gauss’ law in magnetism says the magnetic flux through any closed surface is always zero: B dA 0 Example problems A long, straight wire carries current I. A right-angle bend is made in the middle of the wire. The bend forms an arc of a circle of radius r. Determine the magnetic filed at the center of the arc. Formula to use: Biot-Savart’s Law, or more specifically the results from the discussed two examples: π μo I μo I 2 sin θ 0 For the straight section B 4πr 4πr For the arc The final answer: magnitude B direction B μo I 4πr 2 1 circle 4 ds μo I 2πr μo I 2 4πr 4 8r μo I μo I μo I μo I 2 1 4πr 8r 4πr 4r π 2 pointing into the page.