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Lesson 8 Ampère’s Law and Differential Operators Section 1 Visualizing Ampère’s Law Amperian Loop •An Amperian loop is –any closed loop •Amperian loops include: –a circle –a square –a rubber band •Amperian loops do not include: –a balloon –a piece of string (with two ends) The Field Contour of a Wire Take a wire with current coming out of the screen. The Field Contour of a Wire The field contour is made of half-planes centered on the wire. The Field Contour of a Wire We draw arrows on each plane pointing in the direction of the magnetic field. Amperian Loops We draw an Amperian loop around the wire. Amperian Loops We wish to count the “net number” of field lines pierced by the Amperian loop. Amperian Loops First, we put an arrow on the loop in an overall counterclockwise direction. Counting Surfaces Pierced To count the net number of surfaces pierced by Amperian loop, we add +1 when the loop is “in the direction’ of the plane and −1 when it is “opposite the direction” of the plane. +1 +1 +1 +1 +1 −1 +1 +1 +1 +1 −1 +1 +1 −1 +1 +1 +1 +1 −1 +1 +1 +1 +1 +1 Counting Surfaces Pierced Note there are “+1” appears 20 times and “-1” appears 4 times. +1 +1 +1 −1 +1 +1 +1 +1 +1 +1 +1 +1 +1 −1 +1 −1 +1 +1 −1 +1 +1 +1 +1 +1 Counting Surfaces Pierced The net number of surfaces pierced by the Amperian loop us therefore +16. +1 +1 +1 −1 +1 +1 +1 +1 +1 +1 +1 +1 +1 −1 +1 −1 +1 +1 −1 +1 +1 +1 +1 +1 Other Amperian Loops What is the net number of surfaces pierced by each of these Amperian loops? Other Amperian Loops What is the net number of surfaces pierced by each of these Amperian loops? Other Amperian Loops What is the net number of surfaces pierced by each of these Amperian loops? Other Amperian Loops The net numbers of surfaces pierced by each of these loops is 16. Other Amperian Loops What is the net number of surfaces pierced by this Amperian loop? Other Amperian Loops This time the net number of surfaces pierced by the loop is 0. Why? Other Wires. This is the same loop we saw earlier, but now only 8 surfaces are pierced, since there are only 8 surfaces extending outward from the wire. Other Wires. There are 8 surfaces coming from the wire because the current through the wire is half as much as it was before. Ampère’s Law The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the loop. Sign Convention •Always traverse the Amperian loop in a (generally) counterclockwise direction. •If the number of surfaces pierced N>0, the current comes out of the screen. •If N<0, the current goes into the screen. Section 2 Cylindrically Symmetric Current Density Current Density •There are three kinds of charge density (ρ,σ,λ) •There is one kind of current density (current/unit area) I j A Current Density • The current passing through a small gate of area ΔA is I jA small gate Current Density • The total current passing through the wire is the sum of the current passing through all small gates. I j dA small gates Cylindrically Symmetric Current Distribution The current density, j, can vary with r only. Below, we assume that the current density is greatest near the axis of the wire. Cylindrically Symmetric Current Distribution Outside the distribution, the field contour is composed of surfaces that are half planes, uniformly spaced. Cylindrically Symmetric Current Distribution Inside the distribution, it is difficult to draw perpendicular surfaces, as some surfaces die out as we move inward. – We need to draw many, many surfaces to keep them equally spaced as we move inward. Cylindrically Symmetric Current Distribution But we do know that if we draw enough surfaces, the distribution of the surfaces will be uniform, even inside the wire. Cylindrically Symmetric Current Distribution Let’s draw a circular Amperian loop at radius r. r Cylindrically Symmetric Current Distribution Now we split the wire into two parts – the part outside the Amperian loop and the part inside the Amperian loop. r r Cylindrically Symmetric Current Distribution The total electric field at r will be the sum of the electric fields from the two parts of the wire. r r Inside a Hollow Wire The total number of perpendicular surfaces pierced by the Amperian loop is zero because there is no current passing through it. r How can we get zero net surfaces? 1. We could have all the surfaces pierced twice, one in the positive sense and one in the negative… … but this violates symmetry! How can we get zero net surfaces? 2. We could have some surfaces oriented one way and some the other… … but this violates symmetry, too! How can we get zero net surfaces? 3. Or we could just have no surfaces at all inside the hollow wire. This is the only way it can be done! The Magnetic Field inside a Hollow Wire If the current distribution has cylindrical symmetry, the magnetic field inside a hollow wire must be zero. Cylindrically Symmetric Current Distribution Since the magnetic field inside a hollow wire is zero, the total magnetic field at a distance r from the center of a solid wire is the field of the “core,” the part of the wire within the Amperian loop. r r Cylindrically Symmetric Current Distribution Outside the core, the magnetic field is the same as that of a thin wire that has the same current as the total current passing through the Amperian loop. r Cylindrically Symmetric Current Distribution Inside a cylindrically symmetric current distribution, the magnetic field is: 0ienc B (r ) 2 r r Section 3 Uniform Current Density Example: Uniform Current Distribution A wire of radius R with a uniform current distribution has a total charge i passing through it. What is the magnetic field at r < R ? Example: Uniform Current Distribution A wire of radius R with a uniform current distribution has a total charge i passing through it. What is the magnetic field at r < R ? 0ienc B (r ) 2 r r Example: Uniform Current Distribution A wire of radius R with a uniform current distribution has a total charge i passing through it. What is the magnetic field at r < R ? 0ienc B (r ) 2 r r The current density is uniform, so: ienc i j A Aenc Example: Uniform Current Distribution A wire of radius R with a uniform current distribution has a total charge i passing through it. What is the magnetic field at r < R ? 0ienc B (r ) 2 r Therefore: r ienc Aenc i A Example: Uniform Current Distribution 0ienc B (r ) 2 r Example: Uniform Current Distribution 0ienc B (r ) 2 r ienc Aenc i A Example: Uniform Current Distribution 0ienc B (r ) 2 r r i R2 2 ienc Example: Uniform Current Distribution 0 r i B(r ) 2 2 r R 2 Example: Uniform Current Distribution 0 r i B(r ) 2 2 r R 2 Example: Uniform Current Distribution 0ir B(r ) 2 2 R Section 4 The Line Integral Line Integral We know that the magnetic field is stronger where the perpendicular surfaces are closer together. number of surfaces pierced N Bk k length of field line segment Therefore, the number of surfaces pierced is 1 N B k Let B Line Integral Therefore, the number of surfaces pierced is N k •Λ is called the “line integral” •The line integral is proportional to the number of contours pierced. •Λ=Bℓ if ℓ is a section of a field line and B is constant on ℓ. Line Integral •Λ is called the line integral because it is more generally given by the expression B d d B d B d B d cos Line Integral • Note that this is similar to expression for work you learned in mechanics. Work is the line integral of force along the path an object follows. B d W F d d F dW F d F d cos Line Integral The line integral is a way of measuring the number of contour surfaces pierced by a line segment. B d d B d B d B d cos Line Integral Roughly speaking, it is a measure of how much field lines along a path. B d d B d B d B d cos Ampère’s Law and the Line Integral The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the Amperian loop. Ampère’s Law and the Line Integral The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the Amperian loop. Therefore: B d 0ienc Class 23 Today, we will use Ampere’s law to find the magnetic fields • inside and outside a long, straight wire with radial charge density • of a plane of wires • of a solenoid • of a torus Section 5 Applying Ampère’s Law Ampère’s Law – the Practical Version B 0ienc Ampère’s Law – the Practical Version B 0ienc The number of surfaces pierced Ampère’s Law – the Practical Version B 0ienc The number of surfaces pierced The magnetic field on the Amperian loop – must be a constant over the whole loop. Ampère’s Law – the Practical Version B 0ienc The number of surfaces pierced The magnetic field on the Amperian loop – must be a constant over the whole loop. The length of a the closed Amperian loop (or the part where the field is non-zero). Ampère’s Law – the Practical Version B 0ienc The number of surfaces pierced The magnetic field on the Amperian loop – must be a constant over the whole loop. The length of a the closed Amperian loop (or the part where the field is non-zero). The total current passing through the Amperian loop. Ampère’s Law – the Practical Version B 0ienc 0 jdA This is the line integral This is NOT the line integral Section 6 Ampère’s Law and Cylindrical Wires A Typical Problem • A wire of radius R has current density j magnetic field inside the wire. r .2 Find the • “Inside the wire” means at some point P at a radius r < R. P r R Choosing the Amperian Loop • What shape of Amperian Loop should we choose for current traveling through a cylindrical wire? Choosing the Amperian Loop • Choose a circular Amperian loop. r A wire of radius R has current density j r 2. Find the magnetic field inside the wire. What is the correct expression for the ℓ in the line integral? 2 r Integrating the Current Density • How do we do slice the region inside the Amperian loop to integrate the current density? Integrating the Current Density • We slice the wire into rings. •The current though each ring is dI = j dA A wire of radius R has current density j r 2. Find the magnetic field inside the wire. What is the correct expression for ienc ? r ienc r 2 rdr 2 0 A wire of radius R has current density j r 2. Find the magnetic field inside the wire. r B(r ) 0 4 3 Another Problem • A wire of radius R has current density magnetic field outside the wire. j .r 2 . Find the • “Outside the wire” means at some point P at a radius r > R. P r R A wire of radius R has current density j r 2. Find the magnetic field outside the wire R ienc r 2 rdr 2 0 R B(r ) 0 4r 4 Section 7 Other Applications of Ampère’s Law Ampere’s Law – Plane of Wires The magnetic field from each wire forms circular loops around the wire. r Therefore B is perpendicular to . B1 1 2 3 P Ampere’s Law – Plane of Wires Consider the magnetic field from three wires. B3 B2 B1 d 1 2 3 P Ampere’s Law – Plane of Wires B3 B2 B1 d 1 2 3 P When we add the magnetic fields from each wire, the vector sum points upward. Ampere’s Law – Plane of Wires B 0ienc B Now let’s look at the line integral of the magnetic field around the dotted Amperian loop. Ampere’s Law – Plane of Wires B 0ienc B d Now let’s look at the line integral of the magnetic field around the dotted Amperian loop. Ampere’s Law – Plane of Wires B 0ienc B d The bottom part of the Amperian loop pierces no contour surfaces – so the line integral here is zero. We also know B d but the B field and the path are perpendicular on this segment so Λ=0 for this part of the path. Λ=0 for this top of the path, too. Ampere’s Law – Plane of Wires B d B d Conclusion: The line integral over the top and bottom segments of the Amperian loop is zero because the magnetic field is perpendicular to the path. Ampere’s Law – Plane of Wires B 0ienc B d The line integral over the right side of the Amperian loop is right Bd Ampere’s Law – Plane of Wires B 0ienc B d The line integral over the left side of the Amperian loop is left Bd Ampere’s Law – Plane of Wires B 0ienc B d The total line integral is: B2d The enclosed current is the number of wires in the loop times the current in each wire: ienc Ni Ampere’s Law – Plane of Wires B 0ienc B d B 2d 0 Ni B 0 N 2 d i 0 ni 2 N where n d n is the number of wires per unit length. Ampere’s Law – Two Planes of Wires What would the field be like if there were two planes of wires with currents in opposite directions? Ampere’s Law – Two Planes of Wires Field lines Ampere’s Law – Two Planes of Wires Contour surfaces What can you conclude about the magnetic field? Ampere’s Law – Two Planes of Wires Field lines of the right plane Ampere’s Law – Two Planes of Wires Field lines of the left plane Ampere’s Law – Two Planes of Wires The field of the both planes Ampere’s Law – Two Planes of Wires The field of the both planes Ampere’s Law – Two Planes of Wires B 0ienc What would the field be like if there were two planes of wires with currents in opposite directions? d B B0 1 2 3 B 2 0 ni 2 0 ni Ampere’s Law – Solenoid B 0ienc d B B0 1 2 3 A solenoid is similar to two planes of wires. The magnetic field inside this solenoid points downward. The magnetic field outside the solenoid is zero. Ampere’s Law – Solenoid B 0ienc Bd 0 Ni d B B0 1 2 3 N B 0 i 0 ni d N where n d Right-hand Rule #3 The direction of the magnetic field inside a solenoid is given by a right-hand rule: B Circle the fingers of your right hand in the direction of the current. The magnetic field is in the direction of your thumb. Ampere’s Law – Torus B 0ienc A torus is like a solenoid wrapped around a doughnut-shaped core. The magnetic field inside forms circular loops. The magnetic field outside is zero. r B Ampere’s Law – Torus B 0ienc B 2 r 0 Ni r 0 Ni B 2 r B Ampere’s Law – Torus B 0ienc B 2 r 0 Ni r 0 Ni B 2 r B The wires on the inside of the torus are closer together, so the field is stronger there. Class 24 Today, we will use direct integration to find • electric fields of charged rods and loops • electric potentials of charged rods and loops • magnetic fields of current-carrying wire segments and loop segments (Biot-Savart law) Section 8 Finding Fields by Direct Integration Geometry for Extended Objects r origin to source r origin to P R source to P r R r r P R r The Basic Laws The electric field and potential of a small charge dq: 1 Rdq dE (r ) 40 R 3 1 dq dV (r ) 40 R The magnetic field of a current i in a small length of wire d : 0 i d R dB(r ) 4 R 3 Origin of the Basic Laws Electric field and potential for slowly moving point charges – Coulomb’s Law: E (r ) V (r ) q 1 q rˆ r 2 3 40 r 40 r 1 1 q 40 r The Basic Laws – for dq Electric field and potential for dq. 1 Rdq dE (r ) 40 R 3 dV (r ) 1 dq 40 R Origin of the Basic Laws Remember the geometry from Lesson 2 y head r h line tail rt θ T rh s thread ray rr line ψ line S P P U motion of source x Origin of the Basic Laws An expression for the magnetic field of a point charge: 1 1 B rˆh E rh E c crh 1 s rh rr E crh 1 s rh E because E is parallel to rr crh 1 s E c Origin of the Basic Laws An expression for the magnetic field of a point charge (moving slowly): 1 1 q s r B(r ) s E (r ) 3 c c 40 r 1 q 0 q B(r ) v r v r 2 3 s 3 s 40c r 4 r as c 2 1 0 0 Origin of the Basic Laws Magnetic field for dq 0 dq dB(r ) v r 3 s 4 r d dq dq vs dq d i d dt dt 0 i dB(r ) d r 3 4 r A little sleight of hand, but it’s the same as a more formal proof. The Biot-Savart Law 0 id R dB(r ) 3 4 R The formula for the magnetic field of a wire segment A current i passes through the wire segment The length of the wire segment is dℓ. The direction of dℓ is the direction of the current. r is the vector from the origin to a field point. R is the vector from the segment (the origin) to a field point. Equations for Extended Objects 1 Rdq dE (r ) 3 40 R dV (r ) 1 dq 40 R 0 id R dB(r ) 3 4 R P R r r Now, let’s work some problems… A Charged Rod y P x −L +L The rod has a linear charge density q 2L Find the electric field at P. A Charged Rod y P x −L +L 1 dq dE (r ) R 3 40 R We need to find dq, R, and R. A Charged Rod r −L 1. Find P y +L r A Charged Rod r −L 1. Find P y +L r r yyˆ A Charged Rod P r r −L 1. Find r r yyˆ 2. Choose a slice and find +L r A Charged Rod P r r −L 1. Find r r yyˆ 2. Choose a slice and find +L r r xxˆ Be sure to put primes on the slice variables! A Charged Rod P r r −L 1. Find r r yyˆ 2. Choose a slice and find +L r r xxˆ 3. Find the length and charge of the slice. A Charged Rod dx −L 1. Find r P r r r yyˆ 2. Choose a slice and find +L r r xxˆ 3. Find the length and charge of the slice. dq dx A Charged Rod P r R r −L 1. Find r r yyˆ 2. Choose a slice and find +L r r xxˆ 3. Find the length and charge of the slice. 4. Find R r r dq dx A Charged Rod P r R r −L 1. Find r r yyˆ 2. Choose a slice and find +L r r xxˆ 3. Find the length and charge of the slice. 4. Find R r r R yyˆ x xˆ dq dx A Charged Rod P r R r −L 1. Find r r yyˆ 2. Choose a slice and find +L r r xxˆ 3. Find the length and charge of the slice. 4. Find 5. Find R r r R yyˆ x xˆ R dq dx A Charged Rod P r R r −L 1. Find r r yyˆ 2. Choose a slice and find +L r r xxˆ 3. Find the length and charge of the slice. 4. Find 5. Find R r r R yyˆ x xˆ R R x 2 y 2 dq dx A Charged Rod dE ( r ) 1. Find dq R 3 40 R 1 r Now just substitute! r yyˆ 2. Choose a slice and find r r xxˆ 3. Find the length and charge of the slice. 4. Find 5. Find R r r R yyˆ x xˆ R R x 2 y 2 dq dx A Charged Rod dE (r ) 1. Find dq 1 dx R 3 40 R 40 x 2 y 2 1 r r yyˆ 2. Choose a slice and find yyˆ xxˆ r r xxˆ 3/ 2 3. Find the length and charge of the slice. 4. Find 5. Find R r r R yyˆ x xˆ R R x 2 y 2 dq dx A Charged Rod dE (r ) dq 1 dx R 3 40 R 40 x 2 y 2 1 y E (r ) yˆ 40 1. Find r dx L x L 2 y 2 3/ 2 r yyˆ 2. Choose a slice and find 3/ 2 xˆ 40 r yyˆ xxˆ x L 5. Find 2 y 2 3/ 2 r xxˆ 3. Find the length and charge of the slice. 4. Find x dx L R r r R yyˆ x xˆ R R x 2 y 2 dq dx A Charged Rod dE (r ) dq 1 dx R 3 40 R 40 x 2 y 2 1 y E (r ) yˆ 40 dx L x L 2 y 2 3/ 2 3/ 2 xˆ 40 yyˆ xxˆ x dx L x L 2 y I won’t expect you to evaluate these integrals! E (r ) L 20 y y L 2 2 yˆ 2 3/ 2 A Charged Rod y P x −L +L The rod has a linear charge density q 2L Now find the electric potential at P. A Charged Rod dV (r ) 1. Find r dq 1 dx 40 R 40 x 2 y 2 1 r yyˆ 2. Choose a slice and find r 1/ 2 r xxˆ 3. Find the length and charge of the slice. 4. Find 5. Find R r r R yyˆ x xˆ R R x 2 y 2 dq dx A Charged Rod dV (r ) V (r ) 1. Find r dq 1 dx 40 R 40 x 2 y 2 1 40 dx L x L 2 y 1/ 2 2 1/ 2 r yyˆ 2. Choose a slice and find r r xxˆ 3. Find the length and charge of the slice. 4. Find 5. Find R r r R yyˆ x xˆ R R x 2 y 2 dq dx A Charged Rod dq 1 dx 40 R 40 x2 y 2 1/ 2 L dx V (r ) 2 40 L x y 2 1/ 2 dV (r ) 1 You don’t need to evaluate this integral, either! V (r ) ln 40 L2 y 2 L L2 y 2 L Current in a Wire Segment y P i x −L +L Current i travels to the left along a segment of wire. Find the magnetic field at P. Current in a Wire Segment r −L 1. Find P y i +L r Current in a Wire Segment r −L 1. Find P y i +L r r yyˆ Current in a Wire Segment P r r −L 1. Find r r yyˆ 2. Choose a slice and find i +L r Current in a Wire Segment P r r −L 1. Find i r r yyˆ 2. Choose a slice and find +L r r xxˆ Be sure to put primes on the slice variables! Current in a Wire Segment P r r −L 1. Find i r r yyˆ 2. Choose a slice and find 3. Find d +L r r xxˆ Current in a Wire Segment dx −L 1. Find r P r i r r yyˆ 2. Choose a slice and find r r xx ˆ 3. Find d d dx xˆ dx xˆ +L Current in a Wire Segment P r R r −L 1. Find i r r yyˆ 2. Choose a slice and find r r xx ˆ 3. Find d d dx xˆ dx xˆ 4. Find R r r +L Current in a Wire Segment P r R r −L 1. Find i r r yyˆ 2. Choose a slice and find r r xx ˆ 3. Find d d dx xˆ dx xˆ 4. Find R r r R yyˆ x xˆ +L Current in a Wire Segment P r R r −L 1. Find i r r yyˆ 2. Choose a slice and find r r xx ˆ 3. Find d d dx xˆ dx xˆ 4. Find R r r R yyˆ x xˆ 5. Find R +L Current in a Wire Segment P r R r −L 1. Find i r r yyˆ 2. Choose a slice and find r r xx ˆ 3. Find d d dx xˆ dx xˆ 4. Find R r r R yyˆ x xˆ 5. Find R R x 2 y 2 +L Current in a Wire Segment R −L 1. Find r P r r r yyˆ i 2. Choose a slice and find r r xx ˆ 3. Find d d dx xˆ dx xˆ 4. Find R r r R yyˆ x xˆ 5. Find R 6. Find d R 2 2 R x y +L Current in a Wire Segment R −L 1. Find r P r r r yyˆ i 2. Choose a slice and find r r xx ˆ 3. Find d d dx xˆ dx xˆ 4. Find R r r R yyˆ x xˆ 5. Find R 6. Find d R 2 2 R x y d R ydx zˆ +L Remember: xˆ yˆ zˆ yˆ zˆ xˆ zˆ xˆ yˆ yˆ xˆ zˆ zˆ yˆ xˆ xˆ zˆ yˆ xˆ xˆ 0 yˆ yˆ 0 zˆ zˆ 0 Current in a Wire Segment 0 i d R 0i ydx dB(r ) zˆ 3 4 R 4 x 2 y 2 1. Find r r yyˆ 5. Find R 2 2 R x y 3/ 2 2. Choose a slice and find r r xx ˆ 3. Find d d dx xˆ dx xˆ 4. Find R r r R yyˆ x xˆ 6. Find d R d R ydx zˆ Current in a Wire Segment 0 i d R 0i ydx dB(r ) zˆ 3 4 R 4 x 2 y 2 0 iy L dx B(r ) zˆ 4 L x 2 y 2 3 / 2 3/ 2 1. Find r r yyˆ 5. Find R 2 2 R x y 2. Choose a slice and find r r xx ˆ 3. Find d d dx xˆ dx xˆ 4. Find R r r R yyˆ x xˆ 6. Find d R d R ydx zˆ A Charged Loop Segment y P The rod has a linear charge density q a / 2 Find the electric field at P. a x A Charged Loop Segment y 1. Find r P a x A Charged Loop Segment y 1. Find r r 0 P a x A Charged Loop Segment y r r 1. Find r 0 2. Slice and find r P x a A Charged Loop Segment y r r P x a 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ A Charged Loop Segment y r r P ad x a 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find the charge of the slice. A Charged Loop Segment y r r P ad x a 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find the charge of the slice. dq ad A Charged Loop Segment y r r P ad x a 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find the charge of the slice. dq ad 4. Find R r r A Charged Loop Segment y r r P ad x a 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find the charge of the slice. dq ad 4. Find R r r R a cos xˆ a sin yˆ A Charged Loop Segment y r r P ad x a 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find the charge of the slice. dq ad 4. Find R r r R a cos xˆ a sin yˆ 5. Find R A Charged Loop Segment y r r P ad x a 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find the charge of the slice. dq ad 4. Find R r r R a cos xˆ a sin yˆ Ra 5. Find R A Charged Loop Segment E (r ) /2 1 dq 1 ad a cos xˆ a sin yˆ R 3 3 40 R 40 0 a r 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find the charge of the slice. dq ad 4. Find R r r R a cos xˆ a sin yˆ Ra 5. Find R A Charged Loop Segment E (r ) /2 1 dq 1 ad a cos xˆ a sin yˆ R 3 3 40 R 40 0 a Here we only have to integrate sines and cosines, so the integrals are easy! E (r ) ( xˆ yˆ ) 40 a A Segment of a Current Loop y P i a Current i travels counterclockwise along a segment of a loop of wire. Find the magnetic field at P. x A Segment of a Current Loop y 1. Find r P i a x A Segment of a Current Loop y 1. Find r r 0 P i a x A Segment of a Current Loop i y r r 1. Find r 0 2. Slice and find r P x a A Segment of a Current Loop i y r r x a P 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ A Segment of a Current Loop i y r r ad x a P 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find d A Segment of a Current Loop i y r r ad x a P 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find d d ad sin xˆ ad cos yˆ A Segment of a Current Loop i y r r ad x a P 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find d 4. Find d ad sin xˆ ad cos yˆ R r r A Segment of a Current Loop i y r r ad x a P 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find d 4. Find d ad sin xˆ ad cos yˆ R r r R a cos xˆ a sin yˆ A Segment of a Current Loop i y r r ad x a P 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find d 4. Find 5. Find d ad sin xˆ ad cos yˆ R r r R R a cos xˆ a sin yˆ A Segment of a Current Loop i y r r ad x a P 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find d 4. Find 5. Find d ad sin xˆ ad cos yˆ R r r R a cos xˆ a sin yˆ R Ra A Segment of a Current Loop i y r r ad x a P 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find d 4. Find 5. Find d ad sin xˆ ad cos yˆ R r r R a cos xˆ a sin yˆ R Ra 6. Find d R A Segment of a Current Loop i y r r ad x a P 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find d 4. Find 5. Find d ad sin xˆ ad cos yˆ R r r R a cos xˆ a sin yˆ R Ra 2 2 2 2 2 6. Find d R a sin zˆ a cos d zˆ a d zˆ A Segment of a Current Loop 0i d R 0i / 2 a 2 d B(r ) zˆ 3 4 R 4 0 a 3 r 1. Find r 0 2. Slice and find r r a cos xˆ a sin yˆ 3. Find d 4. Find 5. Find d ad sin xˆ ad cos yˆ R r r R a cos xˆ a sin yˆ R Ra 2 2 2 2 2 6. Find d R a sin zˆ a cos d zˆ a d zˆ A Segment of a Current Loop 0i / 2 a 2 d 0i B(r ) zˆ zˆ 3 4 0 a 8a This is a really easy integral this time! Class 25 Today, we will: • learn the definition of divergence in terms of flux. • learn the definition of curl in terms of the line integral. • • find the gradient, divergence, and curl in terms of derivatives (differential operators) • write Gauss’s laws and Ampere’s law in differential form • work several sample problems Section 9 Gauss’s Law and Divergence Gauss’s Law The net number of electric field lines passing through a Gaussian surface is proportional to the charge enclosed. Gauss’s Law This is true no matter how small the Gaussian surface is. But the number of field lines gets smaller as the volume gets smaller. Divergence Define divergence to be E div E lim v 0 v where div E is the divergence of the E field at point P v is a small volume around point P Divergence Divergence is a scalar field – a scalar defined at every point in space – that tells us if diverging (or converging) field lines are being produced at that point. The larger the divergence, the more field lines are produced. Divergence and Gauss’s Law In a very small volume, charge density is nearly constant. (That is, until we get to the atomic scale where we can start seeing protons and electrons.) Divergence and Gauss’s Law In a very small volume, charge density is nearly constant. E qenc 0 1 0 as v 0 (r )dv (r ) 0 v Divergence and Gauss’s Law In a very small volume, charge density is nearly constant. E qenc 0 1 0 (r )dv as v 0 (r ) E div E lim v 0 v 0 (r ) 0 v Gauss’s Law in Differential Form div E (r ) 0 Divergence and Gauss’s Law Electrical charge produces field lines that tend to spread from (or converge to) a point in space. Divergence is a measure of how much field lines spread from (+) or converge to (-) a point of space. It is a measurement of “spreadingness.” Section 10 Ampère’s Law and Curl Ampère’s Law The net number of perpendicular surfaces pierced by an Amperian loop is proportional to the current passing through the loop. Ampere’s Law This is true no matter how small the Amperian loop is. But the number of surfaces pierced gets smaller as the area of the loop gets smaller. Curl Take a point in space and a line in the x direction passing through the point. The x-component of the curl (around the line) is defined to be: x curl B(r ) x lim a 0 a P x̂ Curl Take a point in space and a line in the y direction passing through the point. The y-component of the curl (around the line) is defined to be: y curl B(r ) y lim a 0 a ŷ P x̂ Curl Take a point in space and a line in the z direction passing through the point. The z-component of the curl (around the line) is defined to be: z curl B(r ) z lim a 0 a ŷ P ẑ x̂ Curl Take a point in space and a line in the x direction passing through the point. The curl (around the line) is defined by: x curl B(r ) x lim a 0 a where curl B(r ) x is the x component of the curl a is the area of the Amperian loop Curl Curl is a vector field, a vector defined a every point in space. The x component of curl tells us if something at the point is producing field lines that loop about a line going in the x direction and passing through the point. Curl and Ampère’s Law In a very small area, current density is nearly constant. Curl and Ampère’s Law In a very small area, current density is nearly constant. 0ienc 0 as a 0 j x (r ) da 0 j x (r )a Curl and Ampère’s Law In a very small volume, the density is nearly a constant. 0ienc 0 j x (r ) da 0 j x (r )a as a 0 x curl B(r ) x lim 0 j x (r ) a 0 a Curl and Ampère’s Law More generally: curl B(r ) 0 j (r ) The curl points in the direction of the current at any point in space. Curl and Ampère’s Law More generally: curl B(r ) 0 j (r ) Curl is a measure of how much field lines ccw (+) or cw to (-) around the direction of the current. It is a measurement of “loopiness.” Curl and Ampère’s Law Electrical current produces magnetic field lines that form loops around the path of the moving charges. Section 11 Differential Operators The Gradient The gradient is a three-dimensional generalization of a slope (derivative). The gradient tells us the direction a scalar field increases the most rapidly and how much it changes per unit distance. The Gradient In terms of derivatives, the gradient is: xˆ yˆ zˆ x y z Electric Field and Electric Potential E ( x, y, z ) V V V V E ( x, y, z ) V xˆ yˆ zˆ x y z The Divergence Operator We can show that another way of representing divergence is in terms of derivatives. Ex E y Ez div E E x y z The Curl Operator We can also express the curl in terms of derivatives. Bz B y Bx Bz B y Bx curlB B xˆ zˆ yˆ y z z x x y xˆ x Bx yˆ y By zˆ z Bz Gauss’s Laws and Ampère’s Law in Differential Form E 0 B 0 B 0 j Some Problems In a region of space, the electric potential is given by the expression V x 2 y where is a constant. Find the electric field. V V V E ( x, y, z ) V xˆ yˆ zˆ x y z 2 ˆ 2xyx x yˆ Some Problems In a region of space, the electric potential is given by the expression V x 2 y where is a constant. Find the charge density. 2 ˆ E ( x, y, z ) 2xyx x yˆ E 0 E y E x 0 E 0 0 2 0y x y Some Problems In a region of space, the magnetic field is given by the expression B( x, y) ( x 2 3 y 2 ) xxˆ (3x 2 y 2 ) yyˆ where and are constants. Find β in terms of α. Some Problems B( x, y ) ( x 2 3 y 2 ) xxˆ (3x 2 y 2 ) yyˆ Bx By B 0 x y ( x 3 y ) x 2 x (3x y ) y 2 y 0 2 2 2 2 (3x 2 3 y 2 ) (3x 2 3 y 2 ) 0 Some Problems In a region of space, the magnetic field is given by the expression B( x, y ) ( x 2 3 y 2 ) xxˆ (3x 2 y 2 ) yyˆ where is a constant. Find the current density (magnitude and direction). 1 B 0 j j B 0 Some Problems B( x, y ) ( x 2 3 y 2 ) xxˆ (3x 2 y 2 ) yyˆ 1 j B 0 1 Bz B y 1 Bx Bz 1 ˆ ˆ ˆ x z y 0 y z 0 z x 0 1 B y Bx zˆ 0 x y 12xy zˆ 6 xy 6 xy zˆ 0 0 B y Bx x y Some Problems B( x, y ) ( x 2 3 y 2 ) xxˆ (3x 2 y 2 ) yyˆ 1 j B 0 1 Bz B y 1 Bx Bz 1 ˆ ˆ ˆ x z y 0 y z 0 z x 0 1 B y Bx zˆ 0 x y 12xy zˆ 6 xy 6 xy zˆ 0 0 B y Bx x y