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Electricity and Magnetism - Physics 121 Lecture 10 - Sources of Magnetic Fields (Currents) Y&F Chapter 28, Sec. 1 - 7 Magnetic fields are due to currents.

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Presentation on theme: "Electricity and Magnetism - Physics 121 Lecture 10 - Sources of Magnetic Fields (Currents) Y&F Chapter 28, Sec. 1 - 7 Magnetic fields are due to currents."— Presentation transcript:

1 Electricity and Magnetism - Physics 121 Lecture Sources of Magnetic Fields (Currents) Y&F Chapter 28, Sec Magnetic fields are due to currents The Biot-Savart Law for wire element and moving charge Calculating field at the centers of current loops Field due to a long straight wire Force between two parallel wires carrying currents Ampere’s Law Solenoids and toroids Field on the axis of a current loop (dipole) Magnetic dipole moment Summary

2 Previously: moving charges and currents feel a force in a magnetic field
Magnets come only as dipole pairs of N and S poles (no monopoles). Magnetic field exerts a force on moving charges (i.e. on currents). The Lorentz force is perpendicular to both the field and the velocity Magnetic force can not change a particle’s speed or KE A charged particle moving in a uniform magnetic field moves in a circle or a spiral. A wire carrying current in a magnetic field feels a force also using cross product. This force is the motor effect. Current loops behave as Magnetic dipoles, with dipole moment, torque, and potential energy given by:

3 Currents Create Magnetic Fields
Oersted : A magnetic compass is deflected by current  Magnetic fields are due to currents (free charges & in wires) Why do the un-magnetized filings line up with the field? In fact, currents are the only way to create magnetic fields. Electromagnetic crane The magnitude of the field created is proportional to

4 Biot-Savart Law (1820): Field due to current in a wire
Same basic role as Coulomb’s Law: magnetic field due to a source Source strength is the “current-length” i ds Field falls off as inverse-square of distance New constant m0 measures “permeability” Direction of B field depends on a cross-product (Right Hand Rule) x P’ “vacuum permeability” m0 = 4px10-7 T.m/A. Unit vector along r - from source to P 10-7 exactly Differential addition to field at P due to distant source current element ids Find total field B by integrating over the whole current region (need lots of symmetry) For a straight wire the magnetic field lines are circles wrapped around it. Another Right Hand Rule shows the direction:

5 Magnetic field due to a moving point charge
Variant of Biot-Savart Law Field source qv has dimensions of current-length Magnitude of B falls off as inverse-square of distance Direction of B via cross-product (Right Hand Rule) Field at a point changes as particle moves past x Unit vector along r - from source to P 10-7 exactly Field at P due to distant source qv snapshot The magnetic field lines are circles wrapped around the current length qv. Velocity into page

6 Direction of Magnetic Field
10 – 1: Which sketch below shows the correct direction of the magnetic field, B, near the point P? A B C D E i P B into page Use RH rule for current segments: thumb along ids - curled fingers show B

7 Example: Magnetic field at the center of a current arc
Circular current arc, radius R Find B at center, point C f is included arc angle, not the cross product angle Cross product angle q is always 900 here dB at center C is toward us ds = Rdf i Integrate on arc angle f’ from 0 to f Thumb points along the current. Curled fingers show direction of B Right hand rule for wire segments Another Right Hand Rule (for current loops only): Curl fingers along current, thumb shows direction of B at center Apply to circular current loop - f = 2p radians: ?? What would formula be for f = 45o, 180o, 4p radians ??

8 Examples: into page same as closed loop into out of page page
FIND B FOR A POINT LINED UP WITH A SHORT STRAIGHT WIRE P Find B AT CENTER OF A HALF LOOP, RADIUS = r into page Find B AT CENTER OF TWO HALF LOOPS OPPOSITE EQUAL CURRENTS PARALLEL EQUAL CURRENTS same as closed loop into out of page page

9 Magnetic field from current loops
10 – 2: The three loops below have the same current. The smaller radius is half of the large one. Rank the loops by the magnitude of magnetic field at the center, greatest first. I II III. I, II, III. III, I, II. II, I, III. III, II, I. II, III, I. Hint: consider radius, direction, arc angle

10 Magnetic field of a long, thin current-carrying straight wire
a is distance perpendicular to wire through P L = length of wire >> a Field of infinitely long wire Above formula derived by direct integration along finite-length wire or using Ampere’s Law Field intensity falls off as 1/a Magnetic field lines are concentric circles around wire RIGHT HAND RULE FOR A WIRE FIELD LINES ARE CIRCLES THEY DO NOT BEGIN OR END

11 Magnetic Field lines near a straight wire carrying current
Two parallel wires carrying current: the magnetic field from one causes a force on the other. . The force is attractive when the currents are parallel. . The force is repulsive when the currents are anti-parallel. Example: Force Magnetic Field lines near a straight wire carrying current i i out of slide

12 L ia d ib Magnitude of the force between two long parallel wires
x ia L d ib Third Law says: Fab = - Fba Use formula for field of infinitely long wire Field of wire a at wire b Into page via RH rule Attractive force for parallel currents Repulsive force for opposed currents Evaluate Fab = force on b due to field of a ibL is normal to Ba Force is toward wire a Fab= - Fba ia ib End View Example: Two parallel wires are 1 cm apart |i1| = |i2| = 100 A.

13 At home: forces on parallel wires carrying currents
10 – 3: Which situation below results in zero force on the center wire? The currents in all the wires have the same magnitude. Zero F if Bnet = 0 A. B. C. D. 1 2 3 4 Answer A Hints: What is the field midway between wires for parallel & opposite currents? What is the net field direction and relative magnitude at center wire

14 passing through the loop
Ampere’s Law Derivable from Biot-Savart Law – A theorem (not proven here) Sometimes a way to find B, given the current that creates it But B is inside an integral  usable only for high symmetry (like Gauss’ Law) Line Integral over an “Amperian loop” - a closed path. Adds up components of B along the loop path. All paths that enclose a certain set of currents have same result. ienc= net current passing through the loop Picture for applications: Only the tangential component of B along ds contributes to the dot product Current outside the loop (i3) creates field but contributions to the path integral cancel around closed path. Another version of RH rule: - curl fingers along Amperian loop - thumb shows + direction for net current

15 Now Ampere’s Law proves it again more simply for a long fat wire.
Example: Find magnetic field outside a long, straight, cylindrical wire carrying current The Biot-Savart Law shows that for a thin infinitely long wire: Now Ampere’s Law proves it again more simply for a long fat wire. End view of wire with radius R R Amperian loop outside R can have any shape Choose a circular loop (of radius r>R) because field lines are circular around a wire. B and ds are parallel and B is constant everywhere on the Amperian path The path integration simplified. ienc is the total current i. Solve for B, get same formula as for the thin wire: R has no effect on the result as long as R < r. (reminiscent of Shell Theorem)

16 Magnetic field inside a long straight wire carrying current, via Ampere’s Law
Assume cylindrical symmetry Assume current density J = i/A is uniform across the wire cross-section. Field lines inside wire are again concentric circles B is cylindrically symmetric again Again, choose a circular Amperian loop path around the axis, of radius r < R. The enclosed current is less than the total current i, because some is outside the Amperian loop. The amount enclosed is r R ~1/r ~r B Apply Ampere’s Law: Outside (r > R), the wire looks like an infinitely thin wire (previous expression) Inside: B(r) grows linearly up to R

17 At home: count the current enclosed by an Amperian Loop
10 – 4: Rank the Amperian paths shown by the value of along each path in descending order, starting with the most positive, taking current direction into account. All of the wires are carrying the same current.. I, II, III, IV, V. II, III, IV, I, V. III, V, IV, II, I. IV, V, III, I, II. I, II, III, V, IV. I. II. III. IV. V. Answer: E

18 Another Ampere’s Law example
Why use COAXIAL CABLE for CATV and other applications? Find B inside and outside the cable Cross section: center wire current i out of sketch shield wire current i into sketch Amperian loop 1 loop 2 Field Inside – use Amperian loop 1: Field Outside – use Amperian loop 2: Outer shield current does not affect field inside Reminiscent of Gauss’s Law Zero field outside due to opposed currents + radial symmetry Losses and interference suppressed

19 Solenoids strengthen fields by using many loops
strong uniform field in center fields from adjacent coils cancel “Long solenoid”  d << L L d Approximation: field is constant inside and zero outside (just like capacitor) FIND FIELD INSIDE IDEAL SOLENOID USING AMPERIAN LOOP abcda Outside B = 0, no contribution from path c-d B is perpendicular to ds on paths d-a and b-c Inside B is uniform and parallel to ds on path a-b inside ideal solenoid only section that has non-zero contribution

20 AMPERIAN LOOP IS A CIRCLE ALONG B
Toroid: A long solenoid bent into a circle Find the magnitude of B inside Choose a different Amperian loop, parallel to the field, circular, with radius r (inside the toroid) The toroid has a total of N turns The Amperian loop encloses current Ni. B is constant on the Amperian path. i outside flows up LINES OF CONSTANT B ARE CIRCLES Inversely proportional to r Same result as for long solenoid N times the result for a long thin wire Find B field outside (ienc=0) Answer AMPERIAN LOOP IS A CIRCLE ALONG B

21 Find B field at point P on z-axis of a current loop (dipole)
Use the Biot-Savart Law directly i is into page f is the angle at the center of the loop. Integrate on azimuth f around the current loop. The field is perpendicular to r but by symmetry the component of B normal to z-axis cancels around the loop - only the part parallel to the z-axis survives. as before recall definition of Dipole moment

22 Same 1/z3 dependence as for field of electrostatic dipole
B field on the axis of a dipole (current loop), continued Far field: suppose z >> R Same 1/z3 dependence as for field of electrostatic dipole Dipole moment vector is normal to loop (RH Rule). For any current loop, on z axis with |z| >> R For charge dipole Dipole-dipole interaction: r 3 Torque depends on Current loops are the elementary sources of magnetic field: Create dipole fields proportional to source strength Dipole 2 feels torque proportional to in external B field

23 Try this at home Answer: B I. II. III.
10-5: The three loops below have the same current. Rank them in terms of the magnitude of magnetic field at the point shown, greatest first. I, II, III. III, I, II. II, I, III. III, II, I. II, III, I. I II III. Hint: consider radius, direction, arc angle Answer: B

24 EXTRA: Magnetic field due to current in a thin wire, finite length
Current i flows to the right along x – axis Wire subtends angles q1 and q2 at P Find B at point P, a distance a from wire. dB points out of page at P for ds anywhere along wire Evaluate dB due to current ids using Biot Savart Law a Angles q, q1, q2 measured CW from –y axis x shown is negative, q positive, q1 positive, q2 negative Integrate on q from q1 to q2: Symmetric Case:

25 Magnetic field near current-carrying thin, straight wire
Special Case: Infinitely long, thin wire. Use result above: FIELD LINES ARE CIRCLES THEY DO NOT BEGIN OR END RIGHT HAND RULE FOR A WIRE a is distance perpendicular to wire through P Example: Field at P due to Semi-Infinite wires A & B: Field is into slide at point P Half the magnitude found for a fully infinite wire Zero Contribution From A A B


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