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Biot-Savart’s Law • Rq P q r R I x R x dB q z q dx r z r dB Our Study of Magnetism • Lorentz Force Eqn F qE qv B mv • Motion in a uniform B-field circular orbit R qB • Forces on charges moving in wires • Magnetic dipole • Today: dF Idl B AI B U B fundamentals of how currents generate magnetic fields • Next Time: Ampere’s Law simplifies calculation Today... • Biot-Savart Law for Calculating B-field • Biot-Savart Law (“brute force”) • Ampere’s Law (“high symmetry”) • B-field of an Straight Wire • Force on Two Parallel Current Carrying Wires • B-field of a Circular Loop Last time:Potential Energy of Dipole U B B x B x x =0 = B X =0 U = -B U=0 U = B negative work 1 positive work B Lecture 14, Act 1 • A circular loop of radius R carries current I as shown in the diagram. A constant 1A magnetic field B exists in the +x direction. Initially the loop is in the x-y plane. – The coil will rotate to which of the following positions? y y w a z 1B R a I b x (c) It will not rotate a b B w (b) (a) y b z What is the potential energy U0 of the loop in its initial position? (a) U0 is minimum (b) U0 is maximum (c) neither Lecture 14, Act 1 • A circular loop of radius R carries current I as shown in the diagram. A constant 1A magnetic field B exists in the +x direction. Initially the loop is in the x-y plane. – The coil will rotate to which of the following positions? y y w a z • • • • R a I b x (c) It will not rotate a b B w (b) (a) y b z The coil will rotate if the torque on it is non-zero: τ μ B The magnetic moment is in +z direction. Therefore the torque is in the +y direction. Therefore the loop will rotate as shown in (b). Lecture 14, Act 1 • A circular loop of radius R carries current I as shown in the diagram. A constant magnetic field B exists in the +x direction. 1B Initially the loop is in the x-y plane. – What is the potential energy U0 of the loop in its initial position? y B R I b a x (a) U0 is minimum (b) U0 is maximum (c) neither – The potential energy of the loop is given by: U μ B – In its initial position, the loop’s magnetic moment vector points in the +z direction, so initial potential energy is ZERO – This does NOT mean that the potential energy is a minimum!!! – When the loop is in the y-z plane and its magnetic moment points in the same direction as the field, its potential energy is NEGATIVE and is in fact the minimum. – Since U0 is not minimum, the coil will rotate, converting potential energy to kinetic energy! Calculation of Electric Field • Two Ways to calculate – Coulomb’s Law dq dE k 2 rˆ r "Brute force" – Gauss’ Law e 0 E dS q "High symmetry" What are the analogous equations for the Magnetic Field? Calculation of Magnetic Field • Two Ways to calculate μ0 I dl rˆ dB 4π r 2 – Biot-Savart Law (“Brute force”) – Ampere’s Law (“High symmetry”) I B dl 0 I –AMPERIAN SURFACE/LOOP These are the analogous equations Biot-Savart Law... ...bits and pieces μ0 I dl rˆ μ0 I dl r dB 2 4π r 4π r 3 dl q r X dB I N 0 4 10 A2 7 The magnetic field “circulates” around the wire Use right-hand rule: thumb along I, fingers curl in direction of B. Preflight 14: A current carrying wire (with no remarkable symmetry) is oriented in the x-y plane. Points A,B, & C lie in the same plane as the wire. The z-axis points out of the screen. 5) In what direction is the magnetic field contribution from the segment dl at point A. Check all non-zero components. +x -x +y -y +z -z 6) In what direction is the magnetic field contribution from the segment dl at point B. Check all non-zero components. +x -x +y -y +z -z 7) In what direction is the magnetic field contribution from the segment dl at point C. Check all non-zero components. +x -x +y -y +z -z dB points in the direction of dl r μ0 I dl r dB 4π r 3 A: dl is to the right, and r is up dB is out of the page B: dl is to the right, and r is up and right dB is out of the page C: dl is to the right, and r is down and right dB is into the page Conclusion: at every point above the wire, dB is . Below the wire , dB is Preflight 14: 9) Would any of your answers for questions 5-7 change if we integrated dl over the whole wire? NO! Why or why not? μ0 I dl r dB 4π r 3 dB points in the direction of dl r At point A: dl is to the right, and r is up dB is out of the page At point B: dl is to the right, and r is up and right dB is out of the page At point C: dl is to the right, and r is down and right dB is into the page For every point in the x-y plane and every piece of wire dl: every dl and every r are always in the x-y plane Since dB must be perpendicular to r and dl, dB is always in the ±z direction! Conclusion: At every point above the wire, the dB due to every piece dl is . Below the wire, the dB due to every piece dl is B-field of Straight Wire • Calculate field at point P using Biot-Savart Law: μ0 I dx r dB 4π r 3 (z ) Which way is B? y P q r R q dx I x μ0 I (dx)r sin θ 3 4 π r B dB This calculation appears in its entirety in the appendix Its result is: μ0 I B 2πR • Units of B are Tesla • 1 Tesla = 104 Gauss • Earth’s B-field ~ 0.5 G Putting it all together • We know that a current-carrying wire can experience force from a B-field. • We know that a a current-carrying wire produces a Bfield. • Therefore: We expect one current-carrying wire to exert a force on another current-carrying wire: Ib d F F • Current goes together wires come together • Current goes opposite wires go opposite Ia F on 2 Parallel CurrentCarrying Wires • Calculate force on length L of wire b due to field of wire a: The field at b due to a is given by: F • Ib d Ia μ0 I a Magnitude 0 I a Ib L Ba F I L B = b b a of F on b 2d 2πd • Calculate force on length L of wire a due to field of wire b: The field at a due to b is given by: Ib F d Ia × μ0 I b Magnitude F I L B 0 I a Ib L Bb = a a b 2 d of F on a 2πd 2 Preflight 14: 2) Two slack wires are carrying current in opposite directions. What will happen to the wires? They will: a) attract b) repel c) twist due to torque 3) Now, two slack wires are carrying current in the same direction. What will happen to the wires? They will: a) attract b) repel c) twist due to torque a) Find B due to one wire at the position of the other wire b) Use F IL B to find the direction of F in each case a: Point your thumb down, your fingers wrap in the direction of B around the wire: The direction of B due to the left wire at the position of the right wire is out of the screen. b: I is up, B is out of the screen, so F is to the right. the force is repulsive. Lecture 14, Act 2A • A current I flows in the +y direction in an infinite wire; a current I also flows in the loop as shown in the diagram. – What is Fx, the net force on the loop in the x-direction? (a) Fx < 0 (b) Fx = 0 I Ftop y Fleft X Fright I (c) Fx > 0 Fbottom x • You may have remembered a previous act in which the net force on a current loop in a uniform B-field is zero, but the B-field produced by an infinite wire is not uniform! • Forces cancel on the top and bottom of the loop. • Forces do not cancel on the left and right sides of the loop. • The left segment is in a larger magnetic field than the right • Therefore, Fleft > Fright Lecture 14, Act 2B I What is the magnitude of the magnetic field at the center of a loop of radius R, carrying current I? R (a) B = 0 (b) B = (0I)/(2R) (c) B = (0I)•(2R) We can immediately guess the right answer must be (b). (a) is wrong, because all parts of the loop contribute to a B going into the paper at the center of the loop. (c) also cannot be correct—it has the wrong units: B Amp meter Lecture 14, Act 2 dl I But how do we calculate B? r We must use Biot-Savart Law to calculate the magnetic field at the center: R 0 I dl r dB 4 r 3 Two nice things about calculating B at the center of the loop: • I dl is always perpendicular to r: dl r r into page • r is constant (r = R) μ0 I (dl ) R B ˜ dB ˜ 4 π R3 μ0 I ˜ dl 2 4πR μ0 I μ0 I (2 π R) 2 4πR 2R Note: B Amp meter Circular Loop, in more detail • Circular loop of radius R carries current I. Calculate B along the axis of the loop: • Rq • Magnitude of dB from element dl: 0 I dl 0 I dl dB = = 4 r 2 4 z2 +R2 r q z R x dB z r dB • What is the direction of the field? • Symmetry B in z-direction. 0 I dl dB z = cosq 2 2 4 z +R cosq = R z2 +R2 0 I R Bz = dl 2 2 3/2 4 (z +R ) Circular Loop, cont. • Rq 0 I R Bz = dl 4 (z 2 +R2 ) 3/2 r q z dl = 2 R R Bz = 0 z r x IR2 Bz • Expressed in terms of the magnetic moment: = dB 2(z 2 +R2 ) 3/2 • Note the form the field takes for z >> R: I R2 dB 0 Bz 2 z3 note the typical dipole field behavior! 0 IR2 2z 3 Circular Loop, anywhere on axis Bz y= 0 IR 2 2 z R 2 0 IR 2 Bz z R 2z3 2 32 Expressed in terms of the magnetic moment I R 2 f(x) 1 0 Bz z R 2 z 3 Bz Note the typical 1/z3 dipole field behavior! 0 0 0 0 z R x= x 3 3 Lecture 14, Act 3 • Equal currents I flow in identical circular loops as shown in the diagram. The loop 3A on the right (left) carries current in the ccw (cw) direction as seen looking along the +z direction. I o I x B A – What is the magnetic field Bz(A) at point A, the midpoint between the two loops? x (a) Bz(A) < 0 3B (b) Bz(A) = 0 o (c) Bz(A) > 0 • What is the magnetic field Bz(B) at point B, just to the right of the right loop? (a) Bz(B) < 0 (b) Bz(B) = 0 (c) Bz(B) > 0 z Lecture 14, Act 3 • Equal currents I flow in identical circular loops as shown in the diagram. The loop 3A on the right (left) carries current in the ccw (cw) direction as seen looking along the +z direction. I o I x B A – What is the magnetic field Bz(A) at point A, the midpoint between the two loops? x (a) Bz(A) < 0 • • • • (b) Bz(A) = 0 o (c) Bz(A) > 0 The right current loop gives rise to Bz <0 at point A. The left current loop gives rise to Bz >0 at point A. From symmetry, the magnitudes of the fields must be equal. Therefore, B(A) = 0 z Lecture 14, Act 3 • Equal currents I flow in identical circular loops as shown in the diagram. The loop on the right (left) carries current in the 3B ccw (cw) direction as seen looking along the +z direction. I o I x B A – What is the magnetic field Bz(B) at point B, just to the right of the right loop? x (a) Bz(B) < 0 (b) Bz(B) = 0 z o (c) Bz(B) > 0 • The signs of the fields from each loop are the same at B as they are at A! • However, point B is closer to the right loop, so its field wins! Summary • Biot-Savart Law for Calculating B-Fields μ0 I dl rˆ μ0 I dl r dB 2 3 4π r 4π r • Current-Carrying Wires Make B-Fields and Are Affected by B-Fields • Current Loop Produces a Dipole Field Appendix A: B-field of Straight Wire y • Calculate field at point P using Biot-Savart Law: μ0 I dx r dB 4π r 3 Which way is B? +z μ0 I (dx)r sin θ 3 4 π r • Rewrite in terms of R,q : B dB r R sin θ R tan θ x 1 therefore, dx R dθ 2 sin θ P q r R q I dx x R cot θ dx(sin θ ) sin θdθ 2 r R x Appendix A: B-field of Straight Wire P π μ0 I dθ B sin θ 4π R 0 q r q dx π μ0 I B sin θdθ 4πR 0 therefore, R I μ0 I π cos θ 0 B 4πR μ0 I B 2πR x