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Lecture 15. Magnetic Fields of Moving Charges and Currents Outline: Magnetic Field of a Moving Charge. Magnetic Field of Currents. Interaction between Two Wires with Current. Lecture 14: Hall Effect. Magnetic Force on a Wire Segment. Torque on a Current-Carrying Loop. 1 Force vs. Torque π2 π₯ πΉβ = πΉπ₯οΏ½ = ππβ = π 2 π₯οΏ½ ππ‘ ππ£β πβ = ππ‘ πβ πβ = πποΏ½ = πΏπβπππ πβπππ ποΏ½ = πΜ × π£οΏ½ πβ π£β π2π = πΏ 2 ποΏ½ ππ‘ ππ π2 π β‘ = πΜ ποΏ½ = 2 ποΏ½ ππ‘ ππ πβ ππ π β‘ πποΏ½ = ποΏ½ ππ π£β πβ × π£β π= 2 π 2 Iclicker Question Imagine that you place a magnet in a uniform magnetic field. Will the magnetic field exert a net force on the magnet? If so, what is the direction of the force (Hint: use the electric dipole analogy). A. yes, in the direction of the magnetic field B. yes, opposite to the direction of the magnetic field C. yes, but the direction depends on the magnetβs orientation D. no, the net force on the magnet is zero. 3 Iclicker Question Imagine that you place a magnet in a uniform magnetic field. Will the magnetic field exert a net force on the magnet? If so, what is the direction of the force (Hint: use the electric dipole analogy). A. yes, in the direction of the magnetic field B. yes, opposite to the direction of the magnetic field C. yes, but the direction depends on the magnetβs orientation D. no, the net force on the magnet is zero. 4 Iclicker Question An elliptical wire loop (3 turns) with current is in the plane of the board. What is the torque on the current loop? 3 turns π΄ = 1π2 π΅ = 1π πΌ = 1π΄ A. 1 Nm into board B. 1 Nm out of board C. 3 Nm up D. 3 Nm down E. Zero π΄ πβ = πβ × π΅ 5 Iclicker Question An elliptical wire loop (3 turns) with current is in the plane of the board. What is the torque on the current loop? 3 turns π΄ = 1π2 π΅ = 1π πΌ = 1π΄ A. 1 Nm into board B. 1 Nm out of board C. 3 Nm up D. 3 Nm down E. Zero π΄ πβ = πβ × π΅ 6 Iclicker Question 7 Iclicker Question 8 Magnetic Field of a Moving Charge Positive charge moving with a constant velocity π βͺ π: π β the speed of light π0 π£β × πΜ π0 π£β × πβ π΅ π = π 2 = π 3 4π π 4π π - the charge is at the origin at this moment 2 ππ π0 = 4π β 10β7 2 πΆ velocity directed into the plane of the page π΅ π2 = 1 π0 π 0 π0 ππ 2 β7 = 1 β 10 4π πΆ2 π0 - the vacuum permeability - a glimpse of a deep sub-structure connecting E and B 9 Iclicker Question A positive point charge is moving directly toward point P. The magnetic field that the point charge produces at point P A. points from the charge toward point P. B. points from point P toward the charge. C. is perpendicular to the line from the point charge to point P. D. is zero. E. The answer depends on the speed of the point charge. π0 π£β × πΜ π΅ π = π 2 4π π 10 Iclicker Question A positive point charge is moving directly toward point P. The magnetic field that the point charge produces at point P A. points from the charge toward point P. B. points from point P toward the charge. C. is perpendicular to the line from the point charge to point P. D. is zero. E. The answer depends on the speed of the point charge. π0 π£β × πΜ π΅ π = π 2 4π π 11 Iclicker Question Two positive point charges move side by side in the same direction with the same velocity. What is the direction of the magnetic force that the upper point charge exerts on the lower one? +q ο² vο +q ο² vο A. toward the upper point charge (the force is attractive) B. away from the upper point charge (the force is repulsive) C. in the direction of the velocity D. opposite to the direction of the velocity E. none of the above π0 π£β × πβ π΅ π = π 3 4π π πΉβ = ππ£β × π΅ 12 Iclicker Question Two positive point charges move side by side in the same direction with the same velocity. What is the direction of the magnetic force that the upper point charge exerts on the lower one? +q ο² vο +q ο² vο A. toward the upper point charge (the force is attractive) B. away from the upper point charge (the force is repulsive) C. in the direction of the velocity D. opposite to the direction of the velocity E. none of the above π0 π£β × πβ π΅ π = π 3 4π π πΉβ = ππ£β × π΅ 13 Magnetic Field of a Moving Charge (contβd) Example: a charge of 10 nC is moving in a straight line at 30 m/s. What is the max magnitude of B produced by the charge at a point 10 cm from its path? What is the max magnitude of E? π΅πππ 30 π0 ππ£ β7 β8 β12 π = = 10 β 10 = 3 β 10 0.12 4π π 2 πΈπππ 10β8 1 π 10 4 π/πΆ = β 10 = 10 0.12 4ππ0 π 2 π΅πππ π£ = π0 π0 π£ = 2 πΈπππ π Magnetic force can be regarded as a relativistic correction to the electrical force. 14 Magnetic Field of Currents Superposition: π£βπ × πΜ π0 π΅ π = ππ π π 4π πβ 2 1820 β first observation of the magnetic field due to currents The charge of carriers in a wire segment π΄π΄π΄: the current segment is at the origin ππ π π = βπ ππ = πππππ πΌ = πππ£π π΄ ππ π π π£βπ = πΌπΌπβ π0 ππβ × πΜ π΅ π = πΌ 4π π2 - proportional to 1/π 2 , as the electric field of a point charge 15 Magnetic Field of a Circular Loop with Current π πΜ O ππβ Find the magnetic field at the center of a circular loop with current. All contribution β in the same direction. Total π΅: Contribution of a small portion of the loop: π0 ππβ × πΜ πΏπ΅ π = πΌ 4π π 2 π0 β ππ οΏ½ πΏπ΅π = πΌ 2 4π π οΏ½ ππ = 2ππ The field at the center of the loop: π0 πΌ π΅ 0 = 2π For N turns with current, the field is N times stronger. 16 Iclicker Question 17 Iclicker Question 18 Iclicker Question A wire consists of two straight sections with a semicircular section between them. If current flows in the wire as shown, what is the direction of the magnetic field at P due to the current? A. to the right B. to the left C. out of the plane of the figure D. into the plane of the figure E. misleading question β the magnetic field at P is zero 19 Iclicker Question A wire consists of two straight sections with a semicircular section between them. If current flows in the wire as shown, what is the direction of the magnetic field at P due to the current? A. to the right B. to the left C. out of the plane of the figure D. into the plane of the figure E. misleading question β the magnetic field at P is zero 20 Magnetic field of an infinitely long straight wire 0 π΅ 0 π Find the magnetic field at a distance π from an infinitely long straight wire with current πΌ. Contribution of a small portion of the wire: πΌππβ All contribution β in the same direction. Total π΅: π0 πΌ π΅ π = 2ππ π0 ππβ × πΜ πΏπ΅ π = πΌ 4π π2 β πΏπ΅π (see Appendix) (the r-dependence resembles that for E(r) of an infinite charged wire) This equation will be obtained next time using Ampereβs Law. 21 Problem Consider a wire bent in the hairpin shape. The wire carries a current πΌ . What is the approximate magnitude of the magnetic field at point a? π πΌ π πΌ Superposition: the field at point a is a superposition of three π΅ fields produced by the current in the semi-circle and two straight wires. Semi-circle: Top wire: Bottom wire: π΅ π = 1 π0 πΌ π0 πΌ = 4π 2 2π π0 πΌ π΅ π = 4ππ π0 πΌ π΅ π = 4ππ (1/2 of the field of a circular loop) (1/2 of the field of an infinite wire) (1/2 of the field of an infinite wire) π0 πΌ π0 πΌ π0 πΌ 2 π΅ π = + = 1+ 4π 2ππ 4π π 22 πΌ2πΌ πΌ1 Interaction between Two Wires with Current πΌ2 πΌ1 The magnetic field due to πΌ1 at the position of the wire with πΌ2 Force per unit length: π0 πΌ1 π΅ π = 2ππ πΉ π0 πΌ1 πΌ2 = πΌ2 π΅ = πΏ 2ππ Parallel (anti-parallel) currents attract (repel) each other. SI unit and definition for electric current: The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2×10-7 newton per meter of length. 23 Iclicker Question The long, straight wire AB carries a 10-A current as shown. The rectangular loop has long edges parallel to AB and carries a clockwise 5-A current. What is the magnitude and direction of the net magnetic force that the straight wire AB exerts on the loop? πΉπ‘π‘π‘ A. C. πΌ2 = 5π΄ π0 2000 2π N to the right N upward (toward AB) B. D. 0.1 π 1π π0 πΌ1 πΌ2 1 1 =πΏ β 2π π1 π2 π0 5000 2π πΌ1 = 10π΄ 0.02 π π0 2000 2π π0 5000 2π N to the left N downward (away from AB) E. misleading question β the net magnetic force is zero 24 Iclicker Question The long, straight wire AB carries a 14-A current as shown. The rectangular loop has long edges parallel to AB and carries a clockwise 5-A current. What is the magnitude and direction of the net magnetic force that the straight wire AB exerts on the loop? πΉπ‘π‘π‘ A. C. πΌ2 = 5π΄ π0 2000 2π N to the right N upward (toward AB) B. D. 0.1 π 1π π0 πΌ1 πΌ2 1 1 =πΏ β 2π π1 π2 π0 5000 2π πΌ1 = 10π΄ 0.02 π π0 2000 2π π0 5000 2π N to the left N downward (away from AB) E. misleading question β the net magnetic force is zero 25 Electrostatics vs. Magnetostatics Elementary source of the static B field: current segment (onedimensional object, vector) Elementary source of the static E field: point charge (zero-dimensional object, scalar) Gaussβ Law: οΏ½ π π’π’π’π’π’π’ πΈ πβ β ππ΄β = πππππ π0 Absence of magnetic monopoles: ππππ - valid in electrostatics only(!), will be modified in electrodynamics. π π’π’π’π’π’π’ π΅ πβ β ππ΄β = 0 - valid in electrodynamics! - valid in electrodynamics! οΏ½ πΈ πβ β ππβ = 0 οΏ½ π΅ π = π0 πΌ 2ππ οΏ½ π΅ πβ β ππβ = π0 πΌ β π ππππ circulation of the field B along the loop π΅ π β For a loop that doesnβt enclose any current, the circulation is 0. In magnetostatics, currents are the only source of the non-zero circulation of B. This will be modified in electrodynamics. 1 π Next time: Lecture 16: Ampereβs Law. §§ 28.6- 28.8 27 Appendix I: Magnetic Force as a Relativistic Correction to the Electric Force In general, electrical repulsion + magnetic attraction. π 1 π2 πΉπΈ = 4ππ0 π 2 Charges at rest (the proton ref. frame), only electric force (repulsion): Charges in motion (the lab. ref. frame, primed), both FE and FB: According to the special theory of relativity: πΉβ₯β² 1 = πΉβ₯ πΎ πΈβ₯β² = πΎπΈβ₯ πΉβ₯β² - force in the lab reference frame πΉβ₯ - force in the proton reference frame π΅β₯β² = πΎπ΅β₯ β² β² πΉβ₯π΅ = πΉβ₯β² β πΉβ₯πΈ πΎ= 1 πΉββ₯ β² = πΉββ₯πΈ β² + πΉββ₯π΅ β π£ 1β π 2 1 πΎ π π£ In the lab ref. frame: πΉβ₯β² 1 1 π2 = πΎ 4ππ0 π 2 1 1 π2 π£ = πΎβ = πΎ π πΎ 4ππ0 π 2 2 1 π2 πΉβ₯πΈ β² = πΎ 4ππ0 π 2 1 π2 π0 ππ = πππΎ = πππ΅β₯β² 2 2 4π π 4ππ0 π Magnetic force could be regarded as a relativistic correction to the electrical force. 28 Appendix II: Integral Form of the Magnetic Field of Currents π ππ ππβ πβ π΅ π πΌ The charge of carriers in a wire segment ππ π π = βπ ππ = πππππ The current: πΌ = π½π΄ = πππ£π π΄ π΄π΄π΄ : ππ π π π£βπ = πΌπΌπβ π½- current density, πΌππβ - current segment π£βπ × πβ β ππ π0 π0 ππβ × πβ β ππ Superposition: π΅ π = ππ π π = πΌ 3 3 4π 4π πβ β ππ πβ β ππ Assuming the current segment is at the origin: The magnetic field of a wire loop with current: π0 ππβ × πβ π0 ππβ × πΜ π΅ π = πΌ = πΌ 4π π3 4π π2 - proportional to 1/π 2 , as an electric field of a point charge ππβ × πβ β ππ π0 π΅ π = πΌοΏ½ 3 4π πβ β ππ 29 Appendix III: Magnetic Field of a Circular Loop with Current Find the magnetic field at the center of a circular loop with current. O π πΌ ππβ ππβ × πβ β ππ π0 π΅ π = πΌοΏ½ 3 4π πβ β ππ Letβs place the origin at the center of the loop π = 0 . πβ β ππ = π ππ = π ππΌ, 0 β€ πΌ β€ 2π 2π π0 ππ π0 πΌ π0 πΌ π΅ 0 = πΌοΏ½ 2 = οΏ½ ππ = 2π 4π 4π π 4π π π 0 The field at the center of the loop: π0 πΌ π΅ 0 = 2π For N turns with current, the field is N times stronger. 30 Appendix IV: Magnetic Field of a Circular Loop with Current ππβ πΌ πβ = π₯ β π₯οΏ½ ππ πβ π πβ β ππ = A wire loop with current has a shape of a circle of radius a. Find the magnetic field at a distance x from its center along the axis of symmetry. ππβ × πβ β ππ π0 π΅ π = πΌοΏ½ 3 4π πβ β ππ π 2 + π₯2 Letβs place the origin at the center of the loop and introduce two angles: πΌ and π. ππ = π ππΌ, 0 β€ πΌ β€ 2π π0 ππ π0 π π΅π₯ π₯ = πΌοΏ½ 2 ππππ = πΌ 4π 4π π 2 + π₯ 2 π + π₯2 π΅π¦ component is zero due to the symmetry. The field at the center of the loop (x=0): 2π ππππ = π π 2 + π₯ 2 π0 π 2 οΏ½ π ππ = πΌ 2 3/2 2 π + π₯2 0 π0 πΌ π΅ 0 = 2π For N turns with current, the field is N times stronger. 3/2 31 Appendix V: Magnetic Field of a Straight Wire Segment π π΅ πβ = 0 π πΌ ππ Find π΅ at a distance π from a straight wire segment carrying πΌ. Letβs choose the origin at the point where we measure π΅ (πβ = 0): ππβ × βππ π0 π΅ πβ = 0 = πΌοΏ½ 3 4π βππ ππβ 1 Express ππβ and ππ in terms of π and πΌ: π = π tanπΌ ππ = π ππΌ cos 2 πΌ 90 + πΌ ππβ × βππ π΅ π = π π πΌ2 π ππ = cosπΌ 2 ππ πππ π = ππ β π β² β sin 90 + πΌ = ππ β π β² β cos πΌ = βπ = cos 2 πΌ cos 2 πΌ 2 3 πΌ2 cos πΌ π0 π π0 πΌ π0 πΌ πΌ οΏ½ ππΌ = οΏ½ cosπΌ ππ = sinπΌ2 β sinπΌ1 2 3 cos πΌ π 4ππ 4ππ 4π πΌ1 πΌ1 Example: Magnetic field of a square loop with current at the center π π of the loop (πΌ2 = , πΌ1 = β ): 4 πΌ 4 π0 πΌ π 4π0 πΌ π΅ π = οΏ½ π΅π = 4 2sin = 4ππ/2 4 2ππ π 32 Magnetic field of an infinitely long straight wire π π΅ πβ = 0 π πΌ ππ 90 + πΌ ππβ π0 πΌ π΅ π = sinπΌ2 β sinπΌ1 4ππ π 2 Infinitely long straight wire: πΌ2 = , πΌ1 = β π0 πΌ π΅ π = 2ππ (the r-dependence resembles that for E(r) of an infinite charged wire) 33 π 2 Problem Consider a wire bent in the hairpin shape. The wire carries a current πΌ . What is the approximate magnitude of the magnetic field at point a? π πΌ π πΌ Superposition: the field at point a is a superposition of three π΅ fields produced by the current in the semi-circle and two straight wires. Semi-circle: Top wire: π΅ π = π0 πΌ π΅ π = 4ππ Bottom wire: π΅ π = 1 π0 πΌ π0 πΌ = 4π 2 2π πΌ2 =0 οΏ½ πΌ1 =βπ/2 πΌ2 =π/2 π0 πΌ οΏ½ 4ππ πΌ1 =0 (1/2 of the field of a circular loop) π0 πΌ cosπΌ ππ = sin0 β sin βπ/2 4ππ π0 πΌ = 4ππ (1/2 of the field of an infinite wire) π0 πΌ cosπΌ ππ = 4ππ π΅ π = π0 πΌ π0 πΌ π0 πΌ 2 + = 1+ 4π 2ππ 4π π 34