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dl. ´. I. x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x. r. a. Examples of Magnetic Field Calculations. Today. Example B-field Calculations: Infinite Straight Wire revisited Inside a Long Straight Wire Infinite Current Sheet
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dl ´ I x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x r a Examples of Magnetic Field Calculations
Today... Example B-field Calculations: • Infinite Straight Wire revisited • Inside a Long Straight Wire • Infinite Current Sheet • Solenoid – with applications! • Toroid
Integral around a path … hopefully a simple one Current “enclosed” by that path ´ I Today is Ampere’s Law Day "High symmetry"
B-field of ¥ Straight Wire, revisited dl ´ I R • Choose loop to becircleof radius R centered on the wire in a plane^ to wire. • Why? • Magnitude of B is constant (function of R only) • Direction of B is parallel to the path. • Evaluate line integral inAmpere’s Law: • ApplyAmpere’s Law: • Calculate field at distance Rfrom wire using Ampere's Law: • Current enclosed by path =I Þ Ampere's Law simplifies the calculation thanks to symmetry of the current! (axial/cylindrical)
Preflight 15: R dl L L dl 2) Evaluate the integral: dl around a circular loop. a) 0 b) 2R c) R2 d) not enough info 3) Evaluate the integral: dl around a square loop. a) 0 b) 4L c) L2 d) not enough info
Preflight 15: A B 4) For which loop is B • dlthe greatest? Two identical loops are placed in proximity to two identical current carrying wires. A B Same
Preflight 15: 5) Now compare loops B and C. For which loop is B • dlthe greatest? C B B C Same
Lecture 15, Act 1 A current Iflows in an infinite straight wire in the+z direction as shown. A concentric infinite cylinder of radius Rcarries current 2Iin the-z direction. What is the magnetic field Bx(a) at point a, just outside the cylinder as shown? y a 2A x b x x 2I x x I x x x (c) Bx(a) > 0 (b) Bx(a) = 0 (a) Bx(a) < 0 x • What is the magnetic field Bx(b) at point b, just inside the cylinder as shown? 2B (a) Bx(b) < 0 (c) Bx(b) > 0 (b) Bx(b) = 0
Lecture 15, Act 1 A currentIflows in an infinite straight wire in the+z direction as shown. A concentric infinite cylinder of radius Rcarries current 2Iin the-z direction. What is the magnetic field Bx(a) at point a, just outside the cylinder as shown? y a 2A x b x x 2I x x I x x x (c) Bx(a) > 0 (b) Bx(a) = 0 (a) Bx(a) < 0 x • This situation has cylindrical symmetry • By applying Ampere’s Law, we see that the field at point • a must just be the field from an infinite wire withcurrent I • flowing in the-z direction! B x B I B B
Lecture 15, Act 1 A currentIflows in an infinite straight wire in the+z direction as shown. A concentric infinite cylinder of radius Rcarries current 2Iin the-z direction. What is the magnetic field Bx(b) at point b, just inside the cylinder as shown? y a 2B x b x x 2I x x I x x x (c) Bx(b) > 0 (b) Bx(b) = 0 (a) Bx(b) < 0 x • This time, the Ampere loop only enclosescurrent Iin the • +z direction— the loop is inside the cylinder! • The current in the cylindrical shell does not contribute to • at pointb.
r a • Bfield is only a function of rÞtake path to be circle of radiusr: Þ • Current passing through circle: Þ • Ampere's Law: B Field Inside a Long Wire x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x • Suppose a total currentIflows through the wire of radiusainto the screen as shown. • CalculateB field as a function ofr, the distance from the center of the wire.
a B r B Field of a Long Wire • Inside the wire: (r < a) • Outside the wire: • ( r > a )
I I a 3a 2a 3a 1A (a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a) • What is the relation between the magnetic field at R = 2afor the two cases (L = left, R = right)? 1B (a) BL(2a)< BR(2a) (b) BL(2a)= BR(2a) (c) BL(2a)> BR(2a) Lecture 15, Act 2 • Two cylindrical conductors each carry currentIinto the screen as shown. The conductor on the left is solid and hasradiusR = 3a. The conductor on the right has a hole in the middle and carries current only betweenR = aandR = 3a. • What is the relation between the magnetic field at R = 6a for the two cases (L= left, R= right)?
I I a 3a 2a 3a 1A (a) BL(6a)< BR(6a) (b) BL(6a)= BR(6a) (c) BL(6a)> BR(6a) Lecture 15, Act 2 • Two cylindrical conductors each carry currentIinto the screen as shown. The conductor on the left is solid and hasradiusR=3a.The conductor on the right has a hole in the middle and carries current only betweenR=a andR=3a. • What is the relation between the magnetic field at R = 6a for the two cases (L=left, R=right)? • Use Ampere’s Law in both cases by drawing a loop in the plane of • the screen atR=6a • Both fields have cylindrical symmetry, so they are tangent to the • loop at all points, thus the field at R=6a only depends oncurrent • enclosed • Ienclosed = I in both cases
I I a 3a 2a 3a Again, field only depends upon current enclosed... 1B LEFT cylinder: RIGHT cylinder: (a) BL(2a)< BR(2a) (b) BL(2a)= BR(2a) (c) BL(2a)> BR(2a) Lecture 15, Act 2 • Two cylindrical conductors each carry currentIinto the screen as shown. The conductor on the left is solid and hasradiusR=3a.The conductor on the right has a hole in the middle and carries current only betweenR=aandR=3a. • What is the relation between the magnetic field at R = 2a for the two cases (L=left, R=right)?
x x x w x x x x x x x • Calculate using Ampere's law for a square of • sidew: x x constant constant • • Þ therefore, B Field of ¥ Current Sheet • Consider an ¥sheet of current described byn wires/lengtheach carryingcurrentiinto the screen as shown.Calculate the B field. • What is the direction of the field? • Symmetry Þvertical direction
L a B Field of a Solenoid • A constant magnetic field can (in principle) be produced byan¥sheet of current. In practice, however, a constant magnetic field is often produced by a solenoid. • A solenoid is defined by a currentiflowing through a wire that is wrappednturns per unit length on a cylinder of radiusa and lengthL. • To correctly calculate the B-field, we should use Biot-Savart, and add up the field from the different loops. • Ifa<< L,the B field is to first order contained within the solenoid, in the axial direction, and of constant magnitude. In this limit, we can calculate the field using Ampere's Law.
x x x x x x x x x x x • To do this, view the ¥solenoid from the • side as 2 ¥current sheets. • • • • • • • • • • • • • • x x x x x x x x • Draw square path of side w: • • • • • • • • • • • Þ Note: B Field of an ¥ Solenoid • To calculate the B field of thesolenoid using Ampere's Law, we need to justify the claim that the B field is nearly 0 outside the solenoid (for an ¥solenoid the B field is exactly 0 outside). • The fields are in the same direction in the region between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid).
Preflight 15: A current carrying wire is wrapped around an iron core, forming an electro-magnet. 6) Which direction does the magnetic field point inside the iron core? a) left b) right c) up d) down e) out of the screen f) into the screen 7) Which side of the solenoid should be labeled as the magnetic north pole? a) right b) left
Use the wrap rule to find the B-field: wrap your fingers in the direction of the current, the B field points in the direction of the thumb (to the left). Since the field lines leave the left end of solenoid, the left end is the north pole.
Solenoids The magnetic field of a solenoid is essentially identical to that of a bar magnet. The big difference is that we can turn the solenoid on and off ! It attracts/repels other permanent magnets; it attracts ferromagnets, etc.
• • • • • x x x • • x x x x x • x • r x x x x • • x x x • • • • B • Þ Apply Ampere’s Law: Toroid • Toroid defined by Ntotal turns with currenti. • B=0outside toroid! (Consider integratingBon circle outside toroid) • To findBinside, consider circle ofradiusr,centered at the center of the toroid.
Magnet off plunger held in place by spring Magnet on plunger expelled strikes bell • Close switch • current magnetic field pulls in plunger closes larger circuit Solenoid Applications Digital [on/off]: • Doorbells • Power door locks • Magnetic cranes • Electronic Switch “relay” • Advantage: • A small current can be used to switch a much larger one • Starter in washer/dryer, car ignition, …
In fact, a typical car has over 20solenoids! Solenoid Applications Analog (deflection µI ): • Variable A/C valves • Speakers Solenoids are everywhere!
Summary Example B-field Calculations: • Inside a Long Straight Wire • Infinite Current Sheet • Solenoid • Toroid • Circular Loop m I r 0 B = 2 2 p a 2 m iR 0 B » 3 z 2z