Chapter 30 - Magnetic Fields and Torque. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University

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1 Chapter 30 - Magnetic Fields and Torque A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007

2 Objectives: After completing this module, you should be able to: Determine the magnitude and direction of the force on a current-carrying carrying wire in a -field. Calculate the magnetic torque on a coil or solenoid of area A,, turns N,, and current I in a given -field. Calculate the magnetic field induced at the center of a loop or coil or at the interior of a solenoid.

Calculate the magnetic torque on a coil or solenoid of area A,, turns N,, and current I in a

3 Force on a Moving Charge Recall that the magnetic field in in teslas (T) was defined in in terms of of the force on on a moving charge: Magnetic Field Intensity : F qv sin F v F v 1 N 1 N 1 T C(m/s) A m N S

4 Force on a Conductor Since a current Iis is charge q moving through a wire, the magnetic force can be be given in in terms of of current. x x x x x x x x F x x x x x x x x x x x x x x x x x x x x x x x L Motion of +q I = q/t Right-hand hand rule: force F is upward. F = qv Since v = L/t, and I = q/t, we can rearrange to find: L q F q L t t The force F on a conductor of length L and current I in perpendicular -field: F = IL

x x x x x x x x F x x x x x x x x x x x x x x x x x x x x x x x L Motion of +q I = q/t Right-hand hand

5 Force Depends on Current Angle Just as for a moving charge, the force on a wire varies with direction. F = IL sin sin v sin I F v Current I in wire: Length L Example 1. A wire of length 6 cm makes an angle of 20 0 with a 3 mt magnetic field. What current is needed to cause an upward force of 1.5 x 10-4 N? I -4 F 1.5 x 10 N -3 0 I I = 2.44 A Lsin (3 x 10 T)(0.06m)sin20

A wire of length 6 cm makes an angle of 20 0 with a 3 mt magnetic field.

6 Forces on a Current Loop Consider a loop of of area A = ab carrying a current I I in in a constant field as as shown below. a b x x x x x x x x x x x x x x x x x x x x x x x 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 N F 1 Torque A Normal vector F 2 n S The right-hand hand rule shows that the side forces cancel each other and the forces F 1 and 1 F 2 cause 2 a torque.

a b x x x x x x x x x x x x x x x x x x x x x x x I x x x x x x x x x x x x x x x x x x x x x x

7 a Torque on Current Loop Recall that torque is product of force and moment arm. b x x x x x x x x x x I x x x x x x x x x x 1 2 The moment arms for F 1 and F 2 are: a 2 sin F 1 = F 2 = Ib ( Ib)( sin ) a 2 ( Ib)( sin ) a 2 a 2 2( Ib)( sin ) I( ab)sin a F 1 I out 2 sin a 2 a 2 a n X 2 sin IAsin In general, for a loop of N turns carrying a current I,, we have: NIAsin I in F 2

F 1 = F 2 = Ib ( Ib)( sin ) a 2 ( Ib)( sin ) a 2 a 2 2( Ib)( sin ) I( ab)sin a F 1 I out 2

8 Example 2: A 200-turn coil of wire has a radius of 20 cm and the normal to the area makes an angle of 30 0 with a 3 mt -field. What is the torque on the loop if the current is 3 A? A NIAsin AR (.2m) 2 2 A = m 2 ; N = 200 turns = 3 mt; = 30 0 ; I = 3 A N N = 200 turns n S = 3 mt; = 30 0 NIAsin (200)(3 A)(0.003T)(0.126 m )sin Resultant torque on loop: = Nm

9 Magnetic Field of a Long Wire When a current II passes through a long straight wire, the magnetic field is is circular as as is is shown by by the pattern of of iron filings below and has the indicated direction. I Iron filings The right-hand hand thumb rule: Grasp wire with right hand; point thumb in in direction of of I. I. Fingers wrap wire in in direction of of the circular -field. I

10 Calculating -field for Long Wire The magnitude of of the magnetic field at at a distance r r from a wire is is proportional to to current I. I. Magnitude of -field for current I at distance r: 0I 2 r The proportionality constant is called the permeability of free space: Permeability: = 4 x 10-7 Tm/A Circular I r X

I. Magnitude of -field for current I at distance r: 0I 2 r The proportionality

11 Example 3: A long straight wire carries a current of 4 A to the right of page. Find the magnitude and direction of the -field at a distance of 5 cm above the wire. r 5 cm =? I = 4 A r = 0.05 m I = 4 A 0I 2 r -7 Tm (4 x 10 A )(4 A) 2 (0.05 m) = 1.60 x T or or T T Right-hand hand thumb rule: Fingers point out of of paper in in direction of of -field. out of paper r I = 4 A

I = 4 A r = 0.05 m I = 4 A 0I 2 r -7 Tm (4 x 10 A )(4 A) 2 (0.05 m) = 1.

12 Example 4: Two parallel wires are separated by 6 cm.. Wire 1 carries a current of 4 A and wire 2 carries a current of 6 A in the same direction. What is the resultant -field at the midpoint between the wires? 3 cm 3 cm I 2 = 6 A =? I 1 = 4 A Resultant is vector sum: R = 0I 2 r 1 is positive 2 is negative 2 2 into paper 1 out of paper 1 x 6 A 4 A

direction. What is the resultant -field at the midpoint between the wires?

13 Example 4 (Cont.): Find resultant at midpoint. 3 cm 3 cm I 2 = 6 A =? I 1 = 4 A Tm (4 x 10 A )(4 A) 2(0.03 m) -7 Tm (4 x 10 A )(6 A) 2(0.03 m) 26.7 T 40.0 T 0I 2 r Resultant is vector sum: R = R = 26.7 T 40 T = T 1 is positive 2 is negative R is into paper: = T

14 Force etween Parallel Wires Recall wire with I 1 creates 1 at P: P 1 0I1 2 d d F 2 d I 2 I 1 Out of paper! Now suppose another wire with current I 2 in same direction is parallel to first wire. Wire 2 experiences force F 2 due to 1. From right-hand hand rule, what is direction of F 2? Force F 22 is is Downward I 2 F 2

Now suppose another wire with current I 2 in same direction is parallel to

15 Parallel Wires (Cont.) Now start with wire 2. I 2 creates 2 at P: 2 0I2 2 d INTO paper! 2 into paper 2 1 d d x I 2 F 1 I 1 P Now the wire with current I 1 in same direction is parallel to first wire. Wire 1 experiences force F 1 due to 2. From right-hand hand rule, what is direction of F 1? Force F 11 is is Upward F 1 I 1

16 Parallel Wires (Cont.) We have seen that two parallel wires with currents in in the same direction are attracted to to each other. d Attraction F 2 I 2 F 1 I 1 Use right-hand hand force rule to to show that oppositely directed currents repel each other. Repulsion F 2 d I 1 F 1 I 2

17 Calculating Force on Wires The field from current in wire 2 is given by: 2 The force F 1 on wire 1 is: 0I2 2 d F 11 = II L 2 1 d Attraction F 2 I 2 F 1 I 1 L I 2 d 0 2 F1 I1 L The force per unit length for for two wires separated by by d is: is: The same equation results when considering F 2 due to 1 F L I I 2 d 0 1 2

1 L I 2 d 0 2 F1 I1 L The force per unit length for for two wires separated by by

18 Example 5: Two wires 5 cm apart carry currents. The upper wire has 4 A north and the lower wire has 6 A south. What is the mutual force per unit length on the wires? 2 Upper wire F 2 I 2 = 4 A d=5 cm F1 I 1 = 6 A 1 L Lower wire F L I I 2 d I 1 = 6 A; I 2 = 4 A; d = 0.05 m Right-hand hand rule applied to either wire shows repulsion. F L (4 x 10 )(6 A)(4 A) -7 Tm A 2 (0.05 m) F L x 10 N/m

What is the mutual force per unit length on the wires?

19 Magnetic Field in a Current Loop Right-hand hand rule shows field directed out of of center. I N I Out Single loop: 0I 2R Coil of N loops: 0NI 2R

20 The Solenoid A solenoid consists of of many turns N of of a wire in in shape of of a helix. The magnetic - - field is is similar to to that of of a bar magnet. The core can be be air air or or any material. Permeability N S If If the core is is air: 4 4x Tm/A The relative permeability r uses this value as a comparison. The relative permeability for for a medium ( r r ): ): r or r0 0

The core can be be air air or or any material.

21 The -field for a Solenoid For a solenoid of of length L, L, with N turns and current I, I, the -field is is given by: Solenoid N L S NI L Such a -field is is called the magnetic induction since it it arises or or is is produced by by the current. It It applies to to the interior of of the solenoid, and its its direction is is given by by the right-hand hand thumb rule applied to to any current coil.

22 Example 6: A solenoid of length 20 cm and 100 turns carries a current of 4 A. The relative permeability of the core is 12,000. What is the magnetic induction of the coil? I = 4 A; N = 100 turns L = 0.20 m; (12000)(4 x10 ) Tm A r 0 7 Tm A N = 100 turns 20 cm I = 4 A ( )(100)(4 A) Tm A m = 30.2 T A ferromagnetic core can significantly increase the -field!!

23 Summary of Formulas The force F on on a wire carrying current I in in a given -field. F = IL sin sin I sin I F v Current I in wire: Length L N F 1 A n S The torque on on a loop or or coil of of N turns and current I in in a -field at at known angle.. F 2 NIAsin

24 Summary (Continued) A circular magnetic field is is induced by by a current in in a wire. The direction is is given by by the right-hand hand thumb rule. Circular I The magnitude depends on the current I and the distance r from the wire. 0I 2 r r X Permeability: = 4 4 x 10-7 Tm/A Permeability: = 4 x 10-7 Tm/A I

25 Summary (Continued) The force per unit length for two wires separated by d is: F L I I 2 d Single loop: 0I 2R Coil of N loops: 0NI 2R For a solenoid of length L, with N turns and current I, the -field is given by: NI L

26 CONCLUSION: Chapter 30 Torque and Magnetic Fields

physics 112N magnetic fields and forces

physics 112N magnetic fields and forces bar magnet & iron filings physics 112N 2 bar magnets physics 112N 3 the Earth s magnetic field physics 112N 4 electro -magnetism! is there a connection between electricity