# Solucionario Capitulo 29 - Paul E. Tippens

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Chapter 29. The Magnetism and the Magnetic Field

Physics, 6th Edition

Chapter 29. Magnetism and the Electric FieldMagnetic Fields29-1. The area of a rectangular loop is 200 cm2 and the plane of the loop makes an angle of 410 with a 0.28-T magnetic field. What is the magnetic flux penetrating the loop? A = 200 cm2 = 0.0200 m2; = 410; B = 0.280 T = 3.67 mWb

= BA sin = (0.280 T)(0.0200 m2) sin 410;

29-2. A coil of wire 30 cm in diameter is perpendicular to a 0.6-T magnetic field. If the coil turns so that it makes an angle of 600 with the field, what is the change in flux? A=

D 2 (0.30 m) 2 ; = 4 4

A = 7.07 x 10-2 m2;

= f -

o

0 = BA sin 900 = (0.6 T)(0.0707 m 2 )(1); f = BA sin 600 = (0.6 T)(0.0707 m 2 )(1); = f - o = 36.7 mWb 42.4 mWb;

0 = 42.4 mWb f = 36.7 mWb = -5.68 mWb

29-3. A constant horizontal field of 0.5 T pierces a rectangular loop 120 mm long and 70 mm wide. Determine the magnetic flux through the loop when its plane makes the following angles with the B field: 00, 300, 600, and 900. [Area = 0.12 m)(0.07 m) = 8.40 x 10-3 m2 ] = BA sin ; BA = (0.5 T)(8.4 x 10-3 m2) = 4.2 x 10-3 T m2 2 = (4.2 x 10-3 T m2) sin 300 = 2.10 mWb; 1 = (4.2 x 10-3 T m2) sin 900 = 4.20 mWb

1 = (4.2 x 10-3 T m2) sin 00 = 0 Wb; 3 = (4.2 x 10-3 T m2) sin 600 = 3.64 mWb;

146

Chapter 29. The Magnetism and the Magnetic Field

Physics, 6th Edition

29-4. A flux of 13.6 mWb penetrates a coil of wire 240 mm in diameter. Find the magnitude of the magnetic flux density if the plane of the coil is perpendicular to the field.

D 2 (0.240 m) 2 ; A= = 4 4B=

A = 4.52 x 10-2 m2;

= BA sin

0.0136 Wb = ; A sin (4.52 x 10-2 m 2 )(1)

B = 0.300 T

29-5. A magnetic flux of 50 Wb passes through a perpendicular loop of wire having an area of 0.78 m2. What is the magnetic flux density? B=

50 x 106 Wb = ; A sin (0.78 m 2 )(1)

B = 64.1 T

29-6. A rectangular loop 25 x 15 cm is oriented so that its plane makes an angle with a 0.6-T B field. What is the angle if the magnetic flux linking the loop is 0.015 Wb? A = (0.25 m)(0.15 m) = 0.0375 m2; = 0.015 Wb = 41.80

= BA sin ;

sin =

0.015 Wb = ; BA (0.6 T)(0.0375 m 2 )

The Force on Moving Charge29-7. A proton (q = +1.6 x 10-19 C) is injected to the right into a B field of 0.4 T directed upward. If the velocity of the proton is 2 x 106 m/s, what are the magnitude and direction of the magnetic force on the proton? F = qvB = (1.6 x 10-19 C)(2 x 106 m/s)(0.4 T) F = 1.28 x 10-13 N, into paper v B Into paper

Right hand screw rule

147

Chapter 29. The Magnetism and the Magnetic Field

Physics, 6th Edition

29-8. An alpha particle (+2e) is projected with a velocity of 3.6 x 106 m/s into a 0.12-T magnetic field. What is the magnetic force on the charge at the instant its velocity is directed at an angle of 350 with the magnetic flux? [ q = 2 (1.6 x 10-19 C) = 3.2 x 10-19 C ] F = qvB sin = (3.2 x 10-19 C)(3.6 x 106 m/s)(0.12 T) sin 350; F = 7.93 x 10-14 N 29-9. An electron moves with a velocity of 5 x 105 m/s at an angle of 600 with an eastward B field. The electron experiences an force of 3.2 x 10-18 N directed into the paper. What are the magnitude of B and the direction the velocity v? In order for the force to be INTO the paper for a NEGATIVE charge, the 600 angle must be S of E. B= N B600

E

= 600 S of EB = 46.3 T

v

F 3.2 x 10-18 N = ; qv sin (1.6 x 10-19 C)(5 x 105 m/s)

29-10. A proton (+1e) is moving vertically upward with a velocity of 4 x 106 m/s. It passes through a 0.4-T magnetic field directed to the right. What are the magnitude and direction of the magnetic force? F = qvB = (1.6 x 10-19 C)(4 x 106 m/s)(0.4 T) ; F = 2.56 x 10-13 N, directed into paper. v B Force is into paper.

29-11. What if an electron replaces the proton in Problem 29-10. What is the magnitude and direction of the magnetic force? The direction of the magnetic force on an electron is opposite to that of the proton, but the magnitude of the force is unchanged since the magnitude of the charge is the same.

148

Chapter 29. The Magnetism and the Magnetic Field Fe = 2.56 x 10-13 N, out of paper.

Physics, 6th Edition

*29-12. A particle having a charge q and a mass m is projected into a B field directed into the paper. If the particle has a velocity v, show that it will be deflected into a circular path of radius: R= mv qB

Draw a diagram of the motion, assuming a positive charge entering the B field from left to right. Hint: The magnetic force provides the necessary centripetal force for the circular motion. FC = mv 2 ; R FB = qvB; mv 2 = qvB R R= mv qB

From which:

The diagram shows that the magnetic force is a centripetal force that acts toward the center causing the charge to move in a counterclockwise circle of radius R. *29-13. A deuteron is a nuclear particle consisting of a proton and a neutron bound together by nuclear forces. The mass of a deuteron is 3.347 x 10-27 kg, and its charge is +1e. A deuteron projected into a magnetic field of flux density 1.2 T is observed to travel in a circular path of radius 300 mm. What is the velocity of the deuteron? See Problem 29-12. FC = mv 2 ; R FB = qvB; mv 2 = qvB; R v= qRB m

v=

qRB (1.6 x 10-19 C)(0.3 m)(1.2 T) = ; m 3.347 x 10-27 kg

v = 1.72 x 107 m/s

Note: This speed which is about 6% of the speed of light is still not fast enough to cause significant effects due to relativity (see Chapter 38.)

149

Chapter 29. The Magnetism and the Magnetic Field

Physics, 6th Edition

Force on a Current-Carrying Conductor29-14. A wire 1 m in length supports a current of 5.00 A and is perpendicular to a B field of 0.034 T. What is the magnetic force on the wire? F = I B l= (5 A)(0.034 T)(1 m); F = 0.170 N

29-15. A long wire carries a current of 6 A in a direction 350 north of an easterly 40-mT magnetic field. What are the magnitude and direction of the force on each centimeter of wire? F = Il B sin = (6 A)(0.040 T)(0.01 m)sin 350 F = 1.38 x 10-3 N, into paper I350

B

The force is into paper as can be seen by turning I into B to advance a screw inward. 29-16. A 12-cm segment of wire carries a current of 4.0 A directed at an angle of 410 north of an easterly B field. What must be the magnitude the B field if it is to produce a 5 N force on this segment of wire? What is the direction of the force? B= F 5.00 N = ; Il sin (4.0 A)(0.12 m) sin 410 B = 15.9 T

The force is directed inward according to the right-hand rule. 29-17. An 80 mm segment of wire is at an angle of 530 south of a westward, 2.3-T B field. What are the magnitude and direction of the current in this wire if it experiences a force of 2 N directed out of the paper? B = 2.30 T; l = 0.080 m; = 53 ; I= F 2.00 N = ; Bl sin (2.3 T)(0.080 m)sin 5300

B F = 2.00 N I = 13.6 A530

I

The current must be directed 530 N of E if I turned into B produces outward force.

150

Chapter 29. The Magnetism and the Magnetic Field

Physics, 6th Edition

*29-18. The linear density of a certain wire is 50.0 g/m. A segment of this wire carries a current of 30 A in a direction perpendicular to the B field. What must be the magnitude of the magnetic field required to suspend the wire by balancing its weight?

=

m ; l

m = l ; W = mg = l g; FB = W = l g FB = I lB

g (0.050 kg/m)(9.8 m/s 2 ) = lg = IlB; B = ; I 30 A

B = 16.3 mT

Calculating Magnetic Fields29-19. What is the magnetic induction B in air at a point 4 cm from a long wire carrying a current of 6 A? B=

0 I (4 x 10-7 T m/A)(6 A) = ; B = 30.0 T 2 l 2 (0.04 m)

29-20. Find the magnetic induction in air 8 mm from a long wire carrying a current of 14.0 A. B=

0 I (4 x 10-7 T m/A)(14 A) = ; B = 350 T 2 l 2 (0.008 m)

29-21. A circular coil having 40 turns of wire in air has a radius of 6 cm and is in the plane of the paper. What current must exist in the coil to produce a flux density of 2 mT at its center? B=

0 NI ; 2r

I=

2rB 0 N I = 4.77 A

I=

2(0.06 m)(0.002 T) ; (4 x 10-7 T m/A)(40)

151

Chapter 29. The Magnetism and the Magnetic Field

Physics, 6th Edition

29-22. If the direction of the current in the coil of Problem 29-21 is clockwise, what is the direction of the magnetic field at the center of the loop? If you grasp the loop with your right hand so that the thumb points in the direction of the current, it is seen that the B field will be directed OUT of the paper at the center of the loop. 29-23. A solenoid of length 30 cm and diameter 4 cm is closely wound with 400 turns of wire around a nonmagnetic material. If the current in the wire is 6 A, determine the magnetic induction along the center of the solenoid. B= I

0 NI (4 x 10-7 T m/A)(400)(6 A) ; = l 0.300 m

B = 10.1 mT

29-24. A circular coil having 60 turns has a radius of 75 mm. What current must exist in the coil to produce a flux density of 300 T at the center of the coil? I= 2rB 2(0.075 m)(300 x 10-6 T) = ; 0 N (4 x 10-7 T m/A)(60) I = 0.597 A

*29-25. A circular loop 240 mm in diameter supports a current of 7.8 A. If it is submerged in a medium of relative permeability 2.0, what is the magnetic induction at the center? r = (240 mm) = 120 mm; = 20 = 8 x 10-7 T m/A B=

NI (8 x 10-7 T m/A)(1)(7.8 A) = ; 2r 2(0.120 m)

B = 81.7 T

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Chapter 29. The Magnetism and the Magnetic Field

Physics, 6th Edition

*29-26. A circular loop of radius 50 mm in the plane of the paper carries a counterclockwise cur