68

BASIC PROBLEMS

1. Since B = A

φ, when A increases, B (flux density) decreases.

2. B = 2m5.0

Wb1500 µ=

A

φ = 3000 µWb/m

2 = 3000 µT

3. B = A

φ

There are 100 centimeters per meter. (1 m/100 cm = 1 m2/10,000 cm

2

A = 150 cm2

2

2

cm 10,000

m 1 = 0.015 m

2

φ = BA = (2500 × 10−6

T)(0.015 m2) = 37.5 µWb

4. 104 G = 1 T

(0.6 G)(1 T/104 G) = 60 T

5. 1 T = 104 G

(100,000 µT)(104 G/T) = 1000 G

6. The compass needle turns 180°.

7. µr = 0µ

µ

µ0 = 4π × 10−7

Wb/At⋅m

µr = m Wb/At104

mWb/At107507-

6

⋅×

⋅× −

π= 597

8. R = 7 2

0.28 m

(150 10 Wb/At m )(0.08 m )

l

A −=

µ × ⋅ = 233 × 10

3 At/Wb

9. Fm = NI = (500 t)(3 A) = 1500 At

CHAPTER 7 MAGNETISM AND ELECTROMAGNETISM

SECTION 7-1 The Magnetic Field

SECTION 7-2 Electromagnetism

69

10. When a solenoid is activated, its plunger is retracted.

11. (a) An electromagnetic force moves the plunger when the solenoid is activated.

(b) A spring force returns the plunger to its at-rest position.

12. The relay connects +9 V to pin 2 turning on lamp 2 and turning off lamp 1.

13. The pointer in a d’Arsonval movement is deflected by the electromagnetic force when there is

current through the coil.

14. Fm = 1500 At

H = m 0.2

At 1500=

l

Fm = 7500 At/m

15. The flux density can be changed by changing the current.

16. (a) H = 500(0.25 A)

0.3 m

mF NI

l l= = = 417 At/m

(b) φ = mF NI

l

A

=

µ

R

µr = 0

µ

µ

µ = µrµ0 = (250)(4π × 10−7

) = 3142 × 10−7

Wb/At⋅m

A = (2 cm)(2 cm) = (0.02 m)(0.02 m) = 0.0004 m2

φ = 6

-7 2

(500 t)(0.25 A) 125 At

2.39 10 At/Wb0.3 m

(3142 10 )(0.0004 m )

=×

×

= 52.3 µWb

(c) B = 2

52.3 Wb

0.0004 mA

φ µ= = 130,750 µWb/m

2

17. Material A has the most retentivity.

18. The induced voltage doubles when the rate of change of the magnetic flux doubles.

19. Iinduced = Ω

=100

mV 100induced

R

V = 1 mA

SECTION 7-3 Electromagnetic Devices

SECTION 7-4 Magnetic Hysteresis

SECTION 7-5 Electromagnetic Induction

70

20. The magnetic field is not changing; therefore, there is no induced voltage.

21. B = 3

2

2

1.24 10 Wb0.172 Wb/m

(0.085 m)A

φ= 0.172 T

v = 44 mV

sin (0.172 T)(0.085 m)(sin 90 )

indv

Bl θ = 3.02 m/s

22. (a) Positive (with respect to other end).

(b) The induced force will oppose the motion; it is downward.

23. efficiency = out

in

0.80P

P

Pin = out 45 W

0.80 0.80

P = 56.3 W

24. IA = IF + IL = 1 A + 12 A = 13 A

25. (a) P = IV = (12 A)(14 V) = 168 W

(b) P = IV = (1.0 A)(14 V) = 14 W

26. (a) P = 0.015Ts = 0.105(3.0 N-m)(1200 rpm) = 378 W

(b) 1 hp

378 W746 W

= 0.51 hp

27. Pin = 62 W; Pout = 50 W. efficiency = out

in

50 W

62 W

P

P = 80.6%

ADVANCED PROBLEMS

28. 60 rev/s × 2 peaks/rev = 120 peaks/s

SECTION 7-7 DC Motors

SECTION 7-6 DC Generators

71

29. The output voltage has a 10 V dc peak with a 120 Hz ripple. See Figure 7-1.

Figure 7-1

30. Upper lamp is open.

31. The design is flawed. 12 V is too little voltage to operate two 12 V relays in series but 24 V is

too much to operate a 12 V lamp. Install a separate 12 V power supply for the lamps and

change the 12 V to 24 V for the relays.

Multisim Troubleshooting Problems