# SOLUTIONS - CHAPTER 30:

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```					EXAMPLE SOLUTIONS – Topic 8: ELECTROMAGNETIC INDUCTION

1) A rod 4 meters in length is placed in a magnetic field at right
angles to the direction of B which has a magnitude of 0.3 Tesla. If               v
the rod is moved with a speed of 3 m/sec in a direction
perpendicular to its length and perpendicular to B, what emf is
induced across the ends of the rod? Draw a figure and indicate the
direction of the induced emf.                                                                  B

Solution: We first determine the direction of the induced Emf in the rod.
We can do this either by Lenz's Law, or in this case by using q v  B. If
we consider a '+' charge in the rod, this is moving to the left. Thus v  B
v            B
is downward, and the bottom of the rod will become the positive end of
the Emf. The Emf produced is given by:
                  
E =    (F
mag
/ q)  d s  (1 / q) Fmag d  v B d                        vx B

Hence: E = B v d = (.3)(3)(4) = 3.6 Volts.

2) A loop of wire of area 0.5 m2 is positioned between the pole faces of a large electromagnet so that the
perpendicular to the plane of the loop makes an angle of 30o with respect to the magnetic field direction.
The magnet is turned on, and a final magnetic induction of 30 Teslas is reached within 200 msec. What is
the average emf induced in the coil during this time?

Solution: We draw the loop (our loop actually has only
one turn). Applying Lenz's Law we have the flux
increasing since the external field is increasing. Hence,                       y
B loop must be opposite B . Applying the right hand rule                                      30
we have the current within the loop from 'A' to 'B'.                                      B
Hence, in an external circuit the current would be from
'B' to 'A'. This makes point 'B' the positive terminal of
A            B loop
the Emf. From Faraday's Law we have:
z   B                                  x

Eave = /t . Since the flux is B A cos 30 then

E ave = A (B/t) cos 30 = (.5)(30)(.866)/(.2) = 65.4 Volts.
3) A rod of length 1 meter is allowed to roll down
the incline shown. The magnetic induction is 0.2 T,
B
and is perpendicular to the plane of the loop. The                                        v
resistance R is 10 ohms. The resistance of the rod
and the rails may be neglected.                                      R                            370
a) In terms of 'v' the speed of the rod down the
plane, determine the emf induced in the rod, and
hence the current produced in the loop.
b) In terms of v determine the magnitude and
direction of the magnetic force exerted on the rod.

c) Let g = 10 m/sec2. If the mass of the rod is 20 gm,
determine the terminal speed of the rod.

Solution: We first apply Lenz's Law to determine the
direction of the induced current. Since the area is
decreasing, the flux through the loop is decreasing.                              v
Hence, B ind must be in the same direction as B.
Applying the right hand rule we have the current as
shown. For a given velocity                                                               I       B

E(v) = B v d where 'd' is the length of the rod.                                 B ind
Thus the current in the loop will also depend on 'v'.

I(v) = E/R = (B v d)/R .

Since there is a current in the rod, then there will be a magnetic
force acting on the rod. Looking at the plane sideways, we
have the current coming out of the rod. Hence:                                            N
y
Fw
x                v
F w = I L  B  acts up the plane.                                                          #
37

From Newton's 2nd Law we then have:                                               I                      B

 Fx = mg sin 37 - F w = m a x          and                                                   mg

 F y = N - mg cos 37 = 0

The magnetic force is F w = I d B sin 90 = I d B = (d B)2 vx/R .

We see that as vx increases, the force up the plane increases as well. Hence, a terminal speed will be reached
when a x  0. That is: F w = mg sin 37  (d B)2 vf /R = mg sin 37 . Solving for v f

v f = m g R sin 37/(B d)2 = (20 x 10-3)(10)(.6)/(.2)2 = 30 m/sec.
4) The coil shown consists of 100 turns of radius 4 cm.                                  y
The magnetic field is constant and has a value of 0.6 T.
The coil is initially situated as shown, and is then
rotated 90o so that the plane of the coil is parallel to the
magnetic field. This rotation takes 0.2 seconds.
B
a) What is the average emf generated in the coil during
A
the rotation?
b) What is the direction of the induced emf?                     z   B                                         x

Solution: We draw the loop (our loop actually has more
turns) looking down the z-axis. Applying Lenz's Law we                   I
have the flux decreasing since the flux is initially a                                       B
maximum amount. Hence, B loop must be the same direction                      A              A
as B . Applying the right hand rule we have the current
within the loop from 'B' to 'A'. Hence, in an external circuit                B
the current would be from 'A' to 'B'. This makes point 'A' the                                       B loop
positive terminal of the Emf. From Faraday's Law we have:

Eave = /t

The final flux is zero, and the initial flux is N B A = (100)(.6)()(4 x 10-2)2 = 0.3 Webers.

E ave = /t = (.3/(.2) = 1.5 Volts.

5) A shunt wound dc motor built to operate on 120 volts will draw 24 amps when started, but only 5 amps
when running at its rated speed. If the field windings have a resistance of 30 ohms, what is the mechanical
power developed by the motor?

The circuit for a shunt wound motor is as shown. The basic
equations for this circuit are:
I
a   o
(a)   I = If + Ia                                                                                 If
DC

(b)   Vab = If Rf                                                             motor                    IA E
Rf
(c)    Vab = Ia Ra + E b                                                                                   RA
b   o

When the motor starts there is no back emf. Hence, the current (24 A) is determined by the combined
resistance in the field and armature windings. For the field windings, we have, using equation (b):

If = V ab/Rf = (120)/(30) = 4 A .

The total resistance of the 2 coils is: R = Vab/I = (120)/(24) = 5 ohms.

Thus: 1/R = 1/Rf + 1/Ra  1/Ra = (1/5) - (1/30) = (1/6) or Ra = 6 ohms.

When the motor is running we then have: Ia = I - If = 5 - 4 = 1 A.

Now applying equation (c): E b = Vab - Ia Ra = 120 - (1)(6) = 144 V.

The mechanical power developed is: Pmech = Ia E b = (1)(114) = 114 W.

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 views: 5 posted: 6/17/2012 language: pages: 4