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== 13.6#h(0.6em)Electric Generators and Back Emf


  A variety of important phenomena and devices can be understood with Faraday’s law. In this section, we examine two of these.


  
    
=== Electric Generators


    #strong[Electric generators] induce an emf by rotating a coil in a magnetic field, as briefly discussed in Motional Emf. We now explore #strong[generators] in more detail. Consider the following example.


    #examplebox("Example 1")[Calculating the Emf Induced in a Generator Coil][
The generator coil shown in  is rotated through one-fourth of a revolution (from #math.equation(block: false, alt: "θ equals 0 °")[$θ = 0 "°"$] to#math.equation(block: false, alt: "θ equals 90 ° close parenthesis")[$θ = 90 "°" \)$] in 15.0 ms. The 200-turn circular coil has a 5.00-cm radius and is in a uniform 0.80-T magnetic field. What is the emf induced?


      #figure(figph[Picture shows a generator coil that is rotated by mechanical means through one-fourth of a revolution.], alt: "Picture shows a generator coil that is rotated by mechanical means through one-fourth of a revolution.", caption: [When this generator coil is rotated through one-fourth of a revolution, the magnetic flux #math.equation(block: false, alt: "Φ sub m")[$"Φ"_("m")$] changes from its maximum to zero, inducing an emf.])


      Strategy
Faraday’s law of induction is used to find the emf induced:


      #math.equation(block: true, alt: "ε equals − N the fraction d Φ sub m over d t .")[$ε = "−" N frac(d "Φ"_("m"), d t) .$]
      We recognize this situation as the same one in . According to the diagram, the projection of the surface normal vector #math.equation(block: false, alt: "n ^")[$hat(n)$] to the magnetic field is initially#math.equation(block: false, alt: "cos θ ,")[$"cos" #h(0.2em) θ ,$] and this is inserted by the definition of the dot product. The magnitude of the magnetic field and area of the loop are fixed over time, which makes the integration simplify quickly. The induced emf is written out using Faraday’s law:


      #math.equation(block: true, alt: "ε equals N B A sin θ the fraction d θ over d t .")[$ε = N B A #h(0.2em) "sin" #h(0.2em) θ frac(d θ, d t) .$]
      Solution
We are given that #math.equation(block: false, alt: "N equals 200 ,")[$N = 200 ,$] #math.equation(block: false, alt: "B equals 0.80 T ,")[$B = 0.80 #h(0.2em) "T" ,$] #math.equation(block: false, alt: "θ equals 90 °")[$θ = 90 "°"$], #math.equation(block: false, alt: "d θ equals 90 ° equals π / 2")[$d θ = 90 "°" = π "/" 2$], and #math.equation(block: false, alt: "d t equals 15.0 ms .")[$d t = 15.0 #h(0.2em) "ms" .$] The area of the loop is


      #math.equation(block: true, alt: "A equals π r squared equals open parenthesis 3.14 close parenthesis open parenthesis 0.0500 m close parenthesis squared equals 7.85 times 10 to the power − 3 m squared .")[$A = π r^(2) = ( 3.14 ) attach(( 0.0500 #h(0.2em) "m" ), t: 2) = 7.85 #h(0.2em) × #h(0.2em) 10^("−" 3) #h(0.2em) "m"^(2) .$]
      Entering this value gives


      #math.equation(block: true, alt: "ε equals open parenthesis 200 close parenthesis open parenthesis 0.80 T close parenthesis open parenthesis 7.85 times 10 to the power −3 m squared close parenthesis sin open parenthesis 90 ° close parenthesis the fraction π / 2 over 15.0 times 10 to the power −3 s equals 131 V .")[$ε = ( 200 ) ( 0.80 #h(0.2em) "T" ) ( 7.85 #h(0.2em) × #h(0.2em) 10^(−3) attach(#h(0.2em) "m", t: 2) ) "sin" #h(0.2em) ( 90 "°" ) #h(0.2em) frac(π "/" 2, 15.0 #h(0.2em) × #h(0.2em) 10^(−3) #h(0.2em) "s") = 131 #h(0.2em) "V" "."$]
      Significance
This is a practical average value, similar to the 120 V used in household power.
]


    The emf calculated in  is the average over one-fourth of a revolution. What is the emf at any given instant? It varies with the angle between the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of time by considering the motional emf on a rotating rectangular coil of width #emph[w] and height #emph[l] in a uniform magnetic field, as illustrated in .


    #figure(figph[Picture shows a single rectangular coil that is rotated at constant angular velocity in a uniform magnetic field.], alt: "Picture shows a single rectangular coil that is rotated at constant angular velocity in a uniform magnetic field.", caption: [A generator with a single rectangular coil rotated at constant angular velocity in a uniform magnetic field produces an emf that varies sinusoidally in time. Note the generator is similar to a motor, except the shaft is rotated to produce a current rather than the other way around.])


    Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in the vertical wires experience forces parallel to the wire, causing currents. But those in the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. We can thus find the induced emf by considering only the side wires. Motional emf is given to be #math.equation(block: false, alt: "ε equals B l v")[$ε = B l v$], where the velocity #emph[v] is perpendicular to the magnetic field #emph[B]. Here the velocity is at an angle #math.equation(block: false, alt: "θ")[$θ$] with #emph[B], so that its component perpendicular to #emph[B] is #emph[v] sin #math.equation(block: false, alt: "θ")[$θ$]. Thus, in this case, the emf induced on each side is #math.equation(block: false, alt: "ε equals B l v sin θ")[$ε = B l v #h(0.2em) "sin" #h(0.2em) θ$], and they are in the same direction. The total emf around the loop is then


    #math.equation(block: true, alt: "ε equals 2 B l v sin θ .")[$ε = 2 B l v #h(0.2em) "sin" #h(0.2em) θ .$]
    This expression is valid, but it does not give emf as a function of time. To find the time dependence of emf, we assume the coil rotates at a constant angular velocity #math.equation(block: false, alt: "ω")[$ω$]. The angle #math.equation(block: false, alt: "θ")[$θ$] is related to angular velocity by #math.equation(block: false, alt: "θ equals ω t ,")[$θ = ω t ,$] so that


    #math.equation(block: true, alt: "ε equals 2 B l v sin open parenthesis ω t close parenthesis .")[$ε = 2 B l v #h(0.2em) "sin" ( ω t ) .$]
    Now, linear velocity #emph[v] is related to angular velocity #math.equation(block: false, alt: "ω")[$ω$] by #math.equation(block: false, alt: "v equals r ω .")[$v = r ω .$] Here, #math.equation(block: false, alt: "r equals w / 2 ,")[$r = w "/" 2 ,$] so that #math.equation(block: false, alt: "v equals open parenthesis w / 2 close parenthesis ω ,")[$v = ( w "/" 2 ) ω ,$] and


    #math.equation(block: true, alt: "ε equals 2 B l the fraction w over 2 ω sin ω t equals open parenthesis l w close parenthesis B ω sin ω t .")[$ε = 2 B l frac(w, 2) ω #h(0.2em) "sin" #h(0.2em) ω t = ( l w ) B ω #h(0.2em) "sin" #h(0.2em) ω t .$]
    Noting that the area of the loop is #math.equation(block: false, alt: "A equals l w ,")[$A = l w ,$] and allowing for #emph[N] loops, we find that


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "ε equals N B A ω sin open parenthesis ω t close parenthesis .")[$ε = N B A ω #h(0.2em) "sin" ( ω t ) .$]
]


    This is the emf induced in a generator coil of #emph[N] turns and area #emph[A] rotating at a constant angular velocity #math.equation(block: false, alt: "ω")[$ω$] in a uniform magnetic field #emph[B]. This can also be expressed as


    #math.equation(block: true, alt: "ε equals ε sub 0 sin ω t ,")[$ε = ε_(0) #h(0.2em) "sin" #h(0.2em) ω t ,$]
    where


    #math.equation(block: true, alt: "ε sub 0 equals N A B ω")[$ε_(0) = N A B ω$]
    is the peak emf, since the maximum value of #math.equation(block: false, alt: "sin open parenthesis w t close parenthesis equals 1")[$sin ( w t ) = 1$]. Note that the frequency of the oscillation is #math.equation(block: false, alt: "f equals ω / 2 π")[$f = ω "/" 2 π$] and the period is #math.equation(block: false, alt: "T equals 1/ f equals 2 π / ω .")[$T = "1/" f = 2 π "/" #h(0.2em) ω .$]  shows a graph of emf as a function of time, and it now seems reasonable that ac voltage is sinusoidal.


    #figure(figph[Picture shows a single rectangular coil that is rotated at constant angular velocity between opposite poles of magnet. Generated current lights the bulb. Graph shows emf plotted as a function of time. Emf has a sinusoidal shape with a period T.], alt: "Picture shows a single rectangular coil that is rotated at constant angular velocity between opposite poles of magnet. Generated current lights the bulb. Graph shows emf plotted as a function of time. Emf has a sinusoidal shape with a period T.", caption: [The emf of a generator is sent to a light bulb with the system of rings and brushes shown. The graph gives the emf of the generator as a function of time, where #math.equation(block: false, alt: "ε sub 0")[$ε_(0)$] is the peak emf. The period is #math.equation(block: false, alt: "T equals 1/ f equals 2 π / ω ,")[$T = "1/" f = 2 π "/" #h(0.2em) ω ,$] where #emph[f] is the frequency.])


    The fact that the peak emf is #math.equation(block: false, alt: "ε sub 0 equals N B A ω")[$ε_(0) = N B A ω$] makes good sense. The greater the number of coils, the larger their area, and the stronger the field, the greater the output voltage. It is interesting that the faster the generator is spun (greater #math.equation(block: false, alt: "ω")[$ω$]), the greater the emf. This is noticeable on bicycle generators—at least the cheaper varieties.


    shows a scheme by which a generator can be made to produce pulsed dc. More elaborate arrangements of multiple coils and split rings can produce smoother dc, although electronic rather than mechanical means are usually used to make ripple-free dc.


    #figure(figph[Picture shows a single rectangular coil that is rotated at constant angular velocity between opposite poles of magnet. Generated current lights the bulb. Graph shows pulsed dc emf output.], alt: "Picture shows a single rectangular coil that is rotated at constant angular velocity between opposite poles of magnet. Generated current lights the bulb. Graph shows pulsed dc emf output.", caption: [Split rings, called commutators, produce a pulsed dc emf output in this configuration.])


    In real life, electric generators look a lot different from the figures in this section, but the principles are the same. The source of mechanical energy that turns the coil can be falling water (hydropower), steam produced by the burning of fossil fuels, or the kinetic energy of wind.  shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator. The generation of electrical energy from mechanical energy is the basic principle of all power that is sent through our electrical grids to our homes.


    #figure(figph[Photograph shows a steam turbine generator.], alt: "Photograph shows a steam turbine generator.", caption: [Steam turbine/generator. The steam produced by burning coal impacts the turbine blades, turning the shaft, which is connected to the generator.])


    Generators illustrated in this section look very much like the motors illustrated previously. This is not coincidental. In fact, a motor becomes a generator when its shaft rotates. Certain early automobiles used their starter motor as a generator. In the next section, we further explore the action of a motor as a generator.


  
  
    
=== Back Emf


    Generators convert mechanical energy into electrical energy, whereas motors convert electrical energy into mechanical energy. Thus, it is not surprising that motors and generators have the same general construction. A motor works by sending a current through a loop of wire located in a magnetic field. As a result, the magnetic field exerts torque on the loop. This rotates a shaft, thereby extracting mechanical work out of the electrical current sent in initially. (Refer to Force and Torque on a Current Loop for a discussion on motors that will help you understand more about them before proceeding.)


    When the coil of a motor is turned, magnetic flux changes through the coil, and an emf (consistent with Faraday’s law) is induced. The motor thus acts as a generator whenever its coil rotates. This happens whether the shaft is turned by an external input, like a belt drive, or by the action of the motor itself. That is, when a motor is doing work and its shaft is turning, an emf is generated. Lenz’s law tells us the emf opposes any change, so that the input emf that powers the motor is opposed by the motor’s self-generated emf, called the #strong[back emf] of the motor .


    #figure(figph[Schematic shows the coil of a dc motor. It consists of driving emf, back emf, resistor, and a switch.], alt: "Schematic shows the coil of a dc motor. It consists of driving emf, back emf, resistor, and a switch.", caption: [The coil of a dc motor is represented as a resistor in this schematic. The back emf is represented as a variable emf that opposes the emf driving the motor. Back emf is zero when the motor is not turning and increases proportionally to the motor’s angular velocity.])


    The generator output of a motor is the difference between the supply voltage and the back emf. The back emf is zero when the motor is first turned on, meaning that the coil receives the full driving voltage and the motor draws maximum current when it is on but not turning. As the motor turns faster, the back emf grows, always opposing the driving emf, and reduces both the voltage across the coil and the amount of current it draws. This effect is noticeable in many common situations. When a vacuum cleaner, refrigerator, or washing machine is first turned on, lights in the same circuit dim briefly due to the #emph[IR] drop produced in feeder lines by the large current drawn by the motor.


    When a motor first comes on, it draws more current than when it runs at its normal operating speed. When a mechanical load is placed on the motor, like an electric wheelchair going up a hill, the motor slows, the back emf drops, more current flows, and more work can be done. If the motor runs at too low a speed, the larger current can overheat it (via resistive power in the coil, #math.equation(block: false, alt: "P equals I squared R close parenthesis ,")[$P = I^(2) R \) ,$] perhaps even burning it out. On the other hand, if there is no mechanical load on the motor, it increases its angular velocity #math.equation(block: false, alt: "ω")[$ω$] until the back emf is nearly equal to the driving emf. Then the motor uses only enough energy to overcome friction.


    Eddy currents in iron cores of motors can cause troublesome energy losses. These are usually minimized by constructing the cores out of thin, electrically insulated sheets of iron. The magnetic properties of the core are hardly affected by the lamination of the insulating sheet, while the resistive heating is reduced considerably. Consider, for example, the motor coils represented in . The coils have an equivalent resistance of #math.equation(block: false, alt: "0.400 Ω")[$0.400 #h(0.2em) "Ω"$] and are driven by an emf of 48.0 V. Shortly after being turned on, they draw a current


    #math.equation(block: true, alt: "I equals V / R equals open parenthesis 48.0 V close parenthesis / open parenthesis 0.400 Ω close parenthesis equals 120 A")[$I = V "/" R = ( 48.0 #h(0.2em) "V" ) "/" ( 0.400 #h(0.2em) "Ω" ) = 120 #h(0.2em) "A"$]
    and thus dissipate #math.equation(block: false, alt: "P equals I squared R equals 5.76 kW")[$P = I^(2) R = 5.76 #h(0.2em) "kW"$] of energy as heat transfer. Under normal operating conditions for this motor, suppose the back emf is 40.0 V. Then at operating speed, the total voltage across the coils is 8.0 V (48.0 V minus the 40.0 V back emf), and the current drawn is


    #math.equation(block: true, alt: "I equals V / R equals open parenthesis 8.0 V close parenthesis / open parenthesis 0.400 Ω close parenthesis equals 20 A .")[$I = V "/" R = ( 8.0 #h(0.2em) "V" ) "/" ( 0.400 #h(0.2em) "Ω" ) = 20 #h(0.2em) "A" #h(0.2em) .$]
    Under normal load, then, the power dissipated is #math.equation(block: false, alt: "P equals I V equals open parenthesis 20 A close parenthesis open parenthesis 8.0 V close parenthesis equals 160 W .")[$P = I V = ( 20 #h(0.2em) "A" ) ( 8.0 #h(0.2em) "V" ) = 160 #h(0.2em) "W" .$] This does not cause a problem for this motor, whereas the former 5.76 kW would burn out the coils if sustained.


    #examplebox("Example 2")[A Series-Wound Motor in Operation][
The total resistance #math.equation(block: false, alt: "open parenthesis R sub f plus R sub a close parenthesis")[$( R_(f) + R_(a) )$] of a series-wound dc motor is #math.equation(block: false, alt: "2.0 Ω")[$2.0 #h(0.2em) "Ω"$] . When connected to a 120-V source (#math.equation(block: false, alt: "ε sub S")[$ε_(S)$]), the motor draws 10 A while running at constant angular velocity. (a) What is the back emf induced in the rotating coil, #math.equation(block: false, alt: "ε sub i ?")[$ε_(i) ?$] (b) What is the mechanical power output of the motor? (c) How much power is dissipated in the resistance of the coils? (d) What is the power output of the 120-V source? (e) Suppose the load on the motor increases, causing it to slow down to the point where it draws 20 A. Answer parts (a) through (d) for this situation.


      #figure(figph[Schematic shows the circuit of a series-wound direct current motor. It consists of two emfs and two resistors.], alt: "Schematic shows the circuit of a series-wound direct current motor. It consists of two emfs and two resistors.", caption: [Circuit representation of a series-wound direct current motor.])


      Strategy
The back emf is calculated based on the difference between the supplied voltage and the loss from the current through the resistance. The power from each device is calculated from one of the power formulas based on the given information.


      Solution


      + The back emf is
#math.equation(block: true, alt: "ε sub i equals ε sub s minus I open parenthesis R sub f plus R sub a close parenthesis equals 120 V minus open parenthesis 10 A close parenthesis open parenthesis 2.0 Ω close parenthesis equals 100 V .")[$ε_(i) = ε_(s) − I ( R_(f) + R_(a) ) = 120 #h(0.2em) "V" − ( 10 #h(0.2em) "A" ) ( 2.0 #h(0.2em) "Ω" ) = 100 #h(0.2em) "V" "."$]
+ Since the potential across the armature is 100 V when the current through it is 10 A, the power output of the motor is
#math.equation(block: true, alt: "P sub m equals ε sub i I equals open parenthesis 100 V close parenthesis open parenthesis 10 A close parenthesis equals 1.0 times 10 cubed W .")[$P_("m") = ε_(i) I = ( 100 #h(0.2em) "V" ) ( 10 #h(0.2em) "A" ) = 1.0 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "W" "."$]
+ A 10-A current flows through coils whose combined resistance is #math.equation(block: false, alt: "2.0 Ω")[$2.0 #h(0.2em) "Ω"$], so the power dissipated in the coils is
#math.equation(block: true, alt: "P sub R equals I squared R equals open parenthesis 10 A close parenthesis squared open parenthesis 2.0 Ω close parenthesis equals 2.0 times 10 squared W .")[$P_(R) = I^(2) R = attach(( 10 #h(0.2em) "A" ), t: 2) ( 2.0 #h(0.2em) "Ω" ) = 2.0 #h(0.2em) × #h(0.2em) 10^(2) #h(0.2em) "W" "."$]
+ Since 10 A is drawn from the 120-V source, its power output is
#math.equation(block: true, alt: "P sub s equals ε sub s I equals open parenthesis 120 V close parenthesis open parenthesis 10 A close parenthesis equals 1.2 times 10 cubed W .")[$P_(s) = ε_(s) #h(0.2em) I = ( 120 #h(0.2em) "V" ) ( 10 #h(0.2em) "A" ) = 1.2 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "W" "."$]
+ Repeating the same calculations with #math.equation(block: false, alt: "I equals 20 A")[$I = 20 #h(0.2em) "A"$], we find 
#math.equation(block: true, alt: "ε sub i equals 80 V , P sub m equals 1.6 times 1 0 cubed W , P sub R equals 8.0 times 10 squared W , and P sub s equals 2.4 times 10 cubed W .")[$ε_("i") = #h(0.2em) 80 #h(0.2em) "V" , P_("m") = #h(0.2em) 1.6 #h(0.2em) × #h(0.2em) 1 0^(3) #h(0.2em) "W" , P_(R) = 8.0 #h(0.2em) × #h(0.2em) 10^(2) #h(0.2em) "W" , #h(0.2em) "and" #h(0.2em) P_(s) = 2.4 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "W" .$]
The motor is turning more slowly in this case, so its power output and the power of the source are larger.


      Significance
Notice that we have an energy balance in part (d): #math.equation(block: false, alt: "1.2 times 10 cubed W equals 1.0 times 10 cubed W plus 2.0 times 10 squared W .")[$1.2 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "W" = 1.0 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "W" + 2.0 #h(0.2em) × #h(0.2em) 10^(2) #h(0.2em) "W" .$]
]


  
  
    
=== Summary


    - An electric generator rotates a coil in a magnetic field, inducing an emf given as a function of time by #math.equation(block: false, alt: "ε equals N B A ω sin open parenthesis ω t close parenthesis")[$ε = N B A ω #h(0.2em) "sin" ( ω t )$] where #emph[A] is the area of an #emph[N]-turn coil rotated at a constant angular velocity #math.equation(block: false, alt: "ω")[$ω$] in a uniform magnetic field #math.equation(block: false, alt: "B → .")[$arrow(B) .$]
- The peak emf of a generator is #math.equation(block: false, alt: "ε sub 0 equals N B A ω")[$ε_(0) = N B A ω$].
- Any rotating coil produces an induced emf. In motors, this is called back emf because it opposes the emf input to the motor.


  
  
    
=== Problems


    
      Design a current loop that, when rotated in a uniform magnetic field of strength 0.10 T, will produce an emf #math.equation(block: false, alt: "ε equals ε sub 0 sin ω t ,")[$ε = ε_(0) #h(0.2em) "sin" #h(0.2em) ω t ,$] where #math.equation(block: false, alt: "ε sub 0 equals 110 V")[$ε_(0) = 110 #h(0.2em) "V"$] and #math.equation(block: false, alt: "ω equals 120 π rad/s .")[$ω = 120 π #h(0.2em) "rad/s" "."$]


    
    #solutionbox[
three turns with an area of 1 m#super[2]
]


    
      A flat, square coil of 20 turns that has sides of length 15.0 cm is rotating in a magnetic field of strength 0.050 T. If the maximum emf produced in the coil is 30.0 mV, what is the angular velocity of the coil?


    
    
      A 50-turn rectangular coil with dimensions #math.equation(block: false, alt: "0.15 m times 0.40 m")[$0.15 #h(0.2em) "m" #h(0.2em) × #h(0.2em) 0.40 #h(0.2em) "m"$] rotates in a uniform magnetic field of magnitude 0.75 T at 3600 rev/min. (a) Determine the emf induced in the coil as a function of time. (b) If the coil is connected to a #math.equation(block: false, alt: "1000 - Ω")[$1000 "-" "Ω"$] resistor, what is the power as a function of time required to keep the coil turning at 3600 rpm? (c) Answer part (b) if the coil is connected to a 2000-#math.equation(block: false, alt: "Ω")[$"Ω"$] resistor.


    
    #solutionbox[
a. #math.equation(block: false, alt: "ω, equals, 120 π rad/s,; ε, equals, 850 sin 120 π t V;")[$ω & = & 120 π #h(0.2em) "rad/s," \ ε & = & 850 #h(0.2em) "sin" #h(0.2em) 120 #h(0.2em) π "t V;"$] #linebreak() 
b. #math.equation(block: false, alt: "P equals 720 sin squared 120 π t W;")[$P = 720 #h(0.2em) "sin"^(2) 120 #h(0.2em) π t #h(0.2em) "W;"$] #linebreak() 
c. #math.equation(block: false, alt: "P equals 360 sin squared 120 π t W")[$P = 360 #h(0.2em) "sin"^(2) 120 #h(0.2em) π t #h(0.2em) "W"$]
]


    
      The square armature coil of an alternating current generator has 200 turns and is 20.0 cm on side. When it rotates at 3600 rpm, its peak output voltage is 120 V. (a) What is the frequency of the output voltage? (b) What is the strength of the magnetic field in which the coil is turning?


    
    
      A flip coil is a relatively simple device used to measure a magnetic field. It consists of a circular coil of #emph[N] turns wound with fine conducting wire. The coil is attached to a ballistic galvanometer, a device that measures the total charge that passes through it. The coil is placed in a magnetic field #math.equation(block: false, alt: "B →")[$arrow(B)$] such that its face is perpendicular to the field. It is then flipped through #math.equation(block: false, alt: "180 ° ,")[$180 "°" ,$] and the total charge #emph[Q] that flows through the galvanometer is measured. (a) If the total resistance of the coil and galvanometer is #emph[R], what is the relationship between #emph[B] and #emph[Q]? Because the coil is very small, you can assume that #math.equation(block: false, alt: "B →")[$arrow(B)$] is uniform over it. (b) How can you determine whether or not the magnetic field is perpendicular to the face of the coil?


    
    #solutionbox[
a. #emph[B] is proportional to #emph[Q]; b. If the coin turns easily, the magnetic field is perpendicular. If the coin is at an equilibrium position, it is parallel.
]


    
      The flip coil of the preceding problem has a radius of 3.0 cm and is wound with 40 turns of copper wire. The total resistance of the coil and ballistic galvanometer is #math.equation(block: false, alt: "0.20 Ω.")[$0.20 #h(0.2em) "Ω."$] When the coil is flipped through #math.equation(block: false, alt: "180 °")[$180 "°"$] in a magnetic field #math.equation(block: false, alt: "B → ,")[$arrow(B) ,$] a change of 0.090 C flows through the ballistic galvanometer. (a) Assuming that #math.equation(block: false, alt: "B →")[$arrow(B)$] and the face of the coil are initially perpendicular, what is the magnetic field? (b) If the coil is flipped through #math.equation(block: false, alt: "90 ° ,")[$90 "°" ,$] what is the reading of the galvanometer?


    
    
      A 120-V, series-wound motor has a field resistance of 80 #math.equation(block: false, alt: "Ω")[$"Ω"$] and an armature resistance of 10 #math.equation(block: false, alt: "Ω")[$"Ω"$]. When it is operating at full speed, a back emf of 75 V is generated. (a) What is the initial current drawn by the motor? When the motor is operating at full speed, where are (b) the current drawn by the motor, (c) the power output of the source, (d) the power output of the motor, and (e) the power dissipated in the two resistances?


    
    #solutionbox[
a. 1.33 A; b. 0.50 A; c. 60 W; d. 37.5 W; e. 22.5W
]


    
      A small series-wound dc motor is operated from a 12-V car battery. Under a normal load, the motor draws 4.0 A, and when the armature is clamped so that it cannot turn, the motor draws 24 A. What is the back emf when the motor is operating normally?