All Solutions
Page 844: Assessment
$$
begin{align*}
varepsilon&=-Ncdotfrac{DeltaPhi}{Delta{t}}\
varepsilon&=-Ncdotfrac{Delta{B}cdot{A}cdotcostheta}{Delta{t}}\
varepsilon&=-Ncdot{A}cdotcosthetacdotfrac{Delta{B}}{Delta{t}}
end{align*}
$$
So the grater change of ratio $frac{Delta{B}}{Delta{t}}$, greater will be $varepsilon$.
In the area $D$ and $F$ there is no change in a magnetic field, so the emf will be zero.
In the area $E$ there is the steepest slope, so the emf will rich the highest value.
After the area $E$ the sharpest slope is in a segment $C$, then $B$ and $A$.
So if we compare and order in increasing order:
$$
boxed{D=F<A<B<C<E}
$$
D=F<A<B<C<E
$$
At position one, the induced current will be clockwise. At that position, the ring entering the magnetic field so the current will flow that way in order to reduce the change of magnetic flux, and due to this the induced current magnetic field will act in the opposite direction.
At position two, there is no change in magnetic flux so there can’t be induced current.
At position three, the induced current will be counterclockwise. As the ring leaves the magnetic field the induced current will try to increase the magnetic flux.
$bold{b)}$
Regarding the upper answer, the best description is:
The induced current is clockwise at 1 to oppose the field, zero at 2 because the field is uniform, and counterclockwise at 3 to maintain the field.
b) The correct answer is $A$.
$$
begin{align*}
B&=0.055text{ T}\
theta&=16text{textdegree}\
R&=3.1text{ cm}=0.031text{ m}
end{align*}
$$
And we need to find the magnitude of the magnetic flux, witch is given by the equation:
$$
begin{align*}
Phi&=Bcdot{A}cdotcostheta\
Phi&=Bcdot{R^2cdotpi}cdotcostheta
end{align*}
$$
Let’s substitute and compute:
$$
begin{align*}
Phi&=0.055text{ T}cdot{(0.031text{ m})^2cdotpi}cdotcos{16text{textdegree}}
end{align*}
$$
$$
boxed{Phi=1.6cdot{10^{-4}} text{T}cdottext{m}^2}
$$
Phi=1.6cdot{10^{-4}} text{T}cdottext{m}^2
$$
$$
begin{align*}
B&=0.025text{ T}\
L&=32.5text{ cm}=0.325text{ m}\
W&=12text{ cm}=0.12text{ m}\
H&=10text{ cm}=0.1text{ m}
end{align*}
$$
And we need to find the magnitude of the magnetic flux through the each side of the box, witch is given by the equation:
$$
begin{align*}
Phi&=Bcdot{A}cdotcostheta
end{align*}
$$
As we see from the picture the sides of the box are parallel with the magnetic field lines. So the angle between the box sides and normal layer to a magnetic field lines is $90text{textdegree}$.
And the angle between the box bottom and the layer normal to a magnetic field lines is $0text{textdegree}$.
So, the flux through the box sides will be:
$$
begin{align*}
Phi_{sides}&=Bcdot{A}cdotcos theta\
Phi_{sides}&=Bcdot{A}cdotcos{ 90}
end{align*}
$$
$$
boxed{Phi_{sides}=0}
$$
And the flux through the box bottom is:
$$
begin{align*}
Phi_{bottom}&=Bcdot{A}cdotcos theta\
Phi_{bottom}&=Bcdot{Wcdot{L}}cdotcos{ 0}\
Phi_{bottom}&=0.025text{ T}cdot{0.12text{ m}cdot{0.325text{ m}}}
end{align*}
$$
$$
boxed{Phi_{bottom}=9.75cdot{10^{-4}}text{ T}cdottext{ m}^2}
$$
$$
Phi_{bottom}=9.75cdot{10^{-4}}text{ T}cdottext{ m}^2
$$
$$
begin{align*}
Phi&=4.8cdot{10^{-5}} text{T}cdottext{m}^2\
theta&=47text{textdegree}\
a&=5.1text{ cm}=0.051text{ m}\
b&=6.8text{ cm}=0.068text{ m}
end{align*}
$$
And we need to find the magnitude of the magnetic field, witch is given by the equation:
$$
begin{align*}
Phi&=Bcdot{A}cdotcos theta\
B&=frac{Phi}{Acdotcos theta}
end{align*}
$$
Where the area of the rectangle is given by:
$$
begin{align*}
A&=acdot{b}
end{align*}
$$
Let’s now substitute all given and compute:
$$
begin{align*}
B&=frac{Phi}{acdot{b}cdotcos theta}\
B&=frac{4.8cdot{10^{-5}} text{T}cdottext{m}^2}{0.051text{ m}cdot{0.068text{ m}}cdotcos{47text{textdegree}}}
end{align*}
$$
$$
boxed{B=2.03cdot{10^{-2}}text{ T}}
$$
B=2.03cdot{10^{-2}}text{ T}
$$
$$
begin{align*}
B_v&=4.2cdot{10^{-5}} text{T}\
B_h&=2.6cdot{10^{-5}} text{T}\
theta_v&=0text{textdegree}\
theta_h&=90text{textdegree}\
a&=22text{ m}\
b&=18text{ m}
end{align*}
$$
And we need to find the magnitude of the magnetic flux through the floor, witch is given by the equation:
$$
begin{align*}
Phi&=Bcdot{A}cdotcos theta
end{align*}
$$
But the complete magnetic flux through the floor will be the sum of vertical and horizontal component, so let’s compute each component separately.
The area of the floor:
$$
begin{align*}
A&=acdot{b}
end{align*}
$$
Let’s first compute the horizontal component:
$$
begin{align*}
Phi_h&=B_hcdot{A}cdotcos theta_h\
Phi_h&=B_hcdot{A}cdotcos{90text{textdegree}}\
Phi_h&=0
end{align*}
$$
And for the vertical component we have:
$$
begin{align*}
Phi_v&=B_vcdot{acdot{b}}cdotcos theta_v\
Phi_v&=4.2cdot{10^{-5}} text{T}cdot{22text{ m}cdot{18text{ m}}}cdotcos{0text{textdegree}}\
Phi_v&=1.66cdot{10^{-2}}text{ Wb}
end{align*}
$$
And the complete magnetic flux through the floor will be:
$$
begin{align*}
Phi&=Phi_v+Phi_h\
Phi&=1.66cdot{10^{-2}}text{ Wb}+0
end{align*}
$$
$$
boxed{Phi=1.66cdot{10^{-2}}text{ Wb}}
$$
Phi=1.66cdot{10^{-2}}text{ Wb}
$$
$$
begin{align*}
B&=1.7text{ T}\
theta&=0text{textdegree}\
l&=2.5text{ m}\
d&=1.2text{ m}
end{align*}
$$
And we need to find the magnitude of the magnetic flux through the solenoid, witch is given by the equation:
$$
begin{align*}
Phi&=Bcdot{A}cdotcos theta
end{align*}
$$
All the lines of the magnetic field always goes through the solenoid, so the angle between the crosssection area of solenoid and the area normal to a direction of the magnetic field lines is zero.
The area of the solenoid crosssection is the area of the circle:
$$
begin{align*}
A&=left(frac{d}{2}right)^2cdotpi
end{align*}
$$
Let’s substitute and compute the magnetic flux:
$$
begin{align*}
Phi&=Bcdot{A}cdotcos theta\
Phi&=Bcdot{left(frac{d}{2}right)^2cdotpi}cdotcos{0text{textdegree}}\
Phi&=1.7text{ T}cdot{left(frac{1.2text{ m}}{2}right)^2cdotpi}
end{align*}
$$
$$
boxed{Phi=1.92text{ Wb}}
$$
Phi=1.92text{ Wb}
$$
$$
begin{align*}
B&=0.45text{ T}\
N&=53\
R&=15text{ cm}=0.15text{ m}\
Delta{t}&=0.12text{ s}
end{align*}
$$
And we need to find the magnitude of the induced emf:
$$
begin{align*}
varepsilon&=-frac{DeltaPhi}{Delta{t}}
end{align*}
$$
Where the change of the magnetic flux is given with:
$$
begin{align*}
DeltaPhi&=Ncdot{A}cdotDelta{B}\
DeltaPhi&=Ncdot{R^2cdotpi}cdotDelta{B}
end{align*}
$$
Let’s substitute and compute the induced emf:
$$
begin{align*}
varepsilon&=-frac{Ncdot{R^2cdotpi}cdotDelta{B}}{Delta{t}}\
varepsilon&=-frac{Ncdot{R^2cdotpi}cdot(B_f-B_i)}{Delta{t}}\
varepsilon&=-frac{53cdot{(0.15text{ m})^2cdotpi}cdot(0-0.45text{ T})}{0.12text{ s}}
end{align*}
$$
$$
boxed{varepsilon=14.05text{ V}}
$$
varepsilon=14.05text{ V}
$$
$$
begin{align*}
N&=1
end{align*}
$$
And we need to find the magnitude of the induced emf:
$$
begin{align*}
varepsilon&=-frac{DeltaPhi}{Delta{t}}
end{align*}
$$
To calculate the induced emf we have to calculate the change of flux at the time sample, so we can take any two arbitrary points at the same slope, but we need to know the magnitude of the magnetic flux and the time at that points. It is easier to take the beginning and the endpoint of the slope.
$bold{a)}$
If we chose the beginning and the endpoint of slope:
At $t_i=0$ the magnetic flux is $Phi_i=0$.
At the $t_f=0.1text{ s}$ the magnetic flux is $Phi_f=10text{ Wb}$.
And the induced emf will be:
$$
begin{align*}
varepsilon&=-frac{Phi_f-Phi_i}{t_f-t_i}\
varepsilon&=-frac{10text{ Wb}-0}{0.1text{ s}-0text{ s}}
end{align*}
$$
$$
boxed{varepsilon=-100text{ V}}
$$
$bold{b)}$
If we chose the beginning and the endpoint of segment:
At $t_i=0.1text{ s}$ the magnetic flux is $Phi_i=10text{ Wb}$.
At the $t_f=0.2text{ s}$ the magnetic flux is $Phi_f=10text{ Wb}$.
And the induced emf will be:
$$
begin{align*}
varepsilon&=-frac{Phi_f-Phi_i}{t_f-t_i}\
varepsilon&=-frac{10text{ Wb}-10text{ Wb}}{0.2text{ s}-0.1text{ s}}
end{align*}
$$
$$
boxed{varepsilon=0}
$$
We can also conclude this because there is no change in flux between two points at the same segment.
If we chose the beginning and the endpoint of slope:
At $t_i=0.2text{ s}$ the magnetic flux is $Phi_i=10text{ Wb}$.
At the $t_f=0.6text{ s}$ the magnetic flux is $Phi_f=-5text{ Wb}$.
And the induced emf will be:
$$
begin{align*}
varepsilon&=-frac{Phi_f-Phi_i}{t_f-t_i}\
varepsilon&=-frac{-5text{ Wb}-10text{ Wb}}{0.6text{ s}-0.2text{ s}}
end{align*}
$$
$$
boxed{varepsilon=37.5text{ V}}
$$
b) $varepsilon=0$
c) $varepsilon=37.5text{ V}$
The magnitude of induced emf is the same for both points because it lays at the same slope segment and the change of the magnetic flux is linear.
$bold{b)}$
The magnitude of the magnetic flux at $t=0.25text{ s}$ and $t=0.55text{ s}$ is:
$$
begin{align*}
varepsilon&=-frac{DeltaPhi}{Delta{t}}\
varepsilon&=-frac{Phi_f-Phi_i}{t_f-t_i}\
varepsilon&=-frac{-5text{ Wb}-10text{ Wb}}{0.6text{ s}-0.2text{ s}}
end{align*}
$$
$$
boxed{varepsilon=37.5text{ V}}
$$
b) $varepsilon=37.5text{ V}$
$$
begin{align*}
B&=0.48text{ T}\
N&=1\
A&=7.4cdot{10^{-2}}text{ m}^2\
R&=110text{ $Omega$}\
I&=0.32text{ A}
end{align*}
$$
And we have to find the rate of field change:
$$
begin{align}
varepsilon&=-frac{DeltaPhi}{Delta{t}}
end{align}
$$
Where the change of the magnetic flux is given with:
$$
begin{align*}
DeltaPhi&=Ncdot{A}cdotDelta{B}
end{align*}
$$
Let’s substitute in first equation and express the ratio of field change:
$$
begin{align*}
varepsilon&=-Ncdot{A}cdotfrac{Delta{B}}{Delta{t}}\
frac{Delta{B}}{Delta{t}}&=-frac{varepsilon}{Ncdot{A}}\
frac{Delta{B}}{Delta{t}}&=-frac{Icdot{R}}{Ncdot{A}}\
frac{Delta{B}}{Delta{t}}&=-frac{0.32text{ A}cdot{110text{ $Omega$}}}{1cdot{7.4cdot{10^{-2}}text{ m}^2}}
end{align*}
$$
$$
boxed{frac{Delta{B}}{Delta{t}}=-475.7 frac{text{T}}{text{s}}}
$$
frac{Delta{B}}{Delta{t}}=-475.7 frac{text{T}}{text{s}}
$$
$$
begin{align*}
l&=0.76text{ m}\
v&=2 frac{text{m}}{text{s}}\
varepsilon&=0.45text{ V}
end{align*}
$$
And we need to compute the magnitude of the magnetic field. The motional emf is given by the equation:
$$
begin{align*}
varepsilon&=vcdot{B}cdot{l}
end{align*}
$$
Let’s express the magnitude of the magnetic field and substitute:
$$
begin{align*}
B&=frac{varepsilon}{vcdot{l}}\
B&=frac{0.45text{ V}}{2 frac{text{m}}{text{s}}cdot{0.76text{ m}}}
end{align*}
$$
$$
boxed{B=0.296text{ T}}
$$
B=0.296text{ T}
$$
$$
begin{align*}
l&=39.9text{ m}\
v&=850 frac{text{km}}{text{s}}\
B&=5cdot{10^{-6}}text{ T}
end{align*}
$$
And we need to compute the induced emf between the tips of airplane wings. The motional emf is given by the equation:
$$
begin{align*}
varepsilon&=vcdot{B}cdot{l}
end{align*}
$$
First we have to convert the speed units:
$$
begin{align*}
v&=850 frac{text{km}}{text{h}}=850 frac{text{km}}{text{h}}cdotfrac{1000 frac{text{m}}{text{km}}}{3600 frac{text{s}}{text{h}}}=236.11 frac{text{m}}{text{s}}
end{align*}
$$
Let’s substitute and compute the emf:
$$
begin{align*}
varepsilon&=236.11 frac{text{m}}{text{s}}cdot{5cdot{10^{-6}}text{ T}}cdot{39.9text{ m}}
end{align*}
$$
$$
boxed{varepsilon=0.047text{ V}}
$$
varepsilon=0.047text{ V}
$$
$$
begin{align*}
varepsilon_{max}&=Ncdot{B}cdot{A}cdotomega
end{align*}
$$
so if doubled the magnitude of the magnetic field and tripled the angular speed, the new maximum of emf will be:
$$
begin{align*}
varepsilon_{max2}&=Ncdot(2cdot{B})cdot{A}cdot(3cdot{omega})\
varepsilon_{max2}&=6cdot{N}cdot{B}cdot{A}cdotomega\
varepsilon_{max2}&=6cdot{varepsilon_{max}}
end{align*}
$$
$$
boxed{varepsilon_{max2}=600text{ V}}
$$
varepsilon_{max2}=600text{ V}
$$
$$
begin{align*}
varepsilon_{max}&=Ncdot{B}cdot{A}cdotomega
end{align*}
$$
so if we halved the magnitude of the magnetic field and tripled the area of the loop, the new maximum of emf will be:
$$
begin{align*}
varepsilon_{max2}&=Ncdot(frac{1}{2}cdot{B})cdot(3cdot{A})cdotomega\
varepsilon_{max2}&=frac{3}{2}cdot{N}cdot{B}cdot{A}cdotomega\
varepsilon_{max2}&=frac{3}{2}cdot{varepsilon_{max}}
end{align*}
$$
$$
boxed{varepsilon_{max2}=150text{ V}}
$$
varepsilon_{max2}=150text{ V}
$$
$$
begin{align}
varepsilon_{max1}&=Ncdot{B}cdot{A}cdotomega_1
end{align}
$$
So to induce the emf of $varepsilon_{max2}=55text{ V}$ the speed of rotor must to be:
$$
begin{align}
omega_2&=frac{varepsilon_{max2}}{Ncdot{B}cdot{A}}
end{align}
$$
Let’s express the numerator from the first equation and substitute in to a second:
$$
begin{align*}
Ncdot{B}cdot{A}&=frac{varepsilon_{max1}}{omega_1}\
omega_2&=frac{varepsilon_{max2}}{Ncdot{B}cdot{A}}\
omega_2&=frac{varepsilon_{max2}}{frac{varepsilon_{max1}}{omega_1}}\
omega_2&=omega_1cdotfrac{varepsilon_{max2}}{varepsilon_{max1}}\
omega_2&=22 frac{text{rad}}{text{s}}cdotfrac{55text{ V}}{45text{ V}}
end{align*}
$$
$$
boxed{omega_2=26.9 frac{text{rad}}{text{s}}}
$$
omega_2=26.9 frac{text{rad}}{text{s}}
$$
$$
begin{align*}
omega&=190 frac{text{rad}}{text{s}}\
a&=25text{ cm}=0.25text{ m}\
b&=35text{ cm}=0.35text{ m}\
N&=120\
varepsilon_{max}&=65text{ V}
end{align*}
$$
The maximum emf of the generator is given by the equation:
$$
begin{align}
varepsilon_{max}&=Ncdot{B}cdot{A}cdotomega
end{align}
$$
And the area of the loop is:
$$
begin{align}
A&=acdot{b}
end{align}
$$
So, let’s express the magnitude of the magnetic field from the first equation and substitute:
$$
begin{align*}
B&=frac{varepsilon_{max}}{Ncdot{A}cdotomega}\
B&=frac{varepsilon_{max}}{Ncdot{acdot{b}}cdotomega}\
B&=frac{65text{ V}}{120cdot{0.25text{ m}cdot{0.35text{ m}}}cdot{190 frac{text{rad}}{text{s}}}}
end{align*}
$$
$$
boxed{B=0.032text{ T}}
$$
B=0.032text{ T}
$$
$$
begin{align*}
l&=1.6text{ m}\
R&=3.2text{ cm}=0.032text{ m}\
B&=0.075text{ T}\
omega&=85text{ rpm}
end{align*}
$$
And we have to compute the maximal induced emf, that is given by the equation:
$$
begin{align}
varepsilon_{max}&=Ncdot{A}cdot{B}cdotomega
end{align}
$$
We do not have given the number of loops, but we have the length of used wire and the radius of a single loop, so we can compute it.
$$
begin{align*}
l&=Ncdot{C}=Ncdot{2}cdot{R}cdotpi\
N&=frac{l}{2cdot{R}cdotpi}\
N&=frac{1.6text{ m}}{2cdot{0.032text{ m}}cdotpi}\
N&=8
end{align*}
$$
Now we have to convert the units of a angular velocity:
$$
begin{align*}
omega&=85text{ rpm}=85text{ rpm}cdotfrac{2 pi frac{text{ rad}}{text{ rev}}}{60 frac{text{min}}{text{s}}}=8.9 frac{text{rad}}{text{s}}
end{align*}
$$
Let’s now substitute in to a first equation and compute:
$$
begin{align*}
varepsilon_{max}&=Ncdot{A}cdot{B}cdotomega\
varepsilon_{max}&=Ncdot(R^2cdotpi)cdot{B}cdotomega\
varepsilon_{max}&=8cdot(0.032text{ m})^2cdotpicdot{0.075text{ T}}cdot{8.9 frac{text{rad}}{text{s}}}
end{align*}
$$
$$
boxed{varepsilon_{max}=1.717cdot{10^{-2}}text{ V}}
$$
varepsilon_{max}=1.717cdot{10^{-2}}text{ V}
$$
begin{align*}
d&=22text{ cm}=0.22text{ m}\
N&=155\
B_h&=3.8cdot{10^{-5}}text{ T}\
B_v&=2.85cdot{10^{-5}}text{ T}\
omega&=1250text{ rpm}
end{align*}
As the coil loops rotate about the vertical axis, only the horizontal component of the Eart magnetic field will induce the emf in coil.
$$
boxed{text{Only the horizontal component of the magnetic field is important.}}
$$
We have to compute the maximal induced emf, that is given by the equation:
$$
begin{align}
varepsilon_{max}&=Ncdot{A}cdot{B_h}cdotomega
end{align}
$$
begin{align*}
omega&=1250text{ rpm}=1250text{ rpm}cdotfrac{2 pi frac{text{ rad}}{text{ rev}}}{60 frac{text{min}}{text{s}}}=130.9 frac{text{rad}}{text{s}}
end{align*}
begin{align*}
varepsilon_{max}&=Ncdot{A}cdot{B_h}cdotomega\
varepsilon_{max}&=Ncdotleft(frac{d}{2}right)^2cdotpicdot{B}cdotomega\
varepsilon_{max}&=155cdotleft(frac{0.22text{ m}}{2}right)^2cdotpicdot{3.8cdot{10^{-5}}text{ T}}cdot{130.9 frac{text{rad}}{text{s}}}
end{align*}
$$boxed{varepsilon_{max}=2.93cdot{10^{-2}}text{ V}}$$
b) $varepsilon_{max}=2.93cdot{10^{-2}}text{ V}$
Regarding the rms value, the values at the samples is first squared so the value will be always positive. Then we calculate the average value of squares and compute the square root, so the value is always positive.
P_{avg}=I_{rms}cdot{V_{rms}}
$$
$$
begin{align*}
frac{V_P}{V_S}&=frac{N_P}{N_S}\
V_S&=V_Pcdotfrac{N_S}{N_P}
end{align*}
$$
For the transformer $A$ we have given:
$$
begin{align*}
V_P&=100text{ V}\
N_S&=100\
N_P&=20
end{align*}
$$
Let’s compute the voltage at the secondary side:
$$
begin{align*}
V_S&=100text{ V}cdotfrac{100}{20}\
V_S&=500text{ V}
end{align*}
$$
For the transformer $B$ we have given:
$$
begin{align*}
V_P&=100text{ V}\
N_S&=20\
N_P&=100
end{align*}
$$
Let’s compute the voltage at the secondary side:
$$
begin{align*}
V_S&=100text{ V}cdotfrac{20}{100}\
V_S&=20text{ V}
end{align*}
$$
For the transformer $C$ we have given:
$$
begin{align*}
V_P&=20text{ V}\
N_S&=50\
N_P&=50
end{align*}
$$
Let’s compute the voltage at the secondary side:
$$
begin{align*}
V_S&=20text{ V}cdotfrac{50}{50}\
V_S&=20text{ V}
end{align*}
$$
$$
begin{align*}
V_P&=50text{ V}\
N_S&=80\
N_P&=400
end{align*}
$$
Let’s compute the voltage at the secondary side:
$$
begin{align*}
V_S&=50text{ V}cdotfrac{80}{400}\
V_S&=100text{ V}
end{align*}
$$
Let’s compare the values of the transformers:
$$
boxed{A>D>B=C}
$$
A>D>B=C
$$
The voltage equation of the transformer is given by:
$$
begin{align*}
frac{V_P}{V_S}&=frac{N_P}{N_S}\
V_S&=V_Pcdotfrac{N_S}{N_P}
end{align*}
$$
So, if the number of loop in secondary coil is decreased, the voltage in secondary coil will also decrease.
$$
boxed{text{The voltage in secondary circuit also decrease.}}
$$
$bold{b)}$
The current equation of the transformer is given by:
$$
begin{align*}
frac{I_P}{I_S}&=frac{N_S}{N_P}\
I_S&=I_Pcdotfrac{N_P}{N_S}
end{align*}
$$
So, if the number of loop in secondary coil is decreased, the current in secondary circuit will increase.
$$
boxed{text{The current in secondary circuit increase.}}
$$
b) The current in secondary circuit increase.
$$
begin{align*}
frac{V_P}{V_S}&=frac{N_P}{N_S}\
V_S&=V_Pcdotfrac{N_S}{N_P}
end{align*}
$$
So for the second transformer, the secondary voltage will be:
$$
begin{align*}
V_{S2}&=V_{P2}cdotfrac{N_{S2}}{N_{P2}}\
V_{S2}&=2cdot{V_P}cdotfrac{2cdot{N_S}}{2cdot{N_P}}\
V_{S2}&=2cdot{V_P}cdotfrac{N_S}{N_P}\
V_{S2}&=2cdot{V_S}
end{align*}
$$
$$
boxed{text{The secondary voltage of the transformer 2 will be $2cdot{V_S}$.}}
$$
V_{S2}=2cdot{V_S}
$$
$$
begin{align*}
frac{I_P}{I_S}&=frac{N_S}{N_P}\
I_S&=I_Pcdotfrac{N_P}{N_S}
end{align*}
$$
So for the second transformer, the current in secondary circuit will be:
$$
begin{align*}
I_{S2}&=I_{P2}cdotfrac{N_{P2}}{N_{S2}}\
I_{S2}&=3cdot{I_P}cdotfrac{2cdot{N_P}}{N_S}\
I_{S2}&=6cdot{I_S}
end{align*}
$$
$$
boxed{text{The current in secondary circuit of the transformer 2 will be $I_{S2}=6cdot{I_S}$.}}
$$
I_{S2}=6cdot{I_S}
$$
$$
begin{align*}
V_{rms}&=frac{V_{max}}{sqrt{2}}\
V_{rms}&=frac{55text{ V}}{sqrt{2}}
end{align*}
$$
$$
boxed{V_{rms}=38.9text{ V}}
$$
V_{rms}=38.9text{ V}
$$
$$
begin{align*}
V_{max}&=sqrt{2}cdot{V_{rms}}\
V_{max}&=sqrt{2}cdot{240text{ V}}
end{align*}
$$
$$
boxed{V_{max}=340text{ V}}
$$
V_{rms}=340text{ V}
$$
$$
begin{align*}
V_{rms}&=120text{ V}\
I_{max}&=2.1text{ A}
end{align*}
$$
The equation for the resistance in an AC circuit is given as:
$$
begin{align*}
R&=frac{V_{max}}{I_{max}}=frac{V_{rms}}{I_{rms}}
end{align*}
$$
So to compute the resistance of the resistor in a circuit we need to compute the maximum value of voltage or rms value of current. Let’s find the maximum value of voltage.
$$
begin{align*}
V_{max}&=sqrt{2}cdot{V_{rms}}\
R&=frac{V_{max}}{I_{max}}\
R&=frac{sqrt{2}cdot{V_{rms}}}{I_{max}}\
R&=frac{sqrt{2}cdot{120text{ V}}}{2.1text{ A}}
end{align*}
$$
$$
boxed{R=80.8text{ $Omega$}}
$$
R=80.8text{ $Omega$}
$$
$$
begin{align*}
R&=150text{ $Omega$}\
I_{rms}&=0.85text{ A}
end{align*}
$$
$bold{a)}$
The average power consumed by the circuit is defined by the equation:
$$
begin{align*}
P_{avg}&=(I_{rms})^2cdot{R}\
P_{avg}&=(0.85text{ A})^2cdot{150text{ $Omega$}}
end{align*}
$$
$$
boxed{P_{avg}=108.4text{ W}}
$$
$bold{b)}$
To compute the maximum power consumption we have adopt equation:
$$
begin{align*}
P_{max}&=(I_{max})^2cdot{R}\
P_{max}&=(sqrt{2}cdot{I_{rms}})^2cdot{R}\
P_{max}&=2cdot(I_{rms})^2cdot{R}\
P_{max}&=2cdot(0.85text{ A})^2cdot{150text{ $Omega$}}
end{align*}
$$
$$
boxed{P_{max}=216.8text{ W}}
$$
b) $P_{max}=216.8text{ W}$
$$
begin{align*}
R&=3.33text{ k$Omega$}\
V_{max}&=141text{ V}
end{align*}
$$
$bold{a)}$
The average power consumed by the circuit is defined by the equation:
$$
begin{align*}
P_{avg}&=frac{(V_{rms})^2}{R}\
P_{avg}&=frac{left(frac{V_{max}}{sqrt{2}}right)^2}{R}\
P_{avg}&=frac{left(V_{max}right)^2}{2cdot{R}}\
P_{avg}&=frac{left(141text{ V}right)^2}{2cdot{3330text{ $Omega$}}}
end{align*}
$$
$$
boxed{P_{avg}=3text{ W}}
$$
$bold{b)}$
To compute the maximum power consumption we have adopt equation:
$$
begin{align*}
P_{max}&=frac{(V_{max})^2}{R}\
P_{max}&=frac{(141text{ V})^2}{3330text{ $Omega$}}
end{align*}
$$
$$
boxed{P_{max}=6text{ W}}
$$
b) $P_{max}=6text{ W}$
$$
begin{align*}
V_P&=110text{ V}\
V_S&=3text{ V}
end{align*}
$$
The primary side of the transformer is connected to a household voltage, and the voltage has to be reduced to a $3text{ V}$. The transforming ratio can be computed from the transformer equation:
$$
begin{align*}
frac{N_P}{N_S}&=frac{V_P}{V_S}\
frac{N_P}{N_S}&=frac{110text{ V}}{3text{ V}}
end{align*}
$$
$$
boxed{frac{N_P}{N_S}=36.67}
$$
frac{N_P}{N_S}=36.67
$$
$$
begin{align*}
V_P&=120text{ V}\
V_S&=9text{ V}\
N_P&=147
end{align*}
$$
$bold{a)}$
The voltage equation of the transformer is given by:
$$
begin{align*}
frac{N_P}{N_S}&=frac{V_P}{V_S}
end{align*}
$$
$$
boxed{text{The number of the loops at secondary side have to be less then 147.}}
$$
$bold{b)}$
Let’s substitute the given:
$$
begin{align*}
frac{N_P}{N_S}&=frac{V_P}{V_S}\
N_S&=frac{9text{ V}}{120text{ V}}cdot{147}
end{align*}
$$
$$
boxed{N_S=11}
$$
b) $N_S=11$
$$
begin{align*}
frac{N_S}{N_P}&=frac{1}{18}\
V_P&=110text{ V}
end{align*}
$$
So, we have to compute the voltage at secondary side of the transformer.
$$
begin{align*}
frac{N_P}{N_S}&=frac{V_P}{V_S}
end{align*}
$$
Let’s express the voltage at secondary side:
$$
begin{align*}
frac{N_P}{N_S}&=frac{V_P}{V_S}\
V_S&=frac{N_S}{N_P}cdot{V_P}\
V_S&=frac{1}{18}cdot{110text{ V}}
end{align*}
$$
$$
boxed{V_S=6.11text{ V}}
$$
V_S=6.11text{ V}
$$
$$
begin{align*}
V_P&=120text{ V}\
V_S&=11000text{ V}
end{align*}
$$
And we have to compute the ratio of loops. The equation of the transformer is given as:
$$
begin{align*}
frac{N_S}{N_P}&=frac{V_S}{V_P}
end{align*}
$$
Let’s express substitute:
$$
begin{align*}
frac{N_S}{N_P}&=frac{11000text{ V}}{120text{ V}}
end{align*}
$$
$$
boxed{frac{N_S}{N_P}=91.7}
$$
frac{N_S}{N_P}=91.7
$$
$$
begin{align*}
P_{avg}&=75text{ W}\
V_{rms}&=120text{ V}
end{align*}
$$
$bold{a)}$
The average power of the resistive element in an AC circuit is given by the equation:
$$
begin{align*}
P_{avg}&=frac{(V_{rms})^2}{R}
end{align*}
$$
Let’s express the resistance from this equation and substitute:
$$
begin{align*}
R&=frac{(V_{rms})^2}{P_{avg}}\
R&=frac{(120text{ V})^2}{75text{ W}}
end{align*}
$$
$$
boxed{R=192text{ $Omega$}}
$$
$bold{b)}$
We have the voltage on the resistor and it’s resistance so we can compute the current through it as:
$$
begin{align*}
I_{max}&=sqrt{2}cdot{I_{rms}}\
I_{max}&=sqrt{2}cdotfrac{V_{rms}}{R}\
I_{max}&=sqrt{2}cdotfrac{120text{ V}}{192text{ $Omega$}}
end{align*}
$$
$$
boxed{I_{max}=0.88text{ A}}
$$
$bold{c)}$
The maximum power at resistor is given by the equation:
$$
begin{align*}
P_{max}&=I_{max}cdot{V_{max}}\
P_{max}&=I_{max}cdot{sqrt{2}cdot{V_{rms}}}\
P_{max}&=0.88text{ A}cdot{sqrt{2}cdot{120text{ V}}}
end{align*}
$$
$$
boxed{P_{max}=149.3text{ W}}
$$
b) $I_{max}=0.88text{ A}$
c) $P_{max}=149.3text{ W}$
So if we hold the loop in that way that the normal is vertical the magnetic flux will be maximum. On the other way, if the normal to a loop is horizontal, the magnetic flux will be almost zero.
$$
boxed{text{If the normal to the loop is horizontal, the flux will be lower.}}
$$
$$
begin{align*}
l&=0.45text{ m}\
R&=12.5text{ $Omega$}\
B&=0.75text{ T}\
I&=0.125text{ A}
end{align*}
$$
The induced moving emf is defined by the equation:
$$
begin{align*}
varepsilon&=Bcdot{l}cdot{v}
end{align*}
$$
And the induced current will be the current through the resistor, so let’s substitute and express the speed:
$$
begin{align*}
Icdot{R}&=Bcdot{l}cdot{v}\
v&=frac{Icdot{R}}{Bcdot{l}}\
v&=frac{0.125text{ A}cdot{12.5text{ $Omega$}}}{0.75text{ T}cdot{0.45text{ m}}}
end{align*}
$$
$$
boxed{v=4.63 frac{text{m}}{text{s}}}
$$
v=4.63 frac{text{m}}{text{s}}
$$
If the current is constant there is no change in magnetic flux, so there is no induced emf.
$$
boxed{text{The current will be zero.}}
$$
$bold{b)}$
The magnetic field produced by the current in the wire will have the direction out from the paper. There will be also an induced magnetic field that tries to oppose the magnetic field of the wire current. The direction of the induced magnetic field will be in the paper. This magnetic field will produce the current in the clockwise direction.
$$
boxed{text{Induced current direction will be clockwise.}}
$$
b) Induced current direction will be clockwise.
The direction of the magnetic field lines that are produced by the magnet will go from the magnet’s north magnetic pole to the south magnetic poles. So when the loop falling down, the magnetic flux will increase. As the magnetic field direction will be toward the rightward, the induced magnetic field will oppose it, so the direction of the induced magnetic field will be leftward. This is possible if:
$$
boxed{text{The direction of the induced current is clockwise.}}
$$
$bold{b)}$
When the loop falling down, the magnetic flux will decrease. As the magnetic field direction will be toward the rightward, the induced magnetic field will oppose it, in order to increase the magnetic flux, so the direction of the induced magnetic field will be rightward. This is possible if:
$$
boxed{text{The direction of the induced current is counterclockwise.}}
$$
b) The direction of the induced current is counterclockwise.
$$
begin{align*}
v&=8cdot{10^3} frac{text{m}}{text{s}}\
B&=2cdot{10^{-10}}text{ T}\
l&=5text{ m}
end{align*}
$$
The potential difference between its ends represent the motional emf that is given by the equation:
$$
begin{align*}
varepsilon&=vcdot{B}cdot{l}
end{align*}
$$
As we have the all given, let’s substitute and compute:
$$
begin{align*}
varepsilon&=8cdot{10^3} frac{text{m}}{text{s}}cdot{2cdot{10^{-10}}text{ T}}cdot{5text{ m}}
end{align*}
$$
$$
boxed{varepsilon=8cdot{10^{-6}}text{ V}}
$$
varepsilon=8cdot{10^{-6}}text{ V}
$$
$$
begin{align*}
a&=5.8text{ cm}=0.058text{ m}\
b&=8.2text{ cm}=0.082text{ m}\
B&=1.3text{ T}\
N&=1\
Delta{T}&=21text{ ms}=0.021text{ s}
end{align*}
$$
The average induced emf is defined by the equation:
$$
begin{align}
varepsilon&=-Ncdotfrac{Delta{Phi}}{Delta{t}}
end{align}
$$
As the loop is rectangular the area of single loop is:
$$
begin{align*}
A&=acdot{b}
end{align*}
$$
We have that the minimal flux during the loop rotation is zero, but we do not have the maximum magnetic flux and we need to compute it. The maximal magnetic flux is achieved when the angle between the normal to loop and and magnetic field is parallel, so the angle have to be $0text{textdegree}$:
$$
begin{align*}
Phi_{f}&=Bcdot{A}cdotcostheta\
Phi_{f}&=Bcdot{A}cdotcos{0text{textdegree}}\
Phi_{f}&=Bcdot{A}
end{align*}
$$
Let’s substitute in the first equation:
$$
begin{align*}
varepsilon&=-Ncdotfrac{Phi_f-Phi_i}{Delta{t}}\
varepsilon&=-Ncdotfrac{Bcdot{A}-0}{Delta{t}}\
varepsilon&=-Ncdotfrac{Bcdot{acdot{b}}}{Delta{t}}\
varepsilon&=-1cdotfrac{Bcdot{acdot{b}}}{Delta{t}}\
varepsilon&=-1cdotfrac{1.3text{ T}cdot{0.058text{ m}cdot{0.082text{ m}}}}{Delta{0.021text{ s}}}
end{align*}
$$
$$
boxed{varepsilon=-0.29text{ V}}
$$
varepsilon=-0.29text{ V}
$$
The holes in the disk will interrupt the eddy’s current so:
$$
boxed{text{Effect of eddy’s current is greater in solid disk}}
$$
$bold{b)}$
The solid disk experiences a greater force because
eddy currents in it flow freely and are not interrupted by
the slots.
$$
boxed{text{The best explanation is $B$.}}
$$
b) The best explanation is $B$.
$$
begin{align*}
N&=1\
frac{Delta{B}}{Delta{t}}&=3cdot{10^4} frac{text{T}}{text{s}}\
A&=1.13cdot{10^{-2}}text{ m}^2
end{align*}
$$
And we have to compute the induced emf, that is given by the equation:
$$
begin{align*}
varepsilon&=-Ncdotfrac{DeltaPhi}{Delta{t}}\
varepsilon&=Ncdotfrac{AcdotDelta{B}}{Delta{t}}\
varepsilon&=Ncdot{A}cdotfrac{Delta{B}}{Delta{t}}
end{align*}
$$
Let’s substitute and compute:
$$
begin{align*}
varepsilon&=-1cdot{1.13cdot{10^{-2}}text{ m}^2}cdot{3}cdot{10^4} frac{text{T}}{text{s}}
end{align*}
$$
$$
boxed{varepsilon=339text{ V}}
$$
varepsilon=339text{ V}
$$
These principles are used in many everyday applications for example the electrical generator, transformer, inductional welding, inductive motor, inductional heaters, wireless chargers, and this is not possible without electromagnetic induction.
The change in magnetic flux in electromagnetic induction is possible to achieve in three ways. By changing the magnitude of the magnetic field (or moving the loop inside the constant magnetic field), changing the area of the loop by deforming it, or by the changing angle between the loop surface and the magnetic field lines.
$$
boxed{text{The induced current direction is counterclockwise.}}
$$
$$
begin{align*}
B_i&=1.2cdot{10^{-5}}text{ T}\
B_f&=2.6cdot{10^{-5}}text{ T}\
Delta{t}&=0.38text{ s}\
R&=0.67text{ m}\
N&=4
end{align*}
$$
and we have to find the induced emf in the loop. The induced emf is defined by the equation:
$$
begin{align*}
varepsilon&=-Ncdotfrac{DeltaPhi}{Delta{t}}=Ncdot{A}cdotfrac{Delta{B}}{Delta{t}}\
varepsilon&=Ncdot{A}cdotfrac{B_f-B_i}{Delta{t}}\
varepsilon&=Ncdot{R^2cdotpi}cdotfrac{B_f-B_i}{Delta{t}}\
varepsilon&=4cdot{(0.67text{ m})^2cdotpi}cdotfrac{2.6cdot{10^{-5}}text{ T}-1.2cdot{10^{-5}}text{ T}}{0.38text{ s}}\
varepsilon&=2.08cdot{10^{-4}}text{ V}
end{align*}
$$
$$
boxed{text{Correct answer is: $C$.}}
$$
$$
begin{align*}
B_i&=1.2cdot{10^{-5}}text{ T}\
Delta{t}&=0.38text{ s}\
R&=0.67text{ m}\
N&=3\
varepsilon&=8.1cdot{10^{-4}}text{ V}
end{align*}
$$
and we have to find the magnitude of the magnetic field in the loop. The induced emf is defined by the equation:
$$
begin{align*}
varepsilon&=-Ncdotfrac{DeltaPhi}{Delta{t}}=Ncdot{A}cdotfrac{Delta{B}}{Delta{t}}
end{align*}
$$
Let’s express the magnitude of magnetic field and substitute:
$$
begin{align*}
Delta{B}&=frac{varepsiloncdotDelta{t}}{Ncdot{A}}\
B_f-B_i&=frac{varepsiloncdotDelta{t}}{Ncdot{A}}\
B_f&=frac{varepsiloncdotDelta{t}}{Ncdot{(R)^2cdotpi}}+B_i\
B_f&=frac{8.1cdot{10^{-4}}text{ V}cdot{0.38text{ s}}}{3cdot{(0.67text{ m})^2cdotpi}}+1.2cdot{10^{-5}}text{ T}\
B_f&=8.5cdot{10^{-5}}text{ T}
end{align*}
$$
$$
boxed{text{Correct answer is: $C$.}}
$$
$$
begin{align*}
B_i&=1.2cdot{10^{-5}}text{ T}\
B_f&=1.9cdot{10^{-5}}text{ T}\
a&=0.75text{ m}\
N&=4\
varepsilon&=1.4cdot{10^{-4}}text{ V}
end{align*}
$$
and we have to find the time needed to induce emf of $1.4cdot{10^{-4}}text{ V}$. The induced emf is defined by the equation:
$$
begin{align*}
varepsilon&=-Ncdotfrac{DeltaPhi}{Delta{t}}=Ncdot{A}cdotfrac{Delta{B}}{Delta{t}}
end{align*}
$$
Let’s express the change in time and substitute:
$$
begin{align*}
Delta{t}&=frac{{N}cdot{A}cdotDelta{B}}{varepsilon}\
Delta{t}&=frac{{N}cdot{a^2}cdot(B_f-B_i)}{varepsilon}\
Delta{t}&=frac{{4}cdot{(0.75text{ m})^2}cdot(1.9cdot{10^{-5}}text{ T}-1.2cdot{10^{-5}}text{ T})}{1.4cdot{10^{-4}}text{ V}}\
Delta{t}&=frac{{4}cdot{(0.75text{ m})^2}cdot(1.9cdot{10^{-5}}text{ T}-1.2cdot{10^{-5}}text{ T})}{1.4cdot{10^{-4}}text{ V}}\
Delta{t}&=0.11text{ s}
end{align*}
$$
$$
boxed{text{Correct answer is: $B$.}}
$$
