All Solutions
Page 849: Standardized Test Prep
$$
begin{align*}
Phi&=Bcdot{A}cdotcostheta\
Phi&=Bcdot{L}cdot{W}cdotcos{0text{textdegree}}\
Phi&=Bcdot{Lcdot{W}}
end{align*}
$$
$$
boxed{text{The correct answer is: $C$.}}
$$
$$
begin{align*}
B&=200text{ T}\
W&=0.1text{ m}\
L&=0.2text{ m}\
v&=0.5 frac{text{m}}{text{s}}\
N&=1\
R&=400text{ $Omega$}
end{align*}
$$
The induced emf force is given by the equation:
$$
begin{align}
varepsilon&=-Ncdotfrac{DeltaPhi}{Delta{t}}
end{align}
$$
Let’s first compute the change in magnetic flux. In initial moment the complete flux is inside the magnetic field, so the magnetic flux through the loop is:
$$
begin{align*}
Phi_i&=Bcdot{A}cdotcos{theta}\
Phi_i&=Bcdot{Wcdot{L}}
end{align*}
$$
At the final moment the loop is just $frac{1}{3}$ inside the magnetic field so the magnetic flux through it will be:
$$
begin{align*}
Phi_f&=Bcdot{A}cdotcos{theta}\
Phi_f&=Bcdotfrac{1}{3}cdot{Wcdot{L}}
end{align*}
$$
And the change in magnetic flux will be:
$$
begin{align*}
DeltaPhi&=Phi_f-Phi_i\
DeltaPhi&=Bcdotfrac{1}{3}cdot{Wcdot{L}}-Bcdot{Wcdot{L}}\
DeltaPhi&=-frac{2}{3}cdot{B}cdot{W}cdot{L}
end{align*}
$$
$$
begin{align*}
v&=frac{Delta{x}}{Delta{t}}\
Delta{t}&=frac{Delta{x}}{v}\
Delta{t}&=frac{frac{2}{3}cdot{L}}{v}\
Delta{t}&=frac{2cdot{L}}{3cdot{v}}
end{align*}
$$
Now, we are able to substitute in a first equation and to compute the induced emf:
$$
begin{align*}
varepsilon&=-Ncdotfrac{DeltaPhi}{Delta{t}}\
varepsilon&=-Ncdotfrac{-frac{2}{3}cdot{B}cdot{W}cdot{L}}{frac{2cdot{L}}{3cdot{v}}}\
varepsilon&=Ncdot{B}cdot{W}cdot{v}
end{align*}
$$
And the current through the loop is:
$$
begin{align*}
I&=frac{varepsilon}{R}\
I&=frac{Ncdot{B}cdot{W}cdot{v}}{R}\
I&=frac{1cdot{200text{ T}}cdot{0.1text{ m}}cdot{0.5 frac{text{m}}{text{s}}}}{400text{ $Omega$}}\
I&=0.025text{ A}
end{align*}
$$
$$
boxed{text{The correct answer is $B$}}
$$
$$
begin{align*}
V_P&=120text{ V}\
I_P&=0.3text{ A}\
V_S&=9text{ V}
end{align*}
$$
The transformer equation is given by:
$$
begin{align*}
frac{V_P}{V_S}&=frac{I_S}{I_P}\
I_S&=frac{V_P}{V_S}cdot{I_P}\
I_S&=frac{120text{ V}}{9text{ V}}cdot{0.3text{ A}}\
I_S&=4text{A}
end{align*}
$$
And the power of the transformer is:
$$
begin{align*}
P&=I_Pcdot{V_P}=I_Scdot{V_S}\
P&=0.3text{ A}cdot{120text{ V}}\
P&=36text{ W}
end{align*}
$$
$$
boxed{text{The correct answer is $A$.}}
$$
This is possible only if the current direction inside the loop is clockwise.
$$
boxed{text{The correct answer is $A$.}}
$$
If we move the loopback and forward the results will be the same. There is no change in magnetic flux and also can’t is the change of current.
In the case of rotation of the loop, the area of the loop will be changed because the angle between the loop surface and the magnetic field line will be changed during rotation, so there will be the change of magnetic flux and also the change in current.
$$
boxed{text{The correct answer is $C$}}
$$
$$
begin{align*}
N&=10\
d&=0.2text{ m}\
R&=10text{ k$Omega$}=10cdot{10^3}text{ $Omega$}\
frac{Delta{B}}{Delta{t}}&=100 frac{text{T}}{text{s}}
end{align*}
$$
The induced emf is given with the equation:
$$
begin{align*}
varepsilon&=-Ncdotfrac{DeltaPhi}{Delta{t}}=Ncdot{A}cdotfrac{Delta{B}}{Delta{t}}
end{align*}
$$
The area of the single loop is the area of the circle of wire:
$$
begin{align*}
A&=R^2cdotpi=left(frac{d}{2}right)^2cdotpi
end{align*}
$$
Let’s substitute and compute the induced emf:
$$
begin{align*}
varepsilon&=Ncdot{A}cdotfrac{Delta{B}}{Delta{t}}\
varepsilon&=Ncdot{left(frac{d}{2}right)^2cdotpi}cdotfrac{Delta{B}}{Delta{t}}\
varepsilon&=10cdot{left(frac{0.2text{ m}}{2}right)^2cdotpi}cdot{100 frac{text{T}}{text{s}}}
end{align*}
$$
$$
boxed{varepsilon=31.4text{ V}}
$$
Now, let’s compute the current through the loop, we know the emf and the resistance of the loop:
$$
begin{align*}
I&=frac{varepsilon}{R}\
I&=frac{31.4text{ V}}{10cdot{10^3}text{ $Omega$}}
end{align*}
$$
$$
boxed{I=3.14text{ mA}}
$$
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