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Page 827: Lesson Check
$$
begin{align*}
abs{varepsilon}=Nabs{frac{Delta Phi}{Delta t}} ,.
end{align*}
$$
The factor $dfrac{Delta Phi}{Delta t}$ is the rate of change of the flux of the magnetic field through a coil. The quicker the magnetic field (its flux through the coil) changes, the absolute value of $dfrac{Delta Phi}{Delta t}$ is greater. Thus, a magnetic field that changes quickly rather than slowly, induces a greater emf.
$$
begin{align*}
Phi=BAcos{theta} ,.
end{align*}
$$
If we double the area of the loop $A$ and halve the magnetic field $B$ the flux becomes
$$
begin{align*}
Phi&=frac{B}{2}2Acos{theta} \
&=BAcos{theta} ,,
end{align*}
$$
which is the same as the starting flux. Thus, the flux doesn’t change when the area of the wire loop is doubled an the magnetic field is halved.
$$
begin{align*}
Phi=BAcos{theta} ,.
end{align*}
$$
The magnitude of the magnetic field $B$ and the area of the loop $A$ are the same for the loops 1 and 2. The angle between the normal to the loop and the magnetic field, $theta$, is equal to $10^{circ}$ and $20^{circ}$ for the loops 1 and 2, respectively. Because
$$
begin{align*}
cos{10^{circ}}>cos{20^{circ}} ,,
end{align*}
$$
the flux through loop 1 is greater than the flux through loop 2.
field, hence there will be no current. So for coil 2 and 4 the current
is zero.
When the coil is moving towards positive $x$ axis, the field is increasing,
so the current should be such that this change is opposed, that is
induced field should point into the page. So, the current should be
clockwise for loop 1.
When the loop is moving towards negative $x$ axis, the field is decreasing.
Hence, to oppose the change, the induced field should be in point
out of the page. So, the current should be counterclockwise for loop
3.
$$
begin{align*}
Phi&=Bcdot{A}cdotcos theta
end{align*}
$$
Where $B$ represents the magnitude of the magnetic field, $A$ stands for the area of the loop, and $theta$ stands for the loop orientation regarding the direction of the magnetic field.
Let’s compute the area of the loop:
$$
begin{align*}
A&=acdot{b}\
A&=0.32text{ m}cdot{0.16text{ m}}\
A&=0.0512text{m}^2
end{align*}
$$
$bold{a)}$
In case that the magnetic field is perpendicular to a plan of the loop, the angle between them is $theta=0text{textdegree}$. So, the flux will be:
$$
begin{align*}
Phi&=Bcdot{A}cdotcos theta\
Phi&=0.77text{ T}cdot{0.0512text{m}^2}cdotcos{0text{textdegree}}
end{align*}
$$
$$
boxed{Phi=0.0394text{ T}cdottext{m}^2}
$$
In case that the magnetic field is paralel to a plan of the loop, the angle between them is $theta=90text{textdegree}$. So, the flux will be:
$$
begin{align*}
Phi&=Bcdot{A}cdotcos theta\
Phi&=0.77text{ T}cdot{0.0512text{m}^2}cdotcos{90text{textdegree}}
end{align*}
$$
$$
boxed{Phi=0}
$$
b) $Phi=0$
$$
begin{align*}
|varepsilon|=Ncdotleft|frac{DeltaPhi}{Delta{t}}right|
end{align*}
$$
As we have the single loop of wire the $N=1$, let’s express the time and substitute:
$$
begin{align*}
|Delta{t}|&=Ncdotleft|frac{DeltaPhi}{varepsilon}right|\
|Delta{t}|&=1cdotleft|frac{0.85text{ T}cdot{ m}^2-0.110text{ T}cdot{ m}^2}{1.48text{ V}}right|\
|Delta{t}|&=1cdotleft|frac{0.74text{ T}cdot{ m}^2}{1.48text{ V}}right|
end{align*}
$$
The time needed for a change in a flux is:
$$
boxed{Delta{t}=0.5text{ s}}
$$
Delta{t}=0.5text{ s}
$$
$$
begin{align*}
|varepsilon|=Ncdotleft|frac{DeltaPhi}{Delta{t}}right|
end{align*}
$$
Where the $N$ stands for the number of loops, $DeltaPhi$ is the magnitude of change in flux, $Delta{t}$ stands for time that is needed for the change in flux, and $varepsilon$ is induced emf.
If we express the change in flux from this equation, we get:
$$
begin{align*}
|DeltaPhi|&=left|frac{varepsiloncdotDelta{t}}{N}right|
end{align*}
$$
Ler’s substitute and compute the magnitude of change in flux:
$$
begin{align*}
|DeltaPhi|&=left|frac{2.6text{ V}cdot{0.35text{ s}}}{25}right|
end{align*}
$$
The magnitude of change in flux is:
$$
boxed{DeltaPhi=0.0364text{ T}cdottext{ m}^2}
$$
DeltaPhi=0.0364text{ T}cdottext{ m}^2
$$
$$
begin{align*}
Phi&=Bcdot{A}cdotcos theta
end{align*}
$$
Where $B$ represents the magnitude of the magnetic field, $A$ stands for the area of the loop, and $theta$ stands for the loop orientation regarding the direction of the magnetic field.
Let’s express the angle from this equation:
$$
begin{align*}
cos theta&=frac{Phi}{Bcdot{A}}\
theta&=arccos {frac{Phi}{Bcdot{A}}}
end{align*}
$$
Let’s substitute and compute:
$$
begin{align*}
theta&=arccos{left(frac{7.1cdot{10^{-4}}text{ T}cdottext{ m}^2}{0.45text{ T}cdot{0.085text{ m}^2}}right)}\
theta&=arccos {left(0.0185621right)}
end{align*}
$$
$$
boxed{theta=88.94text{textdegree}}
$$
theta=88.94text{textdegree}
$$
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