As the bar magnet moves towards the coil of wire, the magnetic flux through the coil increases from $Phi_{text{initial}}=1.3times10^{-5},text{T}cdottext{m}^2$ to $Phi_{text{final}}=3.7times10^{-3},text{T}cdottext{m}^2$ in the time $Delta t=0.25,text{s}$. The change of the magnetic flux during this time is given as

$$
begin{align*}
Delta Phi=Phi_{text{final}}-Phi_{text{initial}},. tag{1}
end{align*}
$$

The magnitude of the induced emf is $abs{varepsilon}=1.5,text{V}$. The expression for the magnitude of the induced emf is given by Faraday’s law:

$$
begin{align*}
abs{varepsilon}=Nabs{frac{Delta Phi}{Delta t}},, tag{2}
end{align*}
$$

where $N$ is the number of loops of wire in the coil.

We want to calculate $N$.

Rearranging equation (2) we get

$$
begin{align*}
N=abs{varepsilon}abs{frac{Delta t}{Delta Phi}} ,.
end{align*}
$$

Putting the expression (1) into the last equation we obtain

$$
begin{align*}
N&=abs{varepsilon}abs{frac{Delta t}{Phi_{text{final}}-Phi_{text{initial}} }} tag{3}\
&=(1.5,text{V})abs{frac{0.25,text{s}}{3.7times10^{-3},text{T}cdottext{m}^2-1.3times10^{-5},text{T}cdottext{m}^2 }} \\
&boxed{N=100},,
end{align*}
$$

where we changed in the values of $abs{varepsilon}$, $Delta t$, $Phi_{text{final}}$ and $Phi_{text{initial}}$ in line (3).

The number of loops of wire in the coil is $N=15$. The coil experiences an induced emf of magnitude $abs{varepsilon}=0.78,text{V}$ when the magnetic flux through the coil is changed in the time $Delta t=0.45,text{s}$. The expression for the magnitude of the induced emf is given by Faraday’s law:

$$
begin{align*}
abs{varepsilon}=Nabs{frac{Delta Phi}{Delta t}},, tag{1}
end{align*}
$$

where $Delta Phi$ is the change of the magnetic flux during the time $Delta t$.

We want to calculate the value of $Delta Phi$.

Taking off the absolute value sign in the factor $abs{dfrac{DeltaPhi}{Delta t}}$ in equation (1) ($Delta t$ is positive and we assume that the flux increases, meaning that the change in flux $Delta Phi$ is also positive), and rearranging equation (1) we get

$$
begin{align*}
Delta Phi&=frac{abs{varepsilon}Delta t}{N} tag{2} \
&=frac{(0.78,text{V})(0.45,text{s})}{15} \\
&boxed{Delta Phi=2.3times10^{-2},text{T}cdottext{m}^2},,
end{align*}
$$

where we changed in the values of $abs{varepsilon}$, $Delta t$ and $N$ in line (2).

$Delta Phi=2.3times10^{-2},text{T}cdottext{m}^2$.

The number of loops of wire in the coil is $N=25$. The magnetic flux through the coil changes by $Delta Phi=2.6times10^{-4},text{T}cdottext{m}^2$ during the time $Delta t$, and produces an induced emf of magnitude $abs{varepsilon}=1.7,text{V}$. The expression for the magnitude of the induced emf is given by Faraday’s law:

$$
begin{align*}
abs{varepsilon}=Nabs{frac{Delta Phi}{Delta t}},, tag{1}
end{align*}
$$

We want to calculate the value of $Delta t$.

Taking off the absolute value sign in the factor $abs{dfrac{DeltaPhi}{Delta t}}$ in equation (1) (both $Delta Phi$ and $Delta t$ are positive), and rearranging equation (1) we get

$$
begin{align*}
Delta t&=frac{NDeltaPhi}{abs{varepsilon}} tag{2} \
&=frac{(25)(2.6times10^{-4},text{T}cdottext{m}^2)}{1.7,text{V}} \\
&boxed{Delta t=3.8,text{ms}},,
end{align*}
$$

where we changed in the values of $N$, $Delta Phi$ and $abs{varepsilon}$ in line (2).