To find the number of turns of wire \( N \) on the solenoid, we can use Faraday's law of electromagnetic induction, which states:
\[ \text{emf} = -N \frac{\Delta \Phi}{\Delta t} \]
Where:
- \(\text{emf}\) is the induced electromotive force (31.54 V in this case),
- \(\Delta \Phi\) is the change in magnetic flux (-7.9 Wb),
- \(\Delta t\) is the time interval (4.76 s),
- \( N \) is the number of turns of the solenoid.
We can rearrange the formula to solve for \( N \):
\[ N = -\frac{\text{emf} \cdot \Delta t}{\Delta \Phi} \]
Now we can plug in the values:
- \(\text{emf} = 31.54\) V
- \(\Delta t = 4.76\) s
- \(\Delta \Phi = -7.9\) Wb (note that we are interested in the absolute value of \(\Delta \Phi\) since it's squared in the calculation).
Now we need to substitute these values into the equation:
\[ N = -\frac{31.54 \cdot 4.76}{-7.9} \]
This simplifies to:
\[ N = \frac{31.54 \cdot 4.76}{7.9} \]
Now we calculate the values:
- Calculate \( 31.54 \cdot 4.76 \):
\[ 31.54 \cdot 4.76 \approx 150.0944 \]
- Now divide by 7.9:
\[ N \approx \frac{150.0944}{7.9} \approx 18.97 \]
Since the number of turns \( N \) must be a whole number, we round to the nearest whole number:
\[ N = 19 \]
Thus, the number of turns of wire on the solenoid is approximately \( \boxed{19} \).
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