The magnetic field between the poles of an electromagnet is \(2.6 \mathrm{T} .\) A coil of wire is placed in this region so that the field is parallel to the axis of the coil. The coil has electrical resistance \(25 \Omega,\) radius $1.8 \mathrm{cm},\( and length \)12.0 \mathrm{cm} .$ When the current supply to the electromagnet is shut off, the total charge that flows through the coil is \(9.0 \mathrm{mC} .\) How many turns are there in the coil?

Faraday's law of electromagnetic induction states that the electromotive force (EMF) induced in a coil is equal to the rate of change of magnetic flux through the coil. It can be represented by the formula: \(\epsilon=-N\frac{d\Phi}{dt}\), where \(\epsilon\) is the induced EMF, N is the number of turns in the coil, and \(\frac{d\Phi}{dt}\) is the rate of change of magnetic flux.

We are given the magnetic field, B = \(2.6 T\). The magnetic flux is given by \(\Phi=BA\), where A is the area of the coil. Since field lines are parallel to the axis of the coil, we only need to consider the area of one circular end. Knowing that the circular area = \(\pi r^2\), we have the formula: \(\Phi=2.6 \times \pi \times (1.8 \times 10^{-2})^2\). We are not given the rate of change of magnetic flux directly, but we can find it using the given total charge that flows through the coil (\(9.0 mC\)). We know that the current is related to the charge by, \(Q=It\), where \(Q\) is the charge, \(I\) is the current, and \(t\) is the time. We also know that the current and EMF are related to the resistance of the coil by Ohm's law: \(I=\frac{\epsilon}{R}\). Combining these two equations, we get: \(Q=\frac{\epsilon}{R}t\). We are given that Q = \(9.0 mC\) and R = \(25\Omega\). So, the rate of change of the magnetic flux is: \(\frac{d\Phi}{dt}=\frac{\epsilon t}{-NR}\).

Now, we have all the information needed to determine the number of turns in the coil, N. By replacing values in Faraday's law, we get: \(\frac{-NR(9.0\times 10^{-3})}{t} = 2.6 \times \pi \times (1.8\times 10^{-2})^2\) \(NR = \frac{2.6 \times \pi \times (1.8\times 10^{-2})^2}{-25 \times (9.0\times 10^{-3})}\) Plug the given values into the equation and solve for N: \(N = \frac{2.6 \times \pi \times (1.8\times 10^{-2})^2}{25 \times (9.0\times 10^{-3})}\) \(N \approx 71\) Since the number of turns in a coil must be an integer, there are approximately 71 turns in the coil.