A particle detector utilizes a solenoid that has 550 turns of wire per centimeter. The wire carries a current of 22 A. A cylindrical detector that lies within the solenoid has an inner radius of \(0.80 \mathrm{~m} .\) Electron and positron beams are directed into the solenoid parallel to its axis. What is the minimum momentum perpendicular to the solenoid axis that a particle can have if it is to be able to enter the detector?

Using Ampere's law, the magnetic field inside a solenoid of length \(l\) is given by the equation \(B = \mu_0 n I\), where \(B\) is the magnetic field strength, \(\mu_0\) is the permeability of free space (\(4 \pi \times 10^{-7}\, \mathrm{T \cdot m / A}\)), \(n\) is the number of turns per length, and \(I\) is the current. In this case, the solenoid has 550 turns per centimeter (or 55000 turns per meter) and carries a current of 22 A. We can use this information to find the magnetic field inside the solenoid.

The magnetic force acting on a charged particle moving in a magnetic field is given by the Lorentz force equation, \(F = q(v \times B)\), where \(F\) is the magnetic force, \(q\) is the charge of the particle, \(v\) is the velocity vector of the particle, and \(B\) is the magnetic field vector. Since we are looking for the minimum perpendicular momentum for a particle to enter the detector, we want the force exerted by the magnetic field to equal the centripetal force needed for a particle to travel in a circular path with a radius equal to the inner radius of the detector. The centripetal force equation is given by \(F_c = \frac{m v^2}{r}\), where \(m\) is the mass of the particle, and \(r\) is the radius of the circle.

To find the minimum perpendicular momentum, we can equate the magnetic force and centripetal force in terms of momentum. Using the relation \(p = mv\) (momentum equals mass times velocity), we can rewrite the centripetal force equation as \(F_c = \frac{p^2}{m r}\). Similarly, the magnetic force in terms of momentum can be written as \(F = q(\frac{p}{m} \times B)\). Equating these two forces, we get the equation: $$\frac{p^2}{m r} = q\left(\frac{p}{m} \times B\right)$$ To find the minimum perpendicular momentum, we can solve this equation for \(p\): $$p = q \cdot B \cdot r$$ Now we can plug in the values for \(q\), \(B\), and \(r\). It's important to note that the charge of both an electron and a positron is \(e = 1.6\times10^{-19}\,\mathrm{C}\).

Using the given values, we can calculate the minimum perpendicular momentum needed for a particle to enter the detector: \(p = e \cdot B \cdot r = (1.6 \times 10^{-19} \,\mathrm{C}) \cdot B \cdot (0.80 \,\mathrm{m})\) We previously found that \(B = \mu_0 n I = (4\pi \times 10^{-7} \,\mathrm{T \cdot m / A}) (55000 \,\mathrm{turns/m}) (22 \,\mathrm{A})\), so we can substitute this value of \(B\) into the equation for \(p\): \(p = (1.6 \times 10^{-19} \,\mathrm{C}) \cdot [(4\pi \times 10^{-7} \,\mathrm{T \cdot m / A}) (55000 \,\mathrm{turns/m}) (22 \,\mathrm{A})] \cdot (0.80 \,\mathrm{m})\) Calculating the value, we find the minimum perpendicular momentum that a particle can have in order to enter the detector:

= 6.06 \times 10^{-23} \,\mathrm{kg \cdot m/s}$