Question

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?

Short answer

Answer: The minimum momentum perpendicular to the solenoid axis that a particle can have if it is to be able to enter the detector is about \(3.04 \times 10^{-24} \, \text{kg m/s}\).

Step-by-step solution

Calculate the magnetic field inside the solenoid

To calculate the magnetic field, we first need to find the number of turns per unit length, n. The number of turns per centimeter is given as 550 turns/cm. Therefore, to find the number of turns per meter, we need to multiply by 100: n = 550 turns/cm * 100 cm/m = 55000 turns/m Now, we can use the formula for the magnetic field inside a solenoid, \(B = \mu_0 nI\): B = (\(4\pi \times 10^{-7} \, \text{T m/A}\) ) * (55000 turns/m) * (22 A) = 0.1519 T The magnetic field inside the solenoid is 0.1519 T (Teslas).

Calculate the angular frequency of the particle in the magnetic field

We can now use the formula \(\omega = \frac{qB}{m}\) to find the angular frequency of the particle in the solenoid. Since we are dealing with electrons and positrons, both have the same magnitude of charge (1.6 x 10^{-19} C) and mass (9.109 x 10^{-31} kg): \(\omega = \frac{1.6 \times 10^{-19} C \times 0.1519 T}{9.109 \times 10^{-31} kg} = 2.679 \times 10^{11} \, \text{rad/s}\) The angular frequency of the particle in the magnetic field is about \(2.679 \times 10^{11} \, \text{rad/s}\).

Calculate the minimum momentum parallel to the solenoid axis

To find the minimum momentum required to enter the detector, we can use the formulas \(p = mvr\) and \(v = \frac{\omega r}{2\pi}\). Replacing the value of v in the momentum equation: \(p = m\frac{\omega r}{2\pi}\) Now, we will plug in the given values: \(p = (9.109 \times 10^{-31} kg) \frac{(2.679 \times 10^{11} \, \text{rad/s})(0.8 \, \text{m})}{2\pi}\) \(p \approx 3.04 \times 10^{-24} \,\text{kg m/s}\) The minimum momentum perpendicular to the solenoid axis that a particle can have if it is to be able to enter the detector is about \(3.04 \times 10^{-24} \, \text{kg m/s}\).