Question

a tightly wound, long solenoid carries a current of 2A. An electron is found to execute

Answer 1

1422 turns m^-1

Answer 2

The provided question "a tightly wound, long solenoid carries a current of 2A. An electron is found to execute" is incomplete. Based on the context of the similar question provided, it is assumed that the question is asking to find the number of turns per meter in the solenoid, given that the electron executes a uniform circular motion with a frequency of $1.00\times {{10}^{8}}rev{{s}^{-1}}$.

Explanation of the solution:

1. Magnetic Field inside a Solenoid: The magnetic field $B$ inside a long, tightly wound solenoid is given by the formula $B = \mu_0 n I$, where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length (turns per meter), and $I$ is the current flowing through the solenoid.

2. Electron's Motion in Magnetic Field: When an electron (charge $q$, mass $m$) executes uniform circular motion in a magnetic field, it implies that its velocity is perpendicular to the magnetic field. In this case, the magnetic force $F_B = qvB$ provides the necessary centripetal force $F_c = \frac{mv^2}{r}$, where $v$ is the speed of the electron and $r$ is the radius of its circular path.

3. Cyclotron Frequency: Equating the magnetic force and centripetal force: $qvB = \frac{mv^2}{r}$ $qB = \frac{mv}{r}$ The frequency of revolution $f$ is related to the speed and radius by $v = 2\pi r f$, which means $\frac{v}{r} = 2\pi f$. Substituting this into the equation: $qB = m(2\pi f)$ So, the magnetic field can also be expressed as $B = \frac{2\pi m f}{q}$.

4. Equating and Solving for $n$: Now, we equate the two expressions for the magnetic field $B$: $\mu_0 n I = \frac{2\pi m f}{q}$ Solving for $n$: $n = \frac{2\pi m f}{\mu_0 q I}$

5. Substitution of Values: Given: Current, $I = 2.00 \text{ A}$ Frequency, $f = 1.00 \times 10^8 \text{ rev s}^{-1}$ Mass of electron, $m = 9.1 \times 10^{-31} \text{ kg}$ Charge of electron, $q = 1.6 \times 10^{-19} \text{ C}$ Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}$

   $n = \frac{2\pi \times (9.1 \times 10^{-31} \text{ kg}) \times (1.00 \times 10^8 \text{ s}^{-1})}{(4\pi \times 10^{-7} \text{ T m A}^{-1}) \times (1.6 \times 10^{-19} \text{ C}) \times (2.00 \text{ A})}$ $n = \frac{2 \times 9.1 \times 10^{-31} \times 10^8}{4 \times 10^{-7} \times 1.6 \times 10^{-19} \times 2}$ $n = \frac{9.1 \times 10^{-23}}{6.4 \times 10^{-26}}$ $n = 1.421875 \times 10^3$ $n \approx 1422 \text{ turns m}^{-1}$

Answer:

Assuming the question asks for the number of turns per meter, the number of turns per meter in the solenoid is approximately $1422 \text{ turns m}^{-1}$.