Question

In the production of polyethylene film, a wide strip is drawn over rollers at a speed of v = 45m/s as shown in the figure. During the processing (mainly due to friction), the surface of the film acquires a uniformly distributed charge. Find the maximum values of magnetic field induction Bmax near the surface of the film, taking into account that at an electric field strength E = 20 kV/cm, an electrical discharge occurs in the air. The value of $B_{max}$ is $\alpha$ x $10^{-9}$ T. The value of $\alpha$ is

Answer 1

1

Answer 2

The polyethylene film acquires a uniformly distributed charge on its surface due to friction. Let the surface charge density be $\sigma$. The film moves at a speed $v$. This moving charge constitutes a current sheet. The linear current density $K$ is given by $K = \sigma v$.

The electric field strength near the surface of a uniformly charged thin sheet with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\epsilon_0}$, where $\epsilon_0$ is the permittivity of free space. The problem states that an electrical discharge occurs in the air when the electric field strength reaches $E_{max} = 20 \, \text{kV/cm}$. This is the maximum electric field the film can sustain, and it corresponds to the maximum surface charge density $\sigma_{max}$. $E_{max} = 20 \, \text{kV/cm} = 20 \times 10^3 \, \text{V} / (10^{-2} \, \text{m}) = 20 \times 10^5 \, \text{V/m}$. From the formula for the electric field, the maximum surface charge density is $\sigma_{max} = 2\epsilon_0 E_{max}$.

The moving charged film can be approximated as an infinite current sheet with linear current density $K_{max} = \sigma_{max} v$. The magnetic field induction near the surface of an infinite current sheet with linear current density $K$ is given by $B = \frac{1}{2}\mu_0 K$, where $\mu_0$ is the permeability of free space. The maximum magnetic field induction $B_{max}$ occurs when the linear current density is maximum, which corresponds to the maximum surface charge density. $B_{max} = \frac{1}{2}\mu_0 K_{max} = \frac{1}{2}\mu_0 (\sigma_{max} v)$. Substitute the expression for $\sigma_{max}$: $B_{max} = \frac{1}{2}\mu_0 (2\epsilon_0 E_{max} v) = \mu_0 \epsilon_0 E_{max} v$.

We know the relationship between $\mu_0$, $\epsilon_0$, and the speed of light in vacuum $c$: $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, which implies $\mu_0 \epsilon_0 = \frac{1}{c^2}$. So, $B_{max} = \frac{1}{c^2} E_{max} v$.

We are given: $v = 45 \, \text{m/s}$ $E_{max} = 20 \times 10^5 \, \text{V/m}$ The speed of light in vacuum is approximately $c = 3 \times 10^8 \, \text{m/s}$.

Now, calculate $B_{max}$: $B_{max} = \frac{(20 \times 10^5 \, \text{V/m}) (45 \, \text{m/s})}{(3 \times 10^8 \, \text{m/s})^2}$ $B_{max} = \frac{20 \times 45 \times 10^5}{9 \times 10^{16}} \, \text{T}$ $B_{max} = \frac{900 \times 10^5}{9 \times 10^{16}} \, \text{T}$ $B_{max} = 100 \times 10^{5-16} \, \text{T}$ $B_{max} = 100 \times 10^{-11} \, \text{T}$ $B_{max} = 1 \times 10^2 \times 10^{-11} \, \text{T}$ $B_{max} = 1 \times 10^{-9} \, \text{T}$.

The problem states that $B_{max} = \alpha \times 10^{-9}$ T. Comparing this with our result, $B_{max} = 1 \times 10^{-9}$ T, we find that $\alpha = 1$.