To manufacture a polyethlene film a wide band is pulled over rollers at `v = 15 m//s`. During this process the film's surface acquires a uniformly distibuted charge `sigma`, mainly because of friction. An electric field of strength 20 kV/cm can cause a discharge in air. Taking into account this fact, maximum possible magnetic field flux density B near the film's surface will be :

Options

A. `3.33 xx 10^(-9) "weber/m"^(2)`
B. `2.22 xx 10^(-9) "weber/m"^(2)`
C. `4.44 xx 10^(-9) "weber/m"^(2)`
D. `5.55xx10^(-9) "weber/m"^(2)`

Answer

A

Explanation

`E_(max)=(sigma_(max))/(2 epsilon_(0))`
`rArr sigma_(max)=2 epsilon_(0) E max`
`B_(max) = (mu_(0) sigma _(max)v)/(2)`
`= mu_(0) epsilon_(0) E_(max)`
`=(v E_(max))/(c^(2))=3.22 xx 10^(-9) "weber/m"^(2)`.