A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of `40pi "rad s"^(-1)` about its axis, perpendicular to its plane. If the magnetic field at its centre is `3.8xx10^(-9)T`, then the charge carried by the ring is close to `(mu_(0)=4pi xx10^(-7)N//A^(2))`

Options

A. `3xx10^(-5)C`
B. `7xx10^(-6)C`
C. `4xx10^(-5)C`
D. `2xx10^(-6)C`

Answer

A

Explanation

`omega =40 pi ` rad/second

`omega =(2pi)/T" "B=(mu_(0))/(4pi) I/r alpha`
`I=q/t" " B=(mu_(0))/(4pi)I/R . 2pi`
`I=q/(2pi//omega)" "B=(mu_(0))/(4pi) . 2pi I/r`
`I=(q omega)/(2pi)" "3.8xx10^(-9)=10^(-7)xx2pixx(q omega)/(2pir)`
`3.8xx10^(-9)=10^(-7)xxqxx(40 pi )/(10xx10^(-2))`
`q=(3.8xx10^(-9)xx10xx10^(-2))/) 40 pi xx10^(-7))`
`q=(3.8xx10^(-10)xx10^(7))/(40xx22/7)`
`q=(3.8xx10^(-3)x7)/(40xx22)`
`q=0.030xx10^(-3)C`
`q=3xx10^(-5)C`