A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant. speed of 1200 r.p.m. about its axis passing through the centre of the ring perpendicular to ity plane. Find the charge carried by the ring, if the magnetic field induction in its centre is `3.8 XX 10^(-9)T.`

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`