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. `2 xx 10^(-6) C`
B. `7 xx 10^(-6) C`
C. `4 xx 10^(-5) C`
D. `3 xx 10^(-5) C`

Answer

d

Explanation

If Q is the charge carried by the ring and co is angular velocity then equivalent current can be written as follows:
` I = Q/(2 pi//omega) = (Q omega)/(2 pi)`
Magnetic field at the center of ring is given by the formula .
`B = (mu_(0)I)/(2R) = (mu_(0)Q omega)/(2R(2 pi))`
`rArr Q = (4 B pi R)/(mu_(0)omega) = (4 xx 3.8 xx 10^(-9) xx pi xx 0.1)/(4 pi xx 10^(-7) xx 40 pi) = 3 xx 10^(-5) C`
Hence option (d) is correct .