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A thin ring of 10 cm radius carries a un...

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))`

Text Solution

Verified by Experts

If q is the charge on the ring then

`i=q/T=(q omega)/(2pi)`
Magnetic field
`B=(mu_(0)i)/(2R)=(mu_(0)((qomega)/(2pi)))/(2R)`
or `3.8xx10^(-9)=((mu_(0))/(4pi))(qomega)/R=(10^(-7))(qxx40pi)/0.10`
`:.q=3xx10^(-5)C`
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