If an electron is moving in a magnetic field of `5.4 xx 10^(-4)T` on a circular path of radius 32 cm having a frequency of 2.5 MHz, then its speed will be

Options

A. `8.56xx10^(6)m s^(-1)`
B. `5.024xx10^(6)m s^(-1)`
C. `8.56xx10^(4)m s^(-1)`
D. `5.024xx10^(4)m s^(-1)`

Answer

B

Explanation

Here, `B=5.4xx10^(-4)T,`
r = 32 cm = `32xx10^(-2)m, upsilon=2.5" MHz = 2.5 "xx10^(6) " Hz"`
The speed of electron on circular path
`v=r xx 2pi upsilon = 32xx10^(-2)xx2xx3.14xx2.5xx10^(6)`
`=502.4xx10^(4)=5.024 xx10^(6)" m s"^(-1)`