Saved Bookmarks
| 1. |
Electron is rotating in circular orbit with radius 5.2xx10^(-11)m and with linear speed 2xx10^(6)ms^(-1) in an hydrogen atom around the proton. Find the magnetic field produced at the centre of the orbit. |
|
Answer» Solution :1. `v=2xx10^(6)ms^(-1),r=5.2xx10^(-11)m` `e=1.6xx10^(-19)C` 2. FREQUENCY of ELECTRON in the orbit f (No. of rotations completed in 1 second) f = `v/(2pir)` 3. ELECTRIC current I = f.e = `v/(2pir)xxe` = `(2xx10^(6))/(2xx3.14xx5.2xx10^(-11))xx1.6xx10^(-19)` = `9.8xx10^(-4)A` 4. Magnetic field produced at the CENTRE of the circular orbit, `B=(mu_(0)I)/(2r)` `thereforeB=(4xx3.14xx10^(-7)xx9.8xx10^(-4))/(2xx5.2xx10^(-11))` `thereforeB=11.8T` |
|