Question

An electron in orbit around a (one proton) nucleus has a speed of about 106 m/s. Deduce the equivalent current due to the motion of the electron and estimate the magnitude of the magnetic dipole moment of this one electron atom.

Answer 1

(a)
`-4.8* 10^(-4)A` (B)
`4.22* 106{-24}A/m^2`

Step-by-step explanation:

We have given the the speed of the electron =10^6 m/sec

The time period of the electron is given by
`T=(2\pi r)/(v)` where r is the Bohr radius, which is equal to
`5.29* 10^(-11)m`

So time period
`T=(2* \pi * 5.29* 10^(-11))/(10^6)=3.324* 10^(-16)m/sec`

(A) We know that current
`i=(q)/(T)=(-e)/(T)=(-1.6* 10^(-19))/(3.324* 10^(-16))=-4.8* 10^(-4)A`

(B) Magnetic moment is given by
`\mu =IA` , where I is current and A is area

So
`\mu =IA=-4.8* 10^(-4)* 3.14* (5.29* 10^(-11))^2=4.22* 10^(-24)A/m^2`