Question

A proton beam in an accelerator carries a current of 106 μa. if the beam is incident on a target, how many protons strike the target in a period of 17.0 s?

Answer 1

The electric current is defined as the charge Q that passes a certain point in a time
`\Delta t`:

`I= (Q)/(\Delta t)`
We know the current,
`I=106 \mu A=106 \cdot 10^(-6) A`, and the time,
`\Delta t=17.0 s`, so the total charge that strikes the target during this time is

`Q=I \Delta t=(106 \cdot 10^(-6)A)(17.0s)=1.8 \cdot 10^(-3)C`

To find the number of proton, we must divide the total charge by the charge of a single proton, which is
`q=1.6 \cdot 10^(-19)C`:

`N= (Q)/(q)= (1.8 \cdot 10^(-3)C)/(1.6 \cdot 10^(-19)C)=1.13 \cdot 10^(16)`
And this is the number of protons that strike the target in 17.0 s.