Question

A beam contains 1.1 × 105 doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of 1.5 × 105 m/s. what is (a) the magnitude of the current density and (b) direction of the current density . (c) what additional quantity do you need to calculate the total current i in this ion beam?

Answer 1

(a) The beam contains
`N=1.1 \cdot 10^5` doubly charged positive ions per cubic centimeter. Each ion has a charge equal to 2 protons, so the total charge contained in 1 cubic centimeter is the number of ions times 2 times the charge of 1 proton:

`Q=1.1 \cdot 10^5 \cdot 2 \cdot (1.6 \cdot 10^(-19)C)=3.5 \cdot 10^(-14)C`
1 cubic centimeter corresponds to
`1 \cdot 10^(-6)m^3`, so the charge density is equal to

`\rho = (3.5 \cdot 10^(-14)C)/(1 \cdot 10^(-6)m^3)=3.5 \cdot 10^(-8)C/m^3`

So now we can calculate the magnitude of the current density, which is given by

`J= \rho v =( 3.5 \cdot 10^(-8)C/m^3)(1.5 \cdot 10^5 m/s)=5.3 \cdot 10^(-3) A/m^2`

(b) Since the charges are positive, then the direction of the current density is equal to the direction of the velocity v.

(c) The current density is also equal to the ratio between the current I and the cross-sectional area A:

`J= (I)/(A)`
So, in order to find the current I, we also need the cross-sectional area A of the beam.