A beam contains 2.0 $ 108 doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of 1.0 $ 105 m/s. What are the (a) magnitude and (b) direction of the current density (c) What additional quantity do you need to calculate the total current i in this ion beam?

a) Magnitude of the current density = 6.4 * 10⁻⁶ A/m²

b) The direction of the current density will be along the direction of the current and opposite to the movement of the positive charge

c) The cross sectional area of the ion beam is required

Explanation:

q = 1.6 * 10⁻¹⁹

Since it is doubly charged, 2q = 3.2 *  10⁻¹⁹

Speed, v = 1.0 * 10⁵ m/s

charge density, n = 2.0 * 10⁸

Current density, J = nqv

J =  2.0 * 10⁸ * 3.2 *  10⁻¹⁹ * 1.0 * 10⁵

J = 6.4 * 10⁻⁶ A/m²

J = 6.4 * 10⁻⁴ A/cm²

b) The direction of the current density will be along the direction of the current and opposite to the movement of the positive charge

c) Additional quantity needed to calculate the total current in the beam is the Cross Sectional Area of the beam since J = IA

Where I = current and A = Cross Sectional Area

ill do it for 100, also this isnt 80 its 40

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.