Question

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?

Answer 1

Answer:

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