The electric field in a current-carrying wire can be modeled as the electric fie

Question

The electric field in a current-carrying wire can be modeled as the electric field at the midpoint between two charged rings. Model a 2.20 mm (diameter) aluminum wire as two 2.20 mm (diameter) rings 2.10 mm apart. What is the current in the wire after 20.0 electrons are transferred from one ring to the other? Please explain.
The electric field in a current-carrying wire can be modeled as the electric field at the midpoint between two charged rings. Model a 2.20 mm (diameter) aluminum wire as two 2.20 mm (diameter) rings 2.10 mm apart. What is the current in the wire after 20.0 electrons are transferred from one ring to the other? What is the current in the wire after 20.0 electrons are transferred from one ring to the other? Please explain.

Explanation / Answer

Given distance between two rings    d = 2.10 mm = 1.05 mm = 0.001 m radius of the rings   =    2.2 / 2 = 1.1 mm = 0.0011 m Electric field at a distance x from the ring of radius carrying a charge q                      E = k q x / ( x ^2 + a^2 ) ^ ( 3/2)                          = ( 9*10^ 9 ) ( 20*1.6*10^ -19 C ) ( 0.001 ) / ( 0.0011 +0.001)^ 3/2                          = 2.992*10^ -7   N /C field due to two rings                      E' = 2 E =   5.985*10^ -7 N /C orce on charg q due tofield E                      F = E q work done moving charge q distance x in field E                     W = 2 F x                          = 2 E q x potential difference involts                    V = W / q = 2 E x                     V = 2 * 5.985*10^ -7 * .001                     = 1.197*10^ -9 V                    I = V / R........(1)                    R = ? L / A     where ? = resistivity of aluminium wire               L = length of the wire                A = area of the wire                       calculate resistance of the wire    and substitue in equation (1)               potential difference involts                    V = W / q = 2 E x                     V = 2 * 5.985*10^ -7 * .001                     = 1.197*10^ -9 V                    I = V / R........(1)                    R = ? L / A     where ? = resistivity of aluminium wire               L = length of the wire                A = area of the wire                       calculate resistance of the wire    and substitue in equation (1)              

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