A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.0 x 10^28 free electrons per cubic meter. (a) What is the current in the wire? (b) What is the magnitude of the drift velocity of the electrons in the wire?

a) 0.0875A

b) 2.06*10^-6m/s

Explanation:

given

Diameter, d = 2.6mm = 0.0026m

Charge, Q = 420C

Time, t = 80mins = 80*60s = 4800s

Concentration of electrons, n = 5*10^28

To find the current, recall that q = it

Therefore, i = Q/t

Current, i = 420/4800

i = 0.0875A

To get the drift velocity, recall that I = n*q*Vd*A

A = πd²/4

A = π*0.0026²/4

A = 5.3*10^-6m²

Then, Vd = I/nqA

Vd = 0.0875/(5*10^28 * 1.6*10^-19 * 5.3*10^-6)

Vd = 0.0875/42400

Vd = 2.06*10^-6m/s

Note, low Vd causes low current.

Therefore, the current is 0.0875A and the Vd is 2.06*10^-6m/s

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach