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

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

6 cos 34, 6 sin 34 trig