Question

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 17.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.7 A , while at 92.0 ∘C it reads 17.3 A . You can ignore any thermal expansion of the rod.

Part A

Find the resistivity and for the material of the rod at 20 ∘C.

Part B

Find the temperature coefficient of resistivity at 20∘C for the material of the rod.

2) An 18-gauge copper wire (diameter 1.02 mm) carries a current with a current density of 1.90×106 A/m2 . Copper has 8.5×1028 free electrons per cubic meter.

Part A

Calculate the current in the wire.

Express your answer using two significant figures.

Part B

Calculate the drift velocity of electrons in the wire.

Express your answer using two significant figures.

Answer 1

1.

Answer:

Part a)

`\rho = 1.35 * 10^(-5)`

Part b)

`\alpha = 1.12 * 10^(-3)`

Step-by-step explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

`i = 18.7 A`

V = 17.0 volts

now we will have

`V = I R`

`17.0 = 18.7 R`

R = 0.91 ohm

now we know that

`R = \rho (L)/(A)`

`0.91 = \rho (1.60)/(\pi(0.275* 10^(-2))^2)`

`\rho = 1.35 * 10^(-5)`

Part b)

Now at higher temperature we have

`V = I R`

`17.0 = 17.3 R`

R = 0.98 ohm

now we know that

`R = \rho (L)/(A)`

`0.98 = \rho' (1.60)/(\pi(0.275* 10^(-2))^2)`

`\rho' = 1.46 * 10^(-5)`

so we will have

`\rho' = \rho(1 + \alpha \Delta T)`

`1.46 * 10^(-5) = 1.35 * 10^(-5)(1 + \alpha (92 - 20))`

`\alpha = 1.12 * 10^(-3)`

2.

Answer:

Part a)

`i = 1.55 A`

Part b)

`v_d = 1.4 * 10^(-4) m/s`

Step-by-step explanation:

Part a)

As we know that current density is defined as

`j = (i)/(A)`

now we have

`i = jA`

Now we have

`j = 1.90 * 10^6 A/m^2`

`A = \pi((1.02 * 10^(-3))/(2))^2`

so we will have

`i = 1.55 A`

Part b)

now we have

`j = nev_d`

so we have

`n = 8.5 * 10^(28)`

`e = 1.6 * 10^(-19) C`

so we have

`1.90 * 10^6 = (8.5 * 10^(28))(1.6 * 10^(-19))v_d`

`v_d = 1.4 * 10^(-4) m/s`