The starter motor of a car engine draws a current of 110 A from the battery. The copper wire to the motor is 4.90 mm in diameter and 1.2 m long. The starter motor runs for 0.800 s until the car engine starts. A) How much charge passes through the starter motor? (C)

B) How far does an electron travel along the wire while the starter motor is on? (mm)

a) Amount of charge that passes through the starter motor = 88 C

b) Distance travelled by the electrons along the wire while the starter motor is on = 0.174 mm

Explanation:

Amount of charge carried that passes through the starter motor, Q = It

I = 110 A

t = 0.8 s

Q = 110 * 0.8 = 88 C

b) now to calculate this velocity, (drift velocity)

I = nqAv

v = I/nqA

v = drift velocity = ?

I = current = 110 A

A = Cross section Area of the wire = πd²/4 =

π (0.0049²) / 4 = 3.77 * 10⁻⁵ m²

q = charge on an electron = 1.602 * 10⁻¹⁹ C

n = number of electrons per m³ for Copper

To calculate this,

Density of Copper = 8.80 * 10³ kg/m³, atomic mass of copper is 63.54 g/mol = 0.06354 kg/mol. We can use these two quantities along with Avogadro's number, 6.02 * 10²³ electrons/mol, to determine n, the number of free electrons per cubic meter.

n = (Density/atomic mass) * avogadro's constant = (8800/0.06354) * 6.02 * 10²³ = 8.342 * 10²⁸ electrons/m³

v = 110/[ (8.342 * 10²⁸) (1.602 * 10⁻¹⁹) (3.77 * 10⁻⁵) ] = 0.000218 m/s = 0.218 mm/s

Since the starter is on for 0.8 s,

Distance travelled by the electrons in that time = 0.218 * 0.8 = 0.174 mm

A.

Q = I*t

Where,

Q = charge

I = current

= 110 A

t = time

= 0.8 s

= 110 * 0.8

= 88 C

B.

The radius of the conductor, r = 4.9/2

= 2.45 mm.

The area of the conductor is:

A = π*r^2

= π * (.00245) ^2

= 1.89 x 10^-5 m^2

Charge on an electron, qe = 1.602 * 10⁻¹⁹ C

Density of Copper = 8.80 * 10³ kg/m³

Molar mass of copper is 63.55 g/mol = 0.06355 kg/mol

Avogadro's constant = 6.02 * 10²³ electrons/mol

Number of moles, n = (Density/molar mass) * avogadro's constant

= (8800/0.06355) * 6.02 * 10²³

= 8.342 * 10²⁸ electrons/m³

I = nqAVd

Vd = I/nqA

Vd = 110/[ (8.342 * 10²⁸) (1.602 * 10⁻¹⁹) (1.89 * 10⁻⁵) ]

= 4.37 x 10^-4 m/s

= 0.437 mm/s

The distance traveled is:

x = v * t

= 0.437 * 0.80

= 0.35 mm.