Question

Three long, straight electrical cables, running north and south, are tightly enclosed in an insulating sheath. One of the cables carries a 22.3 A current southward; the other two carry currents of 18.9 A and 8.6 A northward.

Use Ampere’s law to calculate the magnitude of the magnetic field at a distance of 10.0 \rm m from the cables.

Answer 1

The resultant magnetic field is -1.04x10⁻⁷T

Step-by-step explanation:

Given:

i₁ = current through the first wire = 22.3 A (south)

i₂ = current through the second wire = 18.9 A (north)

i₃ = current through the third wire = 8.6 A (north)

R = distance from the cables = 10 m

The magnetic field through the first wire is equal:

`B_(1) =(\mu _(o)i_(1) )/(2\pi R) =(4\pi x10^(-7)*22.3 )/(2\pi 10) =4.46x10^(-7) T`

The magnetic field through the second wire is equal:

`B_(2) =(\mu _(o)i_(2) )/(2\pi R) =(4\pi x10^(-7)*18.9 )/(2\pi 10) =3.78x10^(-7) T`

The magnetic field through the third wire is equal:

`B_(3) =(\mu _(o)i_(3) )/(2\pi R) =(4\pi x10^(-7)*8.6 )/(2\pi 10) =1.72x10^(-7) T`

The resultant magnetic field is equal:

`B=B_(1) -B_(2) -B_(3) =4.46x10^(-7) -3.78x10^(-7) -1.72x10^(-7) =-1.04x10^(-7) T`

Answer 2

`1.04 * 10^(-7) T`

Step-by-step explanation:

By using Ampere's law formula we have:

`B = (\mu_0 I)/(2\pi r)`

where
`\mu_0 = 4\pi10^(-7) Tm/A` is the permeability constant. r = 10 m is the distance from the wire, and I is the current magnitude. As there are 2 wires carrying currents northward and 1 carrying current southward:

`B = (\mu_0 I_(N1))/(2\pi r) + (\mu_0 I_(N2))/(2\pi r) - (\mu_0 I_(S))/(2\pi r)`

`B = (\mu_0)/(2\pi r)(I_(N1) + I_(N2) - I_S)`

`B = (4 \pi 10^(-7))/(2\pi *10)(18.9 + 8.6 - 22.3) = 1.04 * 10^(-7) T`