Under an Overhead Power Line

solved MagneticFieldOfCurrents

Q. A horizontal overhead power line carries a current of 90 A $I = 90$ A in east to west direction. What is the magnitude of the magnetic field due to the current 1.5 m $d = 1.5$ m below the line, in teslas?

Inputs

current

I

90 A

distance

d

1.50 m

Outputs

field

B_{w}

How it's solved

Solved with MagneticFieldOfCurrents

Given

$I = 90$ A
$d = 1.5$ m

Find $B_{w}$ .

Working

field near a long straight wire(line) $B_{w} = \frac{μ_{0} \cdot I}{2 \cdot π \cdot d}$
$B_{w} = \frac{1.25663706 \times 10^{−6} \cdot 90}{2 \cdot 3.14159265 \cdot 1.5}$
$B_{w} = 1.2 \times 10^{−5}$ T
$B_{w} \approx 1.2 \times 10^{−5}$ Tto 2 significant figures

Conclusion

Below a long straight current the field circles the line; by the right-hand rule, with the current running west, it points south underneath.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :under_an_overhead_power_line do4  kind :numerical5  describes "A horizontal overhead power line carries a current of %{current} in east to west direction. %{field} " \6            "due to the current %{distance} below the line, in teslas?"7  given current: "90 A"8  given distance: "1.5 m"9  asks field: "What is the magnitude of the magnetic field"10end

solution.rb

1require_relative "../questions/under_an_overhead_power_line"2require_relative "../science/moving_charges_and_magnetism/magnetic_field_of_currents"34solve :under_an_overhead_power_line, using: MagneticFieldOfCurrents do5  objects :line67  take current: line.current8  take distance: line.distance_from_the_wire910  field = line.field_near_a_wire11  answer field:, in: "T", figures: 21213  conclusion "Below a long straight current the field circles the line; by the right-hand rule, with the current " \14             "running west, it points south underneath."15end

playground.rb

1require_relative "../solutions/under_an_overhead_power_line"23play_with :under_an_overhead_power_line do4  title "Under an Overhead Power Line"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	MagneticFieldOfCurrents
permeability of free space	$μ_{0}$	N/A²	—	MagneticFieldOfCurrents
current	$I$	A	—	MagneticFieldOfCurrents
turns	$N$	—	—	MagneticFieldOfCurrents
radius	$R$	m	—	MagneticFieldOfCurrents
length of the coil	$L$	m	—	MagneticFieldOfCurrents
turns per length	$n$	m⁻¹	—	MagneticFieldOfCurrents
field at the centre	$B_{c}$	T	—	MagneticFieldOfCurrents
field inside	$B_{s}$	T	—	MagneticFieldOfCurrents
distance from the wire	$d$	m	—	MagneticFieldOfCurrents
field near a wire	$B_{w}$	T	—	MagneticFieldOfCurrents
field at the centre of ground	$B_{c,g}$	T	—
turns of ground	$N_{g}$	—	—
current of ground	$I_{g}$	A	—
radius of ground	$R_{g}$	m	—
field at the centre of line	$B_{c}$	T	—
turns of line	$N$	—	—
current of line	$I$	A	—
radius of line	$R$	m	—
turns per length of ground	$n_{g}$	m⁻¹	—
length of the coil of ground	$L_{g}$	m	—
turns per length of line	$n$	m⁻¹	—
length of the coil of line	$L$	m	—
field near a wire of ground	$B_{w,g}$	T	—
distance from the wire of ground	$d_{g}$	m	—
field near a wire of line	$B_{w}$	T	—
distance from the wire of line	$d$	m	—
field inside of ground	$B_{s,g}$	T	—
field inside of line	$B_{s}$	T	—

B_{c,c} = \frac{μ_{0} \cdot N_{c} \cdot I_{c}}{2 \cdot R_{c}}

field at the centre of a coil

n_{c} = \frac{N_{c}}{L_{c}}

turns on each metre

B_{w,w} = \frac{μ_{0} \cdot I_{w}}{2 \cdot π \cdot d_{w}}

field near a long straight wire

B_{s,c} = μ_{0} \cdot n_{c} \cdot I_{c}

field inside a long solenoid

B_{c,g} = \frac{μ_{0} \cdot N_{g} \cdot I_{g}}{2 \cdot R_{g}}

field at the centre of a coil(ground)

B_{c} = \frac{μ_{0} \cdot N \cdot I}{2 \cdot R}

field at the centre of a coil(line)

n_{g} = \frac{N_{g}}{L_{g}}

turns on each metre(ground)

n = \frac{N}{L}

turns on each metre(line)

B_{w,g} = \frac{μ_{0} \cdot I_{g}}{2 \cdot π \cdot d_{g}}

field near a long straight wire(ground)

B_{w} = \frac{μ_{0} \cdot I}{2 \cdot π \cdot d}

field near a long straight wire(line)

B_{s,g} = μ_{0} \cdot n_{g} \cdot I_{g}

field inside a long solenoid(ground)

B_{s} = μ_{0} \cdot n \cdot I

field inside a long solenoid(line)