Under an Overhead Power Line

Q. A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude of the magnetic field due to the current 1.5 m below the line, in teslas? expects 1.2 × 10⁻⁵ T

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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?

The playground

Inputs

current

I

90 A

distance

d

1.50 m

Outputs

field

B_{w}

The solution

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 Ruby

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

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

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

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)