A Wire Held Up by a Field

solved MagneticForce

Q. A straight wire of mass 200 g $m = 200$ g and length 1.5 m $l = 1.5$ m carries a current of 2 A $I = 2$ A. It is suspended in mid-air $F = m \cdot g$ by a uniform horizontal magnetic field. Taking g as 9.8 m/s² $g = 9.8$ m/s², what is the magnitude of the magnetic field, in teslas?

Inputs

mass

m

200 g

length

l

1.50 m

current

I

2 A

gravity

g

9.8 m/s²

Outputs

field

B

How it's solved

Solved with MagneticForce

Given

$m = 200$ g
$l = 1.5$ m
$I = 2$ A
$g = 9.8$ m/s²
$F = m \cdot g$ from “suspended in mid-air”

Find $B$ .

Working

$m = 200$ g $= 0.2$ kg
suspended in mid-air $F = m \cdot g$
$F = 0.2 \cdot 9.8$
$F = 1.96$ N
force on a wire across a field(wire) $F = I \cdot l \cdot B$
$1.96 = 2 \cdot 1.5 \cdot B$
$B = 0.6533$ T
$B \approx 0.65$ Tto 2 significant figures

Conclusion

The field's push on the current holds up the wire's weight, so the field is the weight over the current times the length.

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 :a_wire_held_up_by_a_field do4  kind :numerical5  describes "A straight wire of mass %{mass} and length %{length} carries a current of %{current}. It is suspended " \6            "in mid-air by a uniform horizontal magnetic field. Taking g as %{gravity}, %{field}, in teslas?"7  given mass: "200 g"8  given length: "1.5 m"9  given current: "2 A"10  given gravity: "9.8 m/s2"11  asks field: "what is the magnitude of the magnetic field"12end

solution.rb

1require_relative "../questions/a_wire_held_up_by_a_field"2require_relative "../science/moving_charges_and_magnetism/magnetic_force"34solve :a_wire_held_up_by_a_field, using: MagneticForce do5  objects :wire67  take mass: wire.mass8  take length: wire.length_in_the_field9  take current: wire.current10  take gravity: g1112  deduce_from "suspended in mid-air", wire.force_on_a_wire => wire.mass * g1314  field = B15  answer field:, in: "T", figures: 21617  conclusion "The field's push on the current holds up the wire's weight, so the field is the weight over " \18             "the current times the length."19end

playground.rb

1require_relative "../solutions/a_wire_held_up_by_a_field"23play_with :a_wire_held_up_by_a_field do4  title "A Wire Held Up by a Field"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	MagneticForce
elementary charge	$e$	C	—	MagneticForce
gravity	$g$	m/s²	—	MagneticForce
magnetic field	$B$	T	—	MagneticForce
mass	$m$	kg	0 … ∞	MagneticForce
charge	$q$	C	—	MagneticForce
speed	$v$	m/s	—	MagneticForce
current	$I$	A	—	MagneticForce
length in the field	$l$	m	—	MagneticForce
force on a wire	$F$	N	—	MagneticForce
force per length	$f$	N/m	—	MagneticForce
radius of the path	$r$	m	—	MagneticForce
angle to the field	$θ$	—	0 … 3.142	MagneticForce
frequency of revolution	$ν$	Hz	—	MagneticForce
force on a wire on ground	$F_{g}$	N	—
current of ground	$I_{g}$	A	—
length in the field of ground	$l_{g}$	m	—
force on a wire on wire	$F$	N	—
current of wire	$I$	A	—
length in the field of wire	$l$	m	—
force per length on ground	$f_{g}$	N/m	—
angle to the field of ground	$θ_{g}$	—	0 … 3.142
force per length on wire	$f$	N/m	—
angle to the field of wire	$θ$	—	0 … 3.142
radius of the path of ground	$r_{g}$	m	—
mass of ground	$m_{g}$	kg	0 … ∞
speed of ground	$v_{g}$	m/s	—
charge of ground	$q_{g}$	C	—
radius of the path of wire	$r$	m	—
mass of wire	$m$	kg	0 … ∞
speed of wire	$v$	m/s	—
charge of wire	$q$	C	—
frequency of revolution of ground	$ν_{g}$	Hz	—
frequency of revolution of wire	$ν$	Hz	—

F_{w} = I_{w} \cdot l_{w} \cdot B

force on a wire across a field

f_{w} = I_{w} \cdot B \cdot \sin θ_{w}

force on each metre of wire

r_{c} = \frac{m_{c} \cdot v_{c}}{q_{c} \cdot B}

a charge circles in a field

v_{c} = 2 \cdot π \cdot r_{c} \cdot ν_{c}

once round the circle

F_{g} = I_{g} \cdot l_{g} \cdot B

force on a wire across a field(ground)

F = I \cdot l \cdot B

force on a wire across a field(wire)

f_{g} = I_{g} \cdot B \cdot \sin θ_{g}

force on each metre of wire(ground)

f = I \cdot B \cdot \sin θ

force on each metre of wire(wire)

r_{g} = \frac{m_{g} \cdot v_{g}}{q_{g} \cdot B}

a charge circles in a field(ground)

r = \frac{m \cdot v}{q \cdot B}

a charge circles in a field(wire)

v_{g} = 2 \cdot π \cdot r_{g} \cdot ν_{g}

once round the circle(ground)

v = 2 \cdot π \cdot r \cdot ν

once round the circle(wire)

F = m \cdot g

suspended in mid-air