A Wire Held Up by a Field

Q. A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field. Taking g as 9.8 m/s², what is the magnitude of the magnetic field, in teslas? expects 0.65 T

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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?

The playground

Inputs

mass

m

200 g

length

l

1.50 m

current

I

2 A

gravity

g

9.8 m/s²

Outputs

field

B

The solution

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 Ruby

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

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

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

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