A Wire Inside a Solenoid

solved MagneticForce

Q. A 3.0 cm $l = 3.0$ cm wire carrying a current of 10 A $I = 10$ A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T $B = 0.27$ T. What is the magnetic force on the wire, in newtons?

Inputs

length

l

3 cm

current

I

10 A

field

B

0.27 T

Outputs

force

F

How it's solved

Solved with MagneticForce

Given

$l = 3.0$ cm
$I = 10$ A
$B = 0.27$ T

Find $F$ .

Working

$l = 3.0$ cm $= 0.03$ m
force on a wire across a field(wire) $F = I \cdot l \cdot B$
$F = 10 \cdot 0.03 \cdot 0.27$
$F = 0.081$ N
$F \approx 0.081$ Nto 2 significant figures

Conclusion

The field inside a solenoid runs along its axis, so a wire across the axis stands square across the field and feels the full force.

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_inside_a_solenoid do4  kind :numerical5  describes "A %{length} wire carrying a current of %{current} is placed inside a solenoid perpendicular to its " \6            "axis. The magnetic field inside the solenoid is given to be %{field}. %{force}, in newtons?"7  given length: "3.0 cm"8  given current: "10 A"9  given field: "0.27 T"10  asks force: "What is the magnetic force on the wire"11end

solution.rb

1require_relative "../questions/a_wire_inside_a_solenoid"2require_relative "../science/moving_charges_and_magnetism/magnetic_force"34solve :a_wire_inside_a_solenoid, using: MagneticForce do5  objects :wire67  take length: wire.length_in_the_field8  take current: wire.current9  take field: B1011  force = wire.force_on_a_wire12  answer force:, in: "N", figures: 21314  conclusion "The field inside a solenoid runs along its axis, so a wire across the axis stands square across " \15             "the field and feels the full force."16end

playground.rb

1require_relative "../solutions/a_wire_inside_a_solenoid"23play_with :a_wire_inside_a_solenoid do4  title "A Wire Inside a Solenoid"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)