A Coil Turned by a Field

solved CurrentLoopInAField

Q. A 100 $N = 100$ turn closely wound circular coil of radius 10 cm $R = 10$ cm carries a current of 3.2 A $I = 3.2$ A. It is free to turn in a uniform magnetic field of 2 T $B = 2$ T, and has turned until its axis is at 90° $θ = 90 °$ to the field. What is the magnitude of the torque on the coil, in N m?

Inputs

turns

N

100

radius

R

10 cm

current

I

3.2 A

field

B

2 T

angle

θ

90.00°

Outputs

torque

τ

How it's solved

Solved with CurrentLoopInAField

Given

$N = 100$
$R = 10$ cm
$I = 3.2$ A
$B = 2$ T
$θ = 90 °$

Find $τ$ .

Working

$R = 10$ cm $= 0.1$ m
moment of a round coil(coil) $m = N \cdot I \cdot π \cdot R^{2}$
$m = 100 \cdot 3.2 \cdot 3.14159265 \cdot {0.1}^{2}$
$m = 10.0530965$ m² A
torque on a coil(coil) $τ = m \cdot B \cdot \sin θ$
$τ = 10.0530965 \cdot 2 \cdot \sin 90 °$
$τ = 20.11$ N m
$τ = 20.106193$ J $= 20.106193$ N m
$τ \approx 20$ N mto 2 significant figures

Conclusion

The coil is a magnet of moment turns times current times area; the field twists it hardest when its axis stands across the field.

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_coil_turned_by_a_field do4  kind :numerical5  describes "A %{turns} turn closely wound circular coil of radius %{radius} carries a current of %{current}. It is " \6            "free to turn in a uniform magnetic field of %{field}, and has turned until its axis is at %{angle} to " \7            "the field. %{torque}, in N m?"8  given turns: "100"9  given radius: "10 cm"10  given current: "3.2 A"11  given field: "2 T"12  given angle: "90 deg"13  asks torque: "What is the magnitude of the torque on the coil"14end

solution.rb

1require_relative "../questions/a_coil_turned_by_a_field"2require_relative "../science/moving_charges_and_magnetism/current_loop_in_a_field"34solve :a_coil_turned_by_a_field, using: CurrentLoopInAField do5  objects :coil67  take turns: coil.turns8  take radius: coil.radius9  take current: coil.current10  take field: B11  take angle: coil.angle_to_the_field1213  torque = coil.torque14  answer torque:, in: "N m", figures: 21516  conclusion "The coil is a magnet of moment turns times current times area; the field twists it hardest " \17             "when its axis stands across the field."18end

playground.rb

1require_relative "../solutions/a_coil_turned_by_a_field"23play_with :a_coil_turned_by_a_field do4  title "A Coil Turned by a Field"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	CurrentLoopInAField
magnetic field	$B$	T	—	CurrentLoopInAField
current	$I$	A	—	CurrentLoopInAField
turns	$N$	—	—	CurrentLoopInAField
radius	$R$	m	0 … ∞	CurrentLoopInAField
area	$A$	m²	—	CurrentLoopInAField
magnetic moment	$m$	m² A	—	CurrentLoopInAField
angle to the field	$θ$	—	0 … 3.142	CurrentLoopInAField
torque	$τ$	J	—	CurrentLoopInAField
magnetic moment of ground	$m_{g}$	m² A	—
turns of ground	$N_{g}$	—	—
current of ground	$I_{g}$	A	—
radius of ground	$R_{g}$	m	0 … ∞
magnetic moment of coil	$m$	m² A	—
turns of coil	$N$	—	—
current of coil	$I$	A	—
radius of coil	$R$	m	0 … ∞
area of ground	$A_{g}$	m²	—
area of coil	$A$	m²	—
torque on ground	$τ_{g}$	J	—
angle to the field of ground	$θ_{g}$	—	0 … 3.142
torque on coil	$τ$	J	—
angle to the field of coil	$θ$	—	0 … 3.142

m_{c} = N_{c} \cdot I_{c} \cdot π \cdot {R_{c}}^{2}

moment of a round coil

m_{c} = N_{c} \cdot I_{c} \cdot A_{c}

moment of a flat coil

τ_{c} = m_{c} \cdot B \cdot \sin θ_{c}

torque on a coil

m_{g} = N_{g} \cdot I_{g} \cdot π \cdot {R_{g}}^{2}

moment of a round coil(ground)

m = N \cdot I \cdot π \cdot R^{2}

moment of a round coil(coil)

m_{g} = N_{g} \cdot I_{g} \cdot A_{g}

moment of a flat coil(ground)

m = N \cdot I \cdot A

moment of a flat coil(coil)

τ_{g} = m_{g} \cdot B \cdot \sin θ_{g}

torque on a coil(ground)

τ = m \cdot B \cdot \sin θ

torque on a coil(coil)