A Square Coil in a Field

solved CurrentLoopInAField

Q. A square coil of side 10 cm $A = 100$ cm² consists of 20 $N = 20$ turns and carries a current of 12 A $I = 12$ A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° $θ = 30 °$ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T $B = 0.80$ T. What is the magnitude of torque experienced by the coil, in N m?

Inputs

turns

N

current

I

12 A

angle

θ

30.00°

field

B

0.8 T

Outputs

torque

τ

How it's solved

Solved with CurrentLoopInAField

Given

$N = 20$
$I = 12$ A
$θ = 30 °$
$B = 0.80$ T
$A = 100$ cm² from “A square coil of side 10 cm”

Find $τ$ .

Working

$A = 100$ cm² $= 0.01$ m²
moment of a flat coil(coil) $m = N \cdot I \cdot A$
$m = 20 \cdot 12 \cdot 0.01$
$m = 2.4$ m² A
torque on a coil(coil) $τ = m \cdot B \cdot \sin θ$
$τ = 2.4 \cdot 0.8 \cdot \sin 30 °$
$τ = 0.96$ N m
$τ = 0.96$ J $= 0.96$ N m
$τ \approx 0.96$ N mto 2 significant figures

Conclusion

A coil's magnetic moment is its turns times its current times its area, whatever its shape, and the field turns it with the sine of the angle between the moment and 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_square_coil_in_a_field do4  kind :numerical5  describes "A square coil of side 10 cm consists of %{turns} turns and carries a current of %{current}. The coil " \6            "is suspended vertically and the normal to the plane of the coil makes an angle of %{angle} with the " \7            "direction of a uniform horizontal magnetic field of magnitude %{field}. %{torque}, in N m?"8  given turns: "20"9  given current: "12 A"10  given angle: "30 deg"11  given field: "0.80 T"12  asks torque: "What is the magnitude of torque experienced by the coil"13end

solution.rb

1require_relative "../questions/a_square_coil_in_a_field"2require_relative "../science/moving_charges_and_magnetism/current_loop_in_a_field"34solve :a_square_coil_in_a_field, using: CurrentLoopInAField do5  objects :coil67  take turns: coil.turns8  take current: coil.current9  take angle: coil.angle_to_the_field10  take field: B1112  deduce_from "A square coil of side 10 cm", coil.area => "100 cm2"1314  torque = coil.torque15  answer torque:, in: "N m", figures: 21617  conclusion "A coil's magnetic moment is its turns times its current times its area, whatever its shape, and " \18             "the field turns it with the sine of the angle between the moment and the field."19end

playground.rb

1require_relative "../solutions/a_square_coil_in_a_field"23play_with :a_square_coil_in_a_field do4  title "A Square Coil in 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)