A Coil Held Against a Field

solved CurrentLoopInAField

Q. A circular coil of 30 $N = 30$ turns and radius 8.0 cm $R = 8.0$ cm carrying a current of 6.0 A $I = 6.0$ A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T $B = 1.0$ T. The field lines make an angle of 60° $θ = 60 °$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning, in N m?

Inputs

turns

N

radius

R

8 cm

current

I

6 A

field

B

1 T

angle

θ

60.00°

Outputs

torque

τ

How it's solved

Solved with CurrentLoopInAField

Given

$N = 30$
$R = 8.0$ cm
$I = 6.0$ A
$B = 1.0$ T
$θ = 60 °$

Find $τ$ .

Working

$R = 8.0$ cm $= 0.08$ m
moment of a round coil(coil) $m = N \cdot I \cdot π \cdot R^{2}$
$m = 30 \cdot 6 \cdot 3.14159265 \cdot {0.08}^{2}$
$m = 3.61911474$ m² A
torque on a coil(coil) $τ = m \cdot B \cdot \sin θ$
$τ = 3.61911474 \cdot 1 \cdot \sin 60 °$
$τ = 3.134$ N m
$τ = 3.1342453$ J $= 3.1342453$ N m
$τ \approx 3.1$ N mto 2 significant figures

Conclusion

To hold the coil still, the counter torque must match the turning the field gives it, which goes with the sine of the angle between the coil's normal 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_coil_held_against_a_field do4  kind :numerical5  describes "A circular coil of %{turns} turns and radius %{radius} carrying a current of %{current} is suspended " \6            "vertically in a uniform horizontal magnetic field of magnitude %{field}. The field lines make an angle " \7            "of %{angle} with the normal of the coil. %{torque}, in N m?"8  given turns: "30"9  given radius: "8.0 cm"10  given current: "6.0 A"11  given field: "1.0 T"12  given angle: "60 deg"13  asks torque: "Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning"14end

solution.rb

1require_relative "../questions/a_coil_held_against_a_field"2require_relative "../science/moving_charges_and_magnetism/current_loop_in_a_field"34solve :a_coil_held_against_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 "To hold the coil still, the counter torque must match the turning the field gives it, which goes " \17             "with the sine of the angle between the coil's normal and the field."18end

playground.rb

1require_relative "../solutions/a_coil_held_against_a_field"23play_with :a_coil_held_against_a_field do4  title "A Coil Held Against 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)