A Bar Magnet Twisted by a Field

solved CurrentLoopInAField

Q. A short bar magnet placed with its axis at 30° $θ = 30 °$ with a uniform external magnetic field of 0.25 T $B = 0.25$ T experiences a torque of magnitude equal to 4.5 × 10⁻² J $τ = 4.5 \times 10^{−2}$ J. What is the magnitude of magnetic moment of the magnet, in J/T?

Inputs

angle

θ

30.00°

field

B

0.25 T

torque

τ

0.045 J

Outputs

moment

m

How it's solved

Solved with CurrentLoopInAField

Given

$θ = 30 °$
$B = 0.25$ T
$τ = 4.5 \times 10^{−2}$ J

Find $m$ .

Working

torque on a coil(magnet) $τ = m \cdot B \cdot \sin θ$
$0.045 = m \cdot 0.25 \cdot \sin 30 °$
$m = 0.36$ J/T
$m = 0.36$ m² A $= 0.36$ J/T
$m \approx 0.36$ J/Tto 2 significant figures

Conclusion

A field twists a magnet towards itself with a torque that grows with the magnet's moment and with the sine of the angle between them, so a measured torque at a known angle gives the moment.

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_bar_magnet_twisted_by_a_field do4  kind :numerical5  describes "A short bar magnet placed with its axis at %{angle} with a uniform external magnetic field of " \6            "%{field} experiences a torque of magnitude equal to %{torque}. %{moment}, in J/T?"7  given angle: "30 deg"8  given field: "0.25 T"9  given torque: "4.5e-2 J"10  asks moment: "What is the magnitude of magnetic moment of the magnet"11end

solution.rb

1require_relative "../questions/a_bar_magnet_twisted_by_a_field"2require_relative "../science/moving_charges_and_magnetism/current_loop_in_a_field"34solve :a_bar_magnet_twisted_by_a_field, using: CurrentLoopInAField do5  objects :magnet67  take angle: magnet.angle_to_the_field8  take field: B9  take torque: magnet.torque1011  moment = magnet.magnetic_moment12  answer moment:, in: "J/T", figures: 21314  conclusion "A field twists a magnet towards itself with a torque that grows with the magnet's moment and with " \15             "the sine of the angle between them, so a measured torque at a known angle gives the moment."16end

playground.rb

1require_relative "../solutions/a_bar_magnet_twisted_by_a_field"23play_with :a_bar_magnet_twisted_by_a_field do4  title "A Bar Magnet Twisted 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 magnet	$m$	m² A	—
turns of magnet	$N$	—	—
current of magnet	$I$	A	—
radius of magnet	$R$	m	0 … ∞
area of ground	$A_{g}$	m²	—
area of magnet	$A$	m²	—
torque on ground	$τ_{g}$	J	—
angle to the field of ground	$θ_{g}$	—	0 … 3.142
torque on magnet	$τ$	J	—
angle to the field of magnet	$θ$	—	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(magnet)

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(magnet)

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(magnet)