A Bar Magnet Twisted by a Field

Q. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. What is the magnitude of magnetic moment of the magnet, in J/T? expects 0.36 J/T

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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?

The playground

Inputs

angle

θ

30.00°

field

B

0.25 T

torque

τ

0.045 J

Outputs

moment

m

The solution

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 Ruby

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

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

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

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)