A Bar Magnet Turned across a Field

Q. A bar magnet of magnetic moment 1.5 J/T lies aligned with the direction of a uniform magnetic field of 0.22 T. An external torque turns the magnet so as to align its magnetic moment normal to the field direction. What is the torque on the magnet in this position, in N m? expects 0.33 N m

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Q. A bar magnet of magnetic moment 1.5 J/T $m = 1.5$ J/T lies aligned with the direction of a uniform magnetic field of 0.22 T $B = 0.22$ T. An external torque turns the magnet so as to align its magnetic moment normal to the field direction $θ = 90 °$ . What is the torque on the magnet in this position, in N m?

The playground

Inputs

moment

m

1.5 J/T

field

B

0.22 T

Outputs

torque

τ

The solution

Solved with CurrentLoopInAField

Given

$m = 1.5$ J/T
$B = 0.22$ T
$θ = 90 °$ from “normal to the field direction”

Find $τ$ .

Working

$m = 1.5$ J/T $= 1.5$ m² A
torque on a coil(magnet) $τ = m \cdot B \cdot \sin θ$
$τ = 1.5 \cdot 0.22 \cdot \sin 90 °$
$τ = 0.33$ N m
$τ = 0.33$ J $= 0.33$ N m
$τ \approx 0.33$ N mto 2 significant figures

Conclusion

Square across the field the magnet feels the largest twist the field can give it, the moment times the field.

The Ruby

1require_relative "../physics"23question :a_bar_magnet_turned_across_a_field do4  kind :numerical5  describes "A bar magnet of magnetic moment %{moment} lies aligned with the direction of a uniform magnetic " \6            "field of %{field}. An external torque turns the magnet so as to align its magnetic moment normal to " \7            "the field direction. %{torque}, in N m?"8  given moment: "1.5 J/T"9  given field: "0.22 T"10  asks torque: "What is the torque on the magnet in this position"11end

1require_relative "../questions/a_bar_magnet_turned_across_a_field"2require_relative "../science/moving_charges_and_magnetism/current_loop_in_a_field"34solve :a_bar_magnet_turned_across_a_field, using: CurrentLoopInAField do5  objects :magnet67  take moment: magnet.magnetic_moment8  take field: B910  deduce_from "normal to the field direction", magnet.angle_to_the_field => "90 deg"1112  torque = magnet.torque13  answer torque:, in: "N m", figures: 21415  conclusion "Square across the field the magnet feels the largest twist the field can give it, the moment " \16             "times the field."17end

1require_relative "../solutions/a_bar_magnet_turned_across_a_field"23play_with :a_bar_magnet_turned_across_a_field do4  title "A Bar Magnet Turned across 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)