A Wire at Thirty Degrees to a Field

solved MagneticForce

Q. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A $I = 8$ A and making an angle of 30° $θ = 30 °$ with the direction of a uniform magnetic field of 0.15 T $B = 0.15$ T, in N/m?

Inputs

current

I

8 A

angle

θ

30.00°

field

B

0.15 T

Outputs

force

f

How it's solved

Solved with MagneticForce

Given

$I = 8$ A
$θ = 30 °$
$B = 0.15$ T

Find $f$ .

Working

force on each metre of wire(wire) $f = I \cdot B \cdot \sin θ$
$f = 8 \cdot 0.15 \cdot \sin 30 °$
$f = 0.6$ N/m
$f \approx 0.6$ N/mto 1 significant figure

Conclusion

Only the part of the field across the wire pushes on its current, so the force goes with the sine of the angle between them.

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_wire_at_thirty_degrees_to_a_field do4  kind :numerical5  describes "%{force} on a wire carrying a current of %{current} and making an angle of %{angle} with the direction " \6            "of a uniform magnetic field of %{field}, in N/m?"7  given current: "8 A"8  given angle: "30 deg"9  given field: "0.15 T"10  asks force: "What is the magnitude of magnetic force per unit length"11end

solution.rb

1require_relative "../questions/a_wire_at_thirty_degrees_to_a_field"2require_relative "../science/moving_charges_and_magnetism/magnetic_force"34solve :a_wire_at_thirty_degrees_to_a_field, using: MagneticForce do5  objects :wire67  take current: wire.current8  take angle: wire.angle_to_the_field9  take field: B1011  force = wire.force_per_length12  answer force:, in: "N/m", figures: 11314  conclusion "Only the part of the field across the wire pushes on its current, so the force goes with the sine " \15             "of the angle between them."16end

playground.rb

1require_relative "../solutions/a_wire_at_thirty_degrees_to_a_field"23play_with :a_wire_at_thirty_degrees_to_a_field do4  title "A Wire at Thirty Degrees to a Field"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	MagneticForce
elementary charge	$e$	C	—	MagneticForce
gravity	$g$	m/s²	—	MagneticForce
magnetic field	$B$	T	—	MagneticForce
mass	$m$	kg	0 … ∞	MagneticForce
charge	$q$	C	—	MagneticForce
speed	$v$	m/s	—	MagneticForce
current	$I$	A	—	MagneticForce
length in the field	$l$	m	—	MagneticForce
force on a wire	$F$	N	—	MagneticForce
force per length	$f$	N/m	—	MagneticForce
radius of the path	$r$	m	—	MagneticForce
angle to the field	$θ$	—	0 … 3.142	MagneticForce
frequency of revolution	$ν$	Hz	—	MagneticForce
force on a wire on ground	$F_{g}$	N	—
current of ground	$I_{g}$	A	—
length in the field of ground	$l_{g}$	m	—
force on a wire on wire	$F$	N	—
current of wire	$I$	A	—
length in the field of wire	$l$	m	—
force per length on ground	$f_{g}$	N/m	—
angle to the field of ground	$θ_{g}$	—	0 … 3.142
force per length on wire	$f$	N/m	—
angle to the field of wire	$θ$	—	0 … 3.142
radius of the path of ground	$r_{g}$	m	—
mass of ground	$m_{g}$	kg	0 … ∞
speed of ground	$v_{g}$	m/s	—
charge of ground	$q_{g}$	C	—
radius of the path of wire	$r$	m	—
mass of wire	$m$	kg	0 … ∞
speed of wire	$v$	m/s	—
charge of wire	$q$	C	—
frequency of revolution of ground	$ν_{g}$	Hz	—
frequency of revolution of wire	$ν$	Hz	—

F_{w} = I_{w} \cdot l_{w} \cdot B

force on a wire across a field

f_{w} = I_{w} \cdot B \cdot \sin θ_{w}

force on each metre of wire

r_{c} = \frac{m_{c} \cdot v_{c}}{q_{c} \cdot B}

a charge circles in a field

v_{c} = 2 \cdot π \cdot r_{c} \cdot ν_{c}

once round the circle

F_{g} = I_{g} \cdot l_{g} \cdot B

force on a wire across a field(ground)

F = I \cdot l \cdot B

force on a wire across a field(wire)

f_{g} = I_{g} \cdot B \cdot \sin θ_{g}

force on each metre of wire(ground)

f = I \cdot B \cdot \sin θ

force on each metre of wire(wire)

r_{g} = \frac{m_{g} \cdot v_{g}}{q_{g} \cdot B}

a charge circles in a field(ground)

r = \frac{m \cdot v}{q \cdot B}

a charge circles in a field(wire)

v_{g} = 2 \cdot π \cdot r_{g} \cdot ν_{g}

once round the circle(ground)

v = 2 \cdot π \cdot r \cdot ν

once round the circle(wire)