A Rod Spun against a Ring

solved MotionalEmf

Q. A 1.0 m $R = 1.0$ m long metallic rod is rotated with an angular frequency of 400 rad/s $ω = 400$ rad/s about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T $B = 0.5$ T parallel to the axis exists everywhere. What is the emf developed between the centre and the ring, in volts?

Inputs

length

R

1.00 m

rate

ω

400 rad/s

field

B

0.5 T

Outputs

emf

ε

How it's solved

Solved with MotionalEmf

Given

$R = 1.0$ m
$ω = 400$ rad/s
$B = 0.5$ T

Find $ε$ .

Working

emf of a rod turning about one end(rod) $ε = \frac{B \cdot ω \cdot R^{2}}{2}$
$ε = \frac{0.5 \cdot 400 \cdot 1^{2}}{2}$
$ε = 100$ V
$ε \approx 100$ Vto 3 significant figures

Conclusion

The far end of the rod sweeps the field fastest and the hinged end not at all, so the rod's emf is the field times the average speed of its points times its length.

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_rod_spun_against_a_ring do4  kind :numerical5  describes "A %{length} long metallic rod is rotated with an angular frequency of %{rate} about an axis normal " \6            "to the rod passing through its one end. The other end of the rod is in contact with a circular " \7            "metallic ring. A constant and uniform magnetic field of %{field} parallel to the axis exists " \8            "everywhere. %{emf}, in volts?"9  given length: "1.0 m"10  given rate: "400 rad/s"11  given field: "0.5 T"12  asks emf: "What is the emf developed between the centre and the ring"13end

solution.rb

1require_relative "../questions/a_rod_spun_against_a_ring"2require_relative "../science/electromagnetic_induction/motional_emf"34solve :a_rod_spun_against_a_ring, using: MotionalEmf do5  objects :rod67  take length: rod.length_of_the_rod8  take rate: rod.angular_speed9  take field: B1011  emf = rod.induced_emf12  answer emf:, in: "V", figures: 31314  conclusion "The far end of the rod sweeps the field fastest and the hinged end not at all, so the rod's " \15             "emf is the field times the average speed of its points times its length."16end

playground.rb

1require_relative "../solutions/a_rod_spun_against_a_ring"23play_with :a_rod_spun_against_a_ring do4  title "A Rod Spun against a Ring"5end

The laws it drew on

quantity	written	units	within	from
magnetic field	$B$	T	—	MotionalEmf
length of the rod	$R$	m	0 … ∞	MotionalEmf
angular speed	$ω$	rad/s	—	MotionalEmf
induced emf	$ε$	V	—	MotionalEmf
induced emf of ground	$ε_{g}$	V	—
angular speed of ground	$ω_{g}$	rad/s	—
length of the rod of ground	$R_{g}$	m	0 … ∞
induced emf of rod	$ε$	V	—
angular speed of rod	$ω$	rad/s	—
length of the rod of rod	$R$	m	0 … ∞

ε_{r} = \frac{B \cdot ω_{r} \cdot {R_{r}}^{2}}{2}

emf of a rod turning about one end

ε_{g} = \frac{B \cdot ω_{g} \cdot {R_{g}}^{2}}{2}

emf of a rod turning about one end(ground)

ε = \frac{B \cdot ω \cdot R^{2}}{2}

emf of a rod turning about one end(rod)