A Rod Turning in a Ring

solved MotionalEmf

Q. A metallic rod of 1 m $R = 1$ m length is rotated with a frequency of 50 rev/s $ω = 50$ rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field of 1 T $B = 1$ T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring, in volts?

Inputs

length

R

1.00 m

rate

ω

50 rev/s

field

B

1 T

Outputs

emf

ε

How it's solved

Solved with MotionalEmf

Given

$R = 1$ m
$ω = 50$ rev/s
$B = 1$ T

Find $ε$ .

Working

$ω = 50$ rev/s $= 314.159265$ rad/s
emf of a rod turning about one end(rod) $ε = \frac{B \cdot ω \cdot R^{2}}{2}$
$ε = \frac{1 \cdot 314.159265 \cdot 1^{2}}{2}$
$ε = 157.1$ V
$ε \approx 157$ Vto 3 significant figures

Conclusion

Each bit of the rod cuts the field at its own speed, faster further out; added along the rod they give half the field times the angular speed times the length squared.

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_turning_in_a_ring do4  kind :numerical5  describes "A metallic rod of %{length} length is rotated with a frequency of %{rate}, with one end hinged at the " \6            "centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis " \7            "passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic " \8            "field of %{field} parallel to the axis is present everywhere. %{emf}, in volts?"9  given length: "1 m"10  given rate: "50 rev/s"11  given field: "1 T"12  asks emf: "What is the emf between the centre and the metallic ring"13end

solution.rb

1require_relative "../questions/a_rod_turning_in_a_ring"2require_relative "../science/electromagnetic_induction/motional_emf"34solve :a_rod_turning_in_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 "Each bit of the rod cuts the field at its own speed, faster further out; added along the rod " \15             "they give half the field times the angular speed times the length squared."16end

playground.rb

1require_relative "../solutions/a_rod_turning_in_a_ring"23play_with :a_rod_turning_in_a_ring do4  title "A Rod Turning in 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)