Electromagnetically induced.

A conducting rod is bent in the shape of a semicircle of radius R , R, and the straight parts along the ends of the diameter of the semicircle are passed through fixed, smooth conducting rings O O and O . O'. A capacitor having capacitance C C is connected to the rings with the help of ideal wires. The system is placed in a uniform magnetic field of strength B B such that the axis of rotation O O OO' is perpendicular to the direction of the magnetic field.

Initially, the plane of the semicircle was normal to the direction of the field. The semicircle is now rotated about the axis O O OO' with a constant angular velocity of ω \omega . Neglect any resistance and self-inductance of the circuit.

The power required to keep the semicircle rotating with a constant angular velocity of ω \omega can be expressed as P = 1 a π b ω c R d C e B f sin ( 2 ω t ) , P = \dfrac{1}{a} \pi^b \omega^c R^d C^e B^f \sin(2\omega t), where letters from a a to f f are positive integers (not necessarily coprime).

Find a + b + c + d + e + f a+b+c+d+e+f .


The answer is 20.

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1 solution

Steven Chase
May 6, 2018

Loop area:

A = A 0 + π R 2 2 c o s ( ω t ) A = A_0 + \frac{\pi R^2}{2} cos(\omega t)

Flux Linkage:

λ = B A = B A 0 + π B R 2 2 c o s ( ω t ) \lambda = B A = B A_0 + \frac{\pi B R^2}{2} cos(\omega t)

Voltage across capacitor (assuming no resistance or self-inductance in circuit):

v C = λ ˙ = π ω B R 2 2 s i n ( ω t ) v_C = \dot{\lambda} = - \frac{\pi \omega B R^2}{2} sin(\omega t)

Current through capacitor:

i C = C v C ˙ = π C ω 2 B R 2 2 c o s ( ω t ) i_C = C \, \dot{v_C} = - \frac{\pi C \omega^2 B R^2}{2} cos(\omega t)

Power associated with capacitor:

P C = v C i C = π 2 ω 3 C B 2 R 4 4 s i n ( ω t ) c o s ( ω t ) = π 2 ω 3 C B 2 R 4 8 s i n ( 2 ω t ) P_C = v_C \, i_C = \frac{\pi^2 \omega^3 C B^2 R^4 }{4} sin(\omega t) \, cos(\omega t) \\ = \frac{\pi^2 \omega^3 C B^2 R^4 }{8} sin(2 \omega t)

The parameters therefore sum to 20 20

Short, simple and very well presented! Very good solution sir!

Tapas Mazumdar - 3 years, 1 month ago

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Thanks very much. Fun problem too

Steven Chase - 3 years, 1 month ago

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Sir what if it had Resistances also? Can u make a qs on that?

Md Zuhair - 2 years, 8 months ago

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@Md Zuhair Yeah, that sounds interesting. @Tapas Mazumdar , is it alright with you if I borrow your graphics for a follow-up problem?

Steven Chase - 2 years, 8 months ago

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@Steven Chase Tapas Mazumdar is studying in college now. Every Indian students after getting into college forgets Brilliant .org and doesn't open it. I Guess it won't be a problem if you borrow his graphic . :P

Md Zuhair - 2 years, 8 months ago

@Steven Chase Yes. By all means feel free to use this image. Btw, @Md Zuhair is right. I don't get sufficient time to open Brilliant nowadays.

Tapas Mazumdar - 2 years, 8 months ago

@Steven Chase If you post the problem. Plz let me know.

Md Zuhair - 2 years, 8 months ago

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@Md Zuhair I think I'll post it within 12 hours. I'll let you know

Steven Chase - 2 years, 8 months ago

@Md Zuhair It's up now https://brilliant.org/problems/crank-generator-power/

Steven Chase - 2 years, 8 months ago

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