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Electricity and Magnetism Level 4

A uniform metallic wire is bent in the form of a parabola and is placed on a horizontal non conducting floor. A vertical uniform magnetic induction B B exists in the region containing the parabolic wire. A straight conducting rod starting from rest at the vertex of the parabola at time t = 0 t = 0 , slides along the parabolic wire with its length perpendicular to the axis of symmetry of the parabola.Take the equation of the parabola to be y = k x 2 y = kx^2 where k k is a constant. Consider that rod always touches wire .The rod moves with a constant speed v v and assume that the rod has resistance λ \lambda per unit lengthb and the parabolic wire has no resistance.Obtain the power P P needed to keep the rod moving with this constant speed v v .

P = a B 2 b λ v c t d k \displaystyle P = \dfrac{aB^2}{b\lambda} \sqrt{\dfrac{v^{c}t^{d}}{k}} for positive constants a , b , c , d a,b,c,d where a , b a,b are in simplest form.Submit your answer as a + b + c + d a+b+c+d .


The answer is 9.

Ayon Ghosh by Ayon Ghosh , 17, India

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1 solution

Ayon Ghosh
Mar 19, 2018

INPhO 2014 P-1.Official solutions (not answers ) unavailable so welcome to my own unreliable solution :P

After time t t rod covers a distance v t vt along parabola.Hence using the equation of parabola co ordinates of end points of parabola are ( − v t k , v t ) ; ( + v t k , v t ) ( - \sqrt{\dfrac{vt}{k}},vt) ; ( +\sqrt{\dfrac{vt}{k}},vt) .The length of the parabola cum rod is now 2 v t k 2\sqrt{\dfrac{vt}{k}} and height simply v t vt .Using integration or simply otherwise area of parabola comes as A ( t ) = 4 3 v 3 t 3 k A(t) = \dfrac{4}{3} \sqrt{\dfrac{v^3t^3}{k}} .Clearly this area is increasing w.r.t time.We can differentiate it to get magnitude of changing flux.

Now simply we use Faraday Law to link the induced EMF with changing flux.Yields E = d ϕ d t = 2 B v 3 t k E =\dfrac{d\phi}{dt} = 2B \sqrt{ \dfrac{v^3t}{k} } . Now resistance will be R = 2 λ v t k R = 2\lambda\sqrt{\dfrac{vt}{k}} .Power provided is simply E 2 R \dfrac{E^2}{R} (Why ? Because all power provided by the external agent appears as heat in the resistor as there is no change in K.E of rod) hence: P = 2 B 2 λ v 5 t k \displaystyle \boxed{P= \dfrac{2B^2}{\lambda} \sqrt{\dfrac{v^{5}t}{k}}}

Edit (21/3/18) : I just discovered an almost same problem in Irodov.Yay !!

1 Helpful 1 Interesting 0 Brilliant 0 Confused

Nice solution.

Akshat Sharda - 3 years ago

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