Question

a conducting loop is painted around the equator of a spherical rubber balloon.a magnetic field B=0.2cos4t is applied perpendicular to the plane of the equator.the balloon is contracting inwardly with a radial velocity v.when the balloon radius is r=0.5m,the rms voltage induced in the loop is 500mV.find the velocity v at this instant?
answer:0.52m/s

Answer 1

E.M.F = -D/Dt( Flux )
Now, Magnetic Flux F = B.A , both B & A vary with time
Here B = BoCos(wt) where Bo = 0.2, w = 4
also, dr/dt = -v , radially inwards

Hence,
V = -D/Dt( B.A) = -d/dt(A* d(B)/Dt + B * dA/dt)
= -D/Dt( BoCostwt * pi * r^2)
(we get 2 terms on differentiating)
= ( -wBoSin(wt). pi*r^2 - BoCoswt*2*pi*r*v)

V = pi*r*Bo (2v Coswt + wr Sinwt)
lets convert this to Cos(A-B) form using CosACosB + SinASinB formula

V = \[ \pi*r*Bo \sqrt{(rw)^2 + (2v)^2} * \cos (wt - \delta)\]

where \[\tan \delta = (r*w/2v)\]

Thus we get our induced voltage,
\[V = Vo \cos (wt - \delta)\]

where

\[Vo = \pi*r*Bo*\sqrt{(r*w)^2 + (2*v)^2}\]

Now, in our question, V(R.M.S) is given = 0.5 Volt
\[Vrms = Vo/\sqrt{2}\]

Thefore -
\[0.5 = \pi*r*Bo*\sqrt{(r*w)^2 + (2*v)^2}/\sqrt{2}\]

solve for v, everything else is known Bo = 0.2, r = 0.5 , w = 4

we get v = 0.5m/s

Answer 2

I have also attached this file which has my original hand written solution. I couldn't write it properly above using the equations editor on this page

Answer 3

i found it by other formula
since in this pblm we have motional emf
we take r=vt.
the formula for emf is
e=v*b*l
e=v*b*2*pi*r(since r=vt)
e=v^2*b*2*pi*t
from this
velocity,v=sqrt(e/2*pi*b*t)------(1)
omega,w=2*pi*f=4(given)
f=2/pi;
t=pi/2=1.57s
e=.5v,b=.2 T,
substituting these values in(1)
we get v=0.51m/s.
anyways,your approach was also very nice

Answer 4

Hey @ananta, thats a nice shortcut you thought of..
one thing I'm wondering about though.. you calculated t(time elapsed until now when radius is 0.5 and velocity = v ) by equating it to the time period of the magnetic field function i.e. t = 1/f = Time period. What is your reasoning for that ?
How would you use this formula, if say, the given radius and consequent EMF were given for a different instant, which would mean the time elapsed would be different and you wouldn't be able to equate it to Magentic Field function's time period ?