1Please solve it fully , it ' ll be better that way - By the way , this is not one of my doubts .
Last question of old NCERT : Chapter - EMI
A line charge " λ " per unit length is lodged uniformly onto the rim of a wheel of mass " M " and radius " R " . The wheel has light non - conducting spokes and is free to rotate without friction about its axis . A uniform magnetic field extends over a circular region within the rim . It is given by -
B = - B 0 K , for " r ≤ a < R " ;
= 0 , otherwise ;
What is the angular velocity of the wheel after the field is suddenly switched off ?
PS - Physics Experts are heartily requested to refrain from solving this problem unless there are no correct answers within a week : )
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3 Answers
11First , we should realise that the torque developed on the ring is due to the induced electric field , not the magnetic field , as it is reduced to zero in an instant .
Having said this , let us find out that electric field .
∫ E . dl = dφBdt
Or , E . 2 π R = d [ B ( t ) π a 2 ]dt = B ' ( t ) π a 2
Hence , E = B ' ( t ) a 22 R
Now , the total torque developed due to this electric field -
Γ ( t ) = ( λ . 2 π R ) . E . R = λ B ' ( t ) π a 2 R
Since the angular momentum of the system remains conserved at all times , hence -
I ω = 0 ∫ t Γ ( t ) dt = λ ( Δ B ) π a 2 R
Or , ω = λ ( Δ B ) π a 2M R
Since , Δ B = B0 - 0 = B0
So , ω = - λ B 0 π a 2M R K
One possible doubt - Why did I use " B ( t ) " , while it is given that " B " is constant ? While the magnetic field is being reduced , the field will obviously depend on time , otherwise how can it be reduced to zero ?