38.Electromagnetic induction

Electromagnetic Induction 50. a) When the speed of wire is V emf developed = B  V

Bv (from b to a) R c) Down ward acceleration of the wire b) Induced current is the wire =

=

mg  F due to the current m

B2 2 V Rm d) Let the wire start moving with constant velocity. Then acceleration = 0 = mg - i  B/m = g –

B 2 2 v mg Rm gRm  Vm  2 2 B  dV e) a dt  



dV mg  B2 2 v / R  dt m dv mg  B2 2 v / R m v mdv



0

mg 

2 2

B  v R

 dt



t

 dt 0

v

2 2    log(mg  B  v  = t  R   B2 2  0 R

m



 mR B2 2

   B2 2 v   loglog mg   log(mg) = t   R    

 B2 2 v  2 2  mg   R    tB   log mg mR      B2 2 v   tB2 2  log1   Rmg  mR 

B2 2 v  1 e Rmg  (1  e B v=

2 2

 / mR

 tB 2  2 mR

)

B 2 2 v Rmg

2 2 Rmg  1  e B  / mR  2 2   B 

 v = v m (1  e  gt / Vm )

Rmg   v m  2 2  B    38.12

a

b