Dear student Emf induced in the half length of the wire is E=Bω(l2)22=Bωl28current in the resistor i=ER=Bωl28RCurrent in half length of the diameter is =i2=Bωl216RForce on half of the length of the wire isF=B×i2×l2=Bl4×Bωl28R=B2l3ω32RTorque on the wire isT=F×l2=B2l3ω32R×l2I×α=B2l4ω64RMl212×α=B2l4ω64Rα=12M×B2l2ω64R=3B2l4ω16MR-dωdt=3B2l2ω16MR∫ωoω-dωω=3B2l216MR∫0tdtln(ωω0)=3B2l2t16MRln(ωω0)=e-3B2l2t16MRω=ω0e-3B2l2t16MRa) current in the circuit isi=Bωl28R=Bl28R(ω0e-3B2l2t16MR)b)-dqdt=Bl28R(ω0e-3B2l2t16MR)dq=Bl28R(ω0e-3B2l2t16MR)dtq=∫0tBl28R(ω0e-3B2l2t16MR)dt=Bl2ω08R∫0te-3B2l2t16MRdtq=-Bl2ω08R×16MR3B2l2(e-3B2l2t16MR)∞0=Bl2ω08R×16MR3B2l2=2ω0M3BFor long questions kindly use different thread for different parts of questionRegards

Solve this:

Solve this: Cl Assignment-Motional emf Date: 20-07-2017 A conducting rod PQ of mass M rotates without friction on a horizontal plane about O as circular rails of diameter The centre O and the periphery are connected by resistance R. The system is located in a uniform magnetic field perpendicular to the plane of the loop. At t = O, PQ starts rotating clockwtse with angular velocity Neglecting the resistance the rails and rod, as well as self inductance, find. irection of current and Its magnitude as a function of time ) Total charge flown through the resistance. c) The heat generated in the circuit by t = cs. R p o final po• A connector AtA2 of mass m and length t is being moved on semi circular rails With a uniform angular velocity (0. The

Dear student

E m f i n d u c e d i n t h e h a l f l e n g t h o f t h e w i r e i s E = \frac{B ω (\frac{l}{2})^{2}}{2} = \frac{B ω l^{2}}{8} c u r r e n t i n t h e r e s i s t o r i = \frac{E}{R} = \frac{B ω l^{2}}{8 R} C u r r e n t i n h a l f l e n g t h o f t h e d i a m e t e r i s = \frac{i}{2} = \frac{B ω l^{2}}{16 R} F o r c e o n h a l f o f t h e l e n g t h o f t h e w i r e i s F = B \times \frac{i}{2} \times \frac{l}{2} = \frac{B l}{4} \times \frac{B ω l^{2}}{8 R} = \frac{B^{2} l^{3} ω}{32 R} T o r q u e o n t h e w i r e i s T = F \times \frac{l}{2} = \frac{B^{2} l^{3} ω}{32 R} \times \frac{l}{2} I \times α = \frac{B^{2} l^{4} ω}{64 R} \frac{M l^{2}}{12} \times α = \frac{B^{2} l^{4} ω}{64 R} α = \frac{12}{M} \times \frac{B^{2} l^{2} ω}{64 R} = \frac{3 B^{2} l^{4} ω}{16 M R} \frac{- d ω}{d t} = \frac{3 B^{2} l^{2} ω}{16 M R} \int_{ω_{o}}^{ω} \frac{- d ω}{ω} = \frac{3 B^{2} l^{2}}{16 M R} \int_{0}^{t} d t \ln (\frac{ω}{ω_{0}}) = \frac{3 B^{2} l^{2} t}{16 M R} \ln (\frac{ω}{ω_{0}}) = e^{- \frac{3 B^{2} l^{2} t}{16 M R}} ω = ω_{0} e^{- \frac{3 B^{2} l^{2} t}{16 M R}} a) c u r r e n t i n t h e c i r c u i t i s i = \frac{B ω l^{2}}{8 R} = \frac{B l^{2}}{8 R} (ω_{0} e^{- \frac{3 B^{2} l^{2} t}{16 M R}}) b) - \frac{d q}{d t} = \frac{B l^{2}}{8 R} (ω_{0} e^{- \frac{3 B^{2} l^{2} t}{16 M R}}) d q = \frac{B l^{2}}{8 R} (ω_{0} e^{- \frac{3 B^{2} l^{2} t}{16 M R}}) d t q = \int_{0}^{t} \frac{B l^{2}}{8 R} (ω_{0} e^{- \frac{3 B^{2} l^{2} t}{16 M R}}) d t = \frac{B l^{2} ω_{0}}{8 R} \int_{0}^{t} e^{- \frac{3 B^{2} l^{2} t}{16 M R}} d t q = - \frac{B l^{2} ω_{0}}{8 R} \times \frac{16 M R}{3 B^{2} l^{2}} (e^{- \frac{3 B^{2} l^{2} t}{16 M R}} {)^{\infty}}_{0} = \frac{B l^{2} ω_{0}}{8 R} \times \frac{16 M R}{3 B^{2} l^{2}} = \frac{2 ω_{0} M}{3 B} F o r l o n g q u e s t i o n s k i n d l y u s e d i f f e r e n t t h r e a d f o r d i f f e r e n t p a r t s o f q u e s t i o n R e g a r d s

View Full Answer