In an AC generator the area vector of a flat coil of area A and N turns makes an

Question

In an AC generator the area vector of a flat coil of area A and N turns makes an angle of theta with respect to a uniform magnetic field directed to the right. (See Figure 3.) The coil is turned counterclockwise with a constant angular speed omega so that theta = omega f. Write down an expression for the magnetic flux phi n through the coil and then use Faraday's law to find an expression for the emf generated by the coil as a function of time. Suppose that the coil is connected to a purely resistive load R (that includes the resistances of the coil, contacts and connecting wires). Use your result from part a to write a formula for the instantaneous current passing through the coil. (This formula will Involve a function of time.) Find a formula for the magnitude of the magnetic dipole moment mu of the coil and then the magnitude of the torque x that must be applied to the coil to keep it rotating with a constant angular speed. Use your result from part c and the result P= tau omega from Physics 140 to find the instantaneous rate at which work is being done to rotate the coil. Find a formula for the instantaneous power dissipated by the load resistance and show that it is the same as your answer to part d above.

Explanation / Answer

Coil of Area = A No.of turns of the coil = N Applied magnetic field = B Ciol makes an angle = with respect to field Constant angular speed of the coil = W = /t (a) Acc. to farady's law for electro magnetic induction ,    we have , Magnetic flux linked with thw coil = B    B = d(BNA) = BN A cos Induced emf in the coil = e = -dB /dt     e = -d(B NA cos ) /dt = -BAN d/dt(cos)         = -BA Nd/dt[cos(Wt) ]      e   =- BANW (-sinWt)   thus, induced emf in the coil is :    e = BAWN sinWt (It is in the function of time) (b)    If R be the resistence of the Coil then ,cuttrrent passing through the coil is : current = I = e / R                  I =( BAWN/R ) sin(Wt) (c)      Magnetic dipole moment :       = I A             = ( BAW/R ) sin(Wt))(A)       =( BA2 WN /R ) sin(Wt)     Torque acting on the coil :       = N I A B sin        = NIAB sin(Wt) (d)      Rate at which work is done :             P = d /dt               = d[NAIB sin(Wt)]     (by using the result form : c)                = N AIB d/dt[sinWt]                = N AB [ BAWN/R ] d/dt[sin(Wt)] (sinWt)                 (As I = =( BAWN/R ) sin(Wt)   )        P =( NABA )2 W /R   d/dt [sin2 (Wt )]             =( NABA )2 W /R d/dt [1- cos 2Wt/2]           P  =    [( NABA W)2 / R ] sin(2Wt) (e)       But , power : P = e2 /R        we can also get the same expression as part (D)                                        

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