1. Suppose we have a uniform magnetic field, B, directed into the plane of the d

Question

1. Suppose we have a uniform magnetic field, B, directed into the plane of the diagram below, whose magnitude varies sinusoidally as indicated. Immersed in this magnetic field is the circuit shown, which consists of a circular loop of radius r, connected to a resistor, R, through a pair of wires that are close together, and essentially out of the region of magnetic field.

a) Derive an expression for the ratio of the amplitude of theinduced magnetic field, compared to . In other words,what is: (BMAX)/B0

b) Now let’s put in some “typical” numbers: Suppose:r = 1 cm.R = 1000 ohms.f = 60 Hz. (typical power line frequency)

c) Since the (varying) induced magnetic field would also produce an induced emf, can weignore this effect in the calculation for induced emf?

d) Answer question c) if the frequency were 6 MHz (6 106 Hz),a typical radio frequency.

Explanation / Answer

      Given:
Magnetic field has the form = B = Bo sinwt
circular loop radious = r
thus,Area of the loop = A= r2
(a)

   It is known by the formual , for
induced emf in the circuit is :
emf = - e = - dB /dt
                   = -d (BA) / dt
                   = A d/dt (B)
                   = - A w Bo coswt
   If R be the resistence connected
   this, induced current = I = e /  R
                                     = (-A wBo cos wt ) /(R )
        thus,induced current : I = (A w Bo /R ) cos wt
    I = r2 (w Bo /R ) cos wt
         (this is in the function in the time )
        Thus, Magnitude of the Magnetic field due this induced
   currrent circulates in the loop given by the formual as ;
   Induced Magnetic field :   B = o I / 2 r
           B =   o I / 2 r          
                  = (  o r2 w B o / 2 r R )   cos wt    

                  = (  o r w Bo / 2 R )   cos wt           
    Maximum induced Magnitude :
      BMAX = (  o A w Bo / 2 r R )   cos wt   
    If time depending term is 1
     i.e coswt = 1
   then ,       BMAX / Bo = (  o r w / 2 R )                   so,          BMAX / Bo = (  o r w / 2 R )                               =  (  o r 2 f / 2 R )                              = (  o 2 r f /   R )     (b)

    Given that :       radious r   = 1 cm = 0.01 m       Resistence : R = 1000 ohms        Frquency    = f = 60 Hz         thus, BMAX / Bo = (  o 2 r f /   R )                                       = 4 x10-7 (3.14)2 (0.01) (60) / (1000)                                        = 0.0743                                        = 0.0743                          ( c )    Yes         (d )     Question wasn't clearly , plz post seperately      
  

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