A closed coil having 100 turns is rotated in a uniform magnetic field B=4.0 x10-4 T about a diameter which is perpendicular to the field. the angular velocity of rotation is 300 revolutions/min.The area of the coil is 25 cm2 and its resistance is 4.0 Ω. Find (a) the average emf devoloped in half a turn from a position where the coil is perpendicular to the magnetic field,(b) the average emf in a full turn and (c) the net charge displaced in part (a).

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Answers

2014-12-23T17:55:00+05:30


Average emf= (2NBA) /t= 2 x 100 x (4.0 x 10^-4) x (25 x 10^-4) / {1/600 x 60} = 2 x 10^-3 V

In full turn, net change of flux will be 0 so average emf will be zero.

Induced Charge= 2NBA/ R = 2 x 100 x (4 x 10^-4) x (25 x 10^-4)/ 4 = 5 x 10^-5 C

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2014-12-23T23:27:36+05:30

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Φ = magnetic flux
B = magnetic field strength  = 4 * 10^-4 Tesla
A = area of circular coil  = 25 cm^2 = 0.0025 m^2
ω = angular velocity = 2 π f = 10 π  rad/sec
f = frequency (rotations per second ) of the coil rotation = 300/60 = 5 /sec
t = time
N = number of turns in the coil = 100
R = resistance in the circuit of the coil = 4 ohms

     Φ = B X A = B * A Cos θ = B * A * Cos ω t = B * A * Cos 2πf t

     ε = - N dΦ/dt = N B A (2πf) Sin 2πf t

  Emf induced in the coil is = ε = 100 * 4 * 10^-4 * 0.0025 * (10π) Sin (10π t) volts
                      ε = π Sin (10π t) milli volts.  --- equation 1
                  this is the instantaneous emf induced.

a)
   time period for one cycle of rotation = T = 1/f = 1/5 sec = 0.2 sec
   time period for 1/2 cycle = 0.1 sec

   time average of a sine wave over half cycle of t = 0 sec to 0.1 sec can be obtained by integrating  Sin (10πt) dt  and dividing by 0.1 sec will be 1/0.1 * 1/(10π) *  [- Cos 10π * 0.1 + Cos 10π * 0]  = 2/π.
   time average (of sin t)  wave  over half cycle is 2/π.  by integration. 

Now using the expression from equation 1, the average emf over time over half cycle
         = π * (2/π) =  2 milli volts.
  
b)  average emf in time for a full cycle will be zero as the sign of the induced emf will change in the second half cycle.  It will cancel the average emf of the first cycle.

c)
         charge = current * time = 2 milli volts / 4 ohms * 0.1 secs
                   = 50 micro coulombs.



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