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A coil of 15 turns and radius 10.0 cm surrounds a long solenoid of radius 2.30 cm and 1.00 x 10^3 turns / meter (see figure below). The current in the solenoid changes as I = 8.00 sin 120 t, where I is in amperes and t is in seconds. 
Find the induced emf (in volts) in the 15-turn coil as a function of time.
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Added Mon, 15 Jun '15
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The field inside the Solenoid :
 B = u_0 * n * l

Therefore, the flux through the large coil :
Phi_b = Integral-of ( B * dA) 
          = B ( N * Pi * r^2) 
          = N * Pi * r^2 * u_0 * n * I

The induced EMF thus is,
E = - dPhi_b  / dt
    = - Pi * u_0 * n * N * r^2 * dI / dt.
    = Pi * (4Pi * 10^-7) * (1.00 * 10^3) * 15 * (2.30 * 10^-2m)^2 * (120) * (8cos120t)
    = 3.00 * 10^-2 * cos(t * 120 rad/s) V
     = 3.00 * 10^-2 * cos(t * 120 rad/s) V * (1000 mV / 1V) 
    = 3.00 * cos(t * 120 rad/s) mV 

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