Figure 31N-2a shows a circuit consisting of an ideal battery with emf xi = 6.00

Question

Figure 31N-2a shows a circuit consisting of an ideal battery with emf xi = 6.00 mu V. a resistance R. and a small wire loop of area 5.0 cm2. For the time interval t = 10 to t = 20 s, an external magnetic field is set up throughout the loop. The field is uniform, its direction is into the page in Fig. 31N-2a, and the field magnitude is given by B = at, where B is in teslas, a is a constant, and t is in seconds. Figure 31N-2b gives the current i in the circuit before, during, and after the external field is set up. Find a.

Explanation / Answer

Given emf of the battery is , = 6.0 V Area of the loop , A = 5.0 cm2 Time interval , t = 10 to t = 20 s Magnitude of the field , B = at From the figure , at t = 0, i = 1.5 mA = 0.0015 A Since i = Vbattery / R            R = 6 *10-6 V / 0.0015 A            R = 0.004 The value of current during 10 to 20 s is 5mA = 0.0005A    (Vbattery + induced) /R = 0.0005 A    induced   = ( 0.0005A * 0.004 ) - 6 *10-6 V = -4 V ------------------------------------------------------------------ The induces emf is , = - dB/ dt = - d/dt(BA)                                = - A a           [where B = at , dB /dt = a]                                a = - /A = - (- 4 *10-6 V) / (5 *10-4 m2)                                a = 0.008 T /s            Time interval , t = 10 to t = 20 s Magnitude of the field , B = at From the figure , at t = 0, i = 1.5 mA = 0.0015 A Since i = Vbattery / R            R = 6 *10-6 V / 0.0015 A            R = 0.004 The value of current during 10 to 20 s is 5mA = 0.0005A    (Vbattery + induced) /R = 0.0005 A    induced   = ( 0.0005A * 0.004 ) - 6 *10-6 V = -4 V ------------------------------------------------------------------ The induces emf is , = - dB/ dt = - d/dt(BA)                                = - A a           [where B = at , dB /dt = a]                                a = - /A = - (- 4 *10-6 V) / (5 *10-4 m2)                                a = 0.008 T /s           

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