Determine the magnetic field (magnitude and direction) B at 1.5 cm from the comm

Question

Determine the magnetic field (magnitude and direction) B at 1.5 cm from the common center. Answer: 2.97 mT. 2.44 mT. 6.54 T. 1.97 mT. Determine only the magnitude of the magnetic field B at 27 cm from the common center. Answer: 7.58 Times 10^(-2)T. 5.38 Times 10^(-5)T. 1.91 Times 10^(-4) T. 1.91 Times 10^(-5) T. Fig.1 (FA-9) shows a metal rod A-C that is forced to move with constant velocity v along two parallel metal rails connected with a strip of metal D-E at one end. A magnitude B = 8.00 mT points in the direction shown perpendicular to the plane of A-C-E-D. If the rails are separated by a distance of 25.0 cm and the speed of the rod is 4.80 cm/s, what is the emf that is generated as the rod moves? Answer: 12.9 mV 2.44 V. 6.54 mV. 10.8 mV. 3.87 mV. What is the current in the rod if it has a resistance of 0.025 milli-Ohm? Answer: 2.97 mA. 4.44 A. 8.54 mA. 1.97 A. 9.87 mA. What is the rate that energy is being transformed in the rod? Answer: 2.58 Times 10^(-2)J. 1.38 Times 10^(-5)W. 8.91 Times 10^(-4)W. 3.91 Times 10^(-5)J. 3.80 Times 10^(-3)W. Use to represent the direction of the direction of the currents shown in the figure and to represent the opposite directions. Let epsilon = 45.0 V, R_1 = 47.0 kilo-Ohm, R_3 - 88.0 kilo -Ohm and l + 15.0 micro Henry. At time t = 0. i.e., immediately after switch S is closed, What is the value of current i_1? 2.97 mA. 2.44 mA. 6.54 mA. 1.97 mA. At time t = 0. i.e., immediately after switch S is closed, what is the value of current i_2? 0.97 mA. 0.44 mA. 1.54 mA. 0.97 mA. For t = infinity (i.e., a long time after the switch is closed) i_1 = 1.97 mA. 1.44 mA. 8.54 mA. 2.97 mA. 5.87 mA. A long time after switch S has been closed it is suddenly opened. Answer this and the following problem. Immediately as the switch is opened, what is the value of i_1? 0.97 mA. 0.44 mA. 4.54 mA. 9.97 mA. 6.87 mA. Immediately as the switch is opened, what the value of i_2? 9.97 mA. 2.94 mA. 7.97 mA. 3.87 mA. An oscillating LC circuit consists of an ideal 150 mH inductao and an ideal 4800 pf capacitor. If the maximum charge on the capacitor is 3.80 Times 10^-15 C, what is the total energy in the circuit? Answer: 50 Times 10^(-21)J. 2.38 Times 10^(-15)W. 8.11 Times 10^(-14)W. 7.91 Times 10^(-15)J. 3.20 Times 10^(-13)J. maximum current in the circuit. Answer: 197 pA. 654 pA. 197 nA. 110 pA.

Explanation / Answer

B = 0.008 T
EMF = d(phi)/dt = d(BA)/dt = BdA/dt = Blv = 0.008*0.25*0.048 = 0.096 mV

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