A magnetic field passes through a stationary wire loop, andits magnitude changes

Question

A magnetic field passes through a stationary wire loop, andits magnitude changes in time according to the graph in thedrawing. The direction of the field remains constant, however.There are three equal time intervals indicated in the graph:0–3.0 s, 3.0–6.0 s, and 6.0–9.0 s. The loopconsists of 50 turns of wire and has an area of 0.15 m2.The magnetic field is oriented parallel to the normal to the loop.For purposes of this problem, this means that = 0° inthe equation: = BA cos

(a) For each interval, determine the induced emf.

Explanation / Answer

a. = B*A*Cos = 0° = B*A The induced electromotive force is given by =-N*(d/dt). is the voltage. N is the number of turns of wire. d/dt is the flux. The negative sign appears in the equation, since the voltagewill try to oppose the change in flux (Lenz's Law). N = 50   (given) d/dt = (d/dt)(B*A) Since the area is constant, the derivative is simple. d/dt = (d/dt)(B*A) = A*(dB/dt) This can be written in another form: A*(B/t) This new form allows us to easily see the connection of thechange in flux (d/dt) with the slope of the graph(B/t). Now, = -N*A*(B/t) =-50*(0.15m2)*(B/t) For the first section of the graph, (B/t) = (0.4T- 0T)/(3s - 0s) = 2/15 T/s = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(2/15 T/s) = -1 T/(m2s) = -1V For the second section of the graph (B/t) = (0.4T- 0.4T)/(6s - 3s) = 0 T/s = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(0 T/s) = 0 T/(m2s)= 0 V (This result shows that for a constant magnetic field, theinduced electromotive force is 0 V). For the third section of the graph (B/t) = (0.2T -0.4T)/(9s - 6s) = -1/15 T/s = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(-1/15 T/s) = 0 T/(m2s)= 0.5 V *Note: The signs on the answers indicate direction ofthe potential (voltage) - positive is counterclockwise andnegative is clockwise. b. R = 0.5 V = IR = IR I = /R For the first section, I = (-1 V)/(0.5) = -2 A For the third section, I = (0.5 V)/(0.5) = 1 A The positive sign indicates a current moving counterclockwise,and the negative sign indicates a current moving clockwise. = B*A The induced electromotive force is given by =-N*(d/dt). is the voltage. N is the number of turns of wire. d/dt is the flux. The negative sign appears in the equation, since the voltagewill try to oppose the change in flux (Lenz's Law). N = 50   (given) d/dt = (d/dt)(B*A) Since the area is constant, the derivative is simple. d/dt = (d/dt)(B*A) = A*(dB/dt) This can be written in another form: A*(B/t) This new form allows us to easily see the connection of thechange in flux (d/dt) with the slope of the graph(B/t). Now, = -N*A*(B/t) =-50*(0.15m2)*(B/t) For the first section of the graph, (B/t) = (0.4T- 0T)/(3s - 0s) = 2/15 T/s = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(2/15 T/s) = -1 T/(m2s) = -1V For the second section of the graph (B/t) = (0.4T- 0.4T)/(6s - 3s) = 0 T/s = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(0 T/s) = 0 T/(m2s)= 0 V (This result shows that for a constant magnetic field, theinduced electromotive force is 0 V). For the third section of the graph (B/t) = (0.2T -0.4T)/(9s - 6s) = -1/15 T/s = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(-1/15 T/s) = 0 T/(m2s)= 0.5 V *Note: The signs on the answers indicate direction ofthe potential (voltage) - positive is counterclockwise andnegative is clockwise. b. R = 0.5 V = IR = IR I = /R For the first section, I = (-1 V)/(0.5) = -2 A For the third section, I = (0.5 V)/(0.5) = 1 A The positive sign indicates a current moving counterclockwise,and the negative sign indicates a current moving clockwise. = B*A The induced electromotive force is given by =-N*(d/dt). is the voltage. N is the number of turns of wire. d/dt is the flux. The negative sign appears in the equation, since the voltagewill try to oppose the change in flux (Lenz's Law). N = 50   (given) d/dt = (d/dt)(B*A) Since the area is constant, the derivative is simple. d/dt = (d/dt)(B*A) = A*(dB/dt) This can be written in another form: A*(B/t) This new form allows us to easily see the connection of thechange in flux (d/dt) with the slope of the graph(B/t). Now, = -N*A*(B/t) =-50*(0.15m2)*(B/t) For the first section of the graph, (B/t) = (0.4T- 0T)/(3s - 0s) = 2/15 T/s = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(2/15 T/s) = -1 T/(m2s) = -1V For the second section of the graph (B/t) = (0.4T- 0.4T)/(6s - 3s) = 0 T/s = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(0 T/s) = 0 T/(m2s)= 0 V = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(0 T/s) = 0 T/(m2s)= 0 V (This result shows that for a constant magnetic field, theinduced electromotive force is 0 V). For the third section of the graph (B/t) = (0.2T -0.4T)/(9s - 6s) = -1/15 T/s = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(-1/15 T/s) = 0 T/(m2s)= 0.5 V *Note: The signs on the answers indicate direction ofthe potential (voltage) - positive is counterclockwise andnegative is clockwise. b. R = 0.5 V = IR = IR I = /R For the first section, I = (-1 V)/(0.5) = -2 A For the third section, I = (0.5 V)/(0.5) = 1 A The positive sign indicates a current moving counterclockwise,and the negative sign indicates a current moving clockwise. = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(-1/15 T/s) = 0 T/(m2s)= 0.5 V *Note: The signs on the answers indicate direction ofthe potential (voltage) - positive is counterclockwise andnegative is clockwise. b. R = 0.5 V = IR = IR I = /R For the first section, I = (-1 V)/(0.5) = -2 A For the third section, I = (0.5 V)/(0.5) = 1 A The positive sign indicates a current moving counterclockwise,and the negative sign indicates a current moving clockwise. = -50*(0.15m2)*(B/t) =-50*(0.15m2)*(-1/15 T/s) = 0 T/(m2s)= 0.5 V *Note: The signs on the answers indicate direction ofthe potential (voltage) - positive is counterclockwise andnegative is clockwise. b. R = 0.5 V = IR = IR I = /R For the first section, I = (-1 V)/(0.5) = -2 A For the third section, I = (0.5 V)/(0.5) = 1 A The positive sign indicates a current moving counterclockwise,and the negative sign indicates a current moving clockwise.

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