familiari nf3795 Induced Currents yeazell 58395 1 This print out should have 11 questions Multiple choice questions may continue on the next column or page nd all choices before answering 001 10 0 points A circular coil is made of N turns of copper wire as shown in the gure When viewed is wound counter from the right the coil clockwise A resistor R is inserted in the copper wire Initially a uniform magnetic eld of magnitude Bi points horizontally from left to right through the perpendicular plane of the coil clockwise as the wire goes from the right to left terminals The current must enter the loop from the right terminal and exit at the left terminal Since the current is continuous the current must ow through the resistor in the left to right direction 002 part 1 of 2 10 0 points Consider the circuit shown R P E R L S R Magnetic Field B t During a time interval t the eld uniformly changes at a constant rate until a reversed eld is reached equal in magnitude to the initial eld The current in the resistor R 1 is zero 2 ows from right to left 3 ows in a direction that cannot be deter mined from the information given 4 ows from left to right correct Explanation As the left to right magnetic eld decreases and eventually ipping sign and increasing in magnitude it follows from Lenz s law op position to the change in magnetic eld will tend to keep the current constant and ow ing in the same direction that the induced emf will produce a left to right magnetic eld arising from induced currents in the coil By the right hand rule the induced current ows counter clockwise when viewed from the right and the coils are wound counter What is the instantaneous current through the upper resistor and what is the instanta neous current at point P immediately after the switch is closed 4 IR 0 IP 0 0 correct 1 IR 0 0 IP 0 2 IR 0 IP 0 3 IR 0 0 IP 0 0 5 IR 0 IP 0 6 IR 0 0 IP 0 7 IR 0 0 IP 0 8 IR 0 IP 0 E R E R E R E R E 2 R E R E 2 R E R L E R E R L Explanation The current in L has to change gradually so immediately after the switch is closed there is no current going through point P 003 part 2 of 2 10 0 points familiari nf3795 Induced Currents yeazell 58395 2 When the switch has been closed for a long time what is the energy stored in the induc tor 1 UL 2 UL 3 UL 4 UL 5 UL 6 UL 7 UL 8 UL 9 UL L E 8 R L E 3 R L R2 2 E 2 L E 16 R L E 32 R L E 2 4 R2 L E 2 R L E 4 R L E 2 2 R2 correct E 2 R2 4 L 10 UL Explanation After the switch has been closed for a long time the current in L does not change any more so there is no voltage increase nor de crease across L Therefore the current going through L is I stored in L as which gives the energy E R 1 2 UL L I 2 L E 2 2 R2 004 part 1 of 4 10 0 points Consider the circuit The switch is closed at t 0 The current I through the inductor takes the form I 1 e t x E Rx cid 16 cid 17 where Rx and x are to be determined L R2 R1 I I2 E S I1 Find I immediately after the circuit is closed E R1 E 1 I 2 I R1 R2 3 I 0 correct 4 I E R2 Explanation Before the circuit is closed no current is owing When we have just closed the circuit we are at t 0 a mathematical nota tion meaning a very short time after t 0 Nothing happens in the circuit at t 0 only immediately after when the switch is indeed closed However this is just a mathematical detail There are two loops in the prob lem one with E R1 R2 and one with E R1 L So at t 0 the battery wants to drive a current through both loops The rst loop presents no problem since there is no in ductance working against us a current will immediately be set up The second loop how ever has an inductor which tries to prevent any change in the current going through it and so goes up smoothly from I 0 as can be seen in the given solution just put t 0 to nd I 0 Therefore at this instant the inductor L carries no current and we can ne glect it when we nd the current through R2 The equivalent resistance is Req R1 R2 so from E Req I we nd E I2 R1 R2 familiari nf3795 Induced Currents yeazell 58395 3 005 part 2 of 4 10 0 points Find I2 immediately after the circuit is closed 006 part 3 of 4 10 0 points Find I after the circuit is closed for a long time so the time constant is 1 I2 E R2 2 I2 0 3 I2 correct E R1 R2 E 4 I2 R1 Explanation See previous explanation 1 I 0 E R2 E 2 I 3 I R1 R2 E R1 4 I correct Explanation When we wait a long time the current I 0 though the inductor levels out i e The voltage over the inductor is d I dt VL L d I dt at all times so VL 0 as t and we can replace the inductor in the gure by a straight wire When the current now comes from R1 to the junction where I1 splits into I2 and I there is no resistance in the I path but a nonzero resistance R2 in the other Naturally the current takes the path with no resistance Since we do pass through R1 in any case the equivalent resistance is now R1 At t I and I2 0 E R1 Alternate Solution An equation of the form L dI dt R I E 0 has a solution which is composed of a ho mogeneous solution and a particular solution The particular solution corresponds to the rst term in the I in the problem statement the homogeneous corresponds to the second term The homogeneous solution is therefore the only one of relevance to the time con stant x and we can disregard any batteries E This provides a simple solution to Part 3 Just ignore the battery and nd the equiva lent resistance connected to L in this case 1 Req 1 R1 1 R2 L L Req 1 R1 1 R2 cid 18 cid 19 as before Nobody seems to have proven that ignoring the batteries always works so the previous solution is more …