EMI 3c: Circular Loops within a Solenoid ................................................................... 2 EMI3c—RT1: Circular Loop Within a Solenoid ............................................................................................. 3 EMI3c—WBT1: Circular Loop Within a Solenoid .......................................................................................... 4 EMI3c—WWT1: Circular Loop Within a Solenoid ......................................................................................... 5 EMI3c—TT1: Circular Loop Within a Solenoid.............................................................................................. 6 EMI3c—CCT1: Circular Loop Within a Solenoid........................................................................................... 7 EMI3c—CRT1: Circular Loop Within a Solenoid........................................................................................... 8 EMI3c—PET1: Circular Loop Within a Solenoid ........................................................................................... 9 EMI3c—QRT1: Circular Loop Within a Solenoid......................................................................................... 10 EMI3c—M/MCT1: Circular Loop Within a Solenoid .................................................................................... 11 1 of 11 EMI3c key 6_08_02 EMI 3C: CIRCULAR LOOPS WITHIN A SOLENOID 2 of 11 EMI3c key 6_08_02 EMI3C—RT1: CIRCULAR LOOP WITHIN A SOLENOID The figures below show six situations where circular loops of wire are placed inside six solenoids. The solenoids, which have uniform fields in their interiors, are concentric with the loops but have larger diameters. The circular loops vary in diameter, which of course means they have different areas. We are told the areas for the loops as well as the INDUCED current in each loop as a result of a change in the strength of the field within the solenoid. (Assume all loops have the same resistance.) Rank these situations, from greatest to least, on the basis of the change in current per unit time in the solenoids that produced the induced currents in the circular loops. A B C A = 24 cm2 A = 24 cm2 A = 32 cm2 I = 8 mA I = 6 mA I = 8 mA D E F A = 16 cm2 A = 32 cm2 A = 16 cm2 I = 8 mA I = 6 mA I = 4 mA Greatest 1 ___D__ 2 ___A__ 3 _B C F_ 4 ______ 5 ______ 6 ___E___ Least OR, The change in current per unit time is the same for ALL SIX solenoids. _______ OR, There is no change in current in any of the solenoids. _______ Please carefully explain your reasoning. In each case, the area of the loop is fixed; thus, the change of flux is due to a change in the magnetic field. The change in the magnetic field is due to a change in current in the solenoid. Since we are giving the induced currents, the change in current per unit time in the solenoid is the induced current divided by the area or Iloop = Emfinduced dφ d ( B ⋅ Aloop ) dB di di ∝ Emfinduced = = = Aloop ⋅ = Aloop ⋅ µ0 ⋅ nsolenoid ⋅ solenoid ∝ Aloop ⋅ solenoid Rloop dt dt dt dt dt I disolenoid ∝ loop since the B = µ0 ⋅ n ⋅ isolenoid dt Aloop How sure were you of your ranking? (circle one) Basically Guessed Sure 1 2 3 4 5 6 7 3 of 11 EMI3c key 8 9 Very Sure 10 6_08_02 EMI3C—WBT1: CIRCULAR LOOP WITHIN A SOLENOID Construct a physical situation to which the following relation could apply. 12 mA = turns )( I2 − I1 ) 1 cm (π )( 4 cm)2 3 msec 6Ω ( µo )(800 Comparing it to the equation for the current induced in a loop placed in within the magnetic field of a solenoid that is changing the currentIloop = ∆i Emfinduced = Aloop ⋅ µ0 ⋅ nsolenoid ⋅ solenoid ∆t Rloop Rloop = µ0 ⋅ nsolenoid ⋅ ∆isolenoid 2 ⋅π ⋅ rloop ⋅ Rloop ∆t We see that one physical situation that could be constructed is to place a circular loop with radius of 4 cm and a resistance of 6 ohms inside a solenoid such that the centers are concentric and aligned parallel. The solenoid has 800 turns per centimeter and the current is changing from I1 to I2 in 3 milliseconds. 4 of 11 EMI3c key 6_08_02 EMI3C—WWT1: CIRCULAR LOOP WITHIN A SOLENOID What, if anything, is wrong with the following situation? A circular wire loop that is inside, and concentric with, a solenoid will have an induced current if the current in the solenoid changes. Looks ok. 5 of 11 EMI3c key 6_08_02 EMI3C—TT1: CIRCULAR LOOP WITHIN A SOLENOID Something is wrong with the situation described below. Identify the problem and explain how to correct it. A circular wire loop that is inside, and concentric with, a solenoid has an induced current in it. Consequently, we know that the current in the solenoid must be increasing. This could be correct, but the changing current could also be decreasing. 6 of 11 EMI3c key 6_08_02 EMI3C—CCT1: CIRCULAR LOOP WITHIN A SOLENOID The statements below are about a situation where a circular wire loop is inside and concentric with a large solenoid. Student A: “If the current in the solenoid changes, there will be an induced current in the wire loop because the magnetic field will change.” Student B: “Changing the current in the solenoid will not produce an induced current since the magnetic field will still be uniform.” Student C: “Changing the current in the solenoid will have no effect on any current in the wire if they are not connected together.” With which, if any, of these students do you agree? Student A __X__ Student B ______ Student C_______ Disagree with all_______ Carefully explain your reasoning. The magnetic field of the solenoid produces a magnetic flux through the loop. If the current in the solenoid changes, this changes the magnetic flux through the loop. This magnetic flux change produces an induced emf that produces a current in the loop. 7 of 11 EMI3c key 6_08_02 EMI3C—CRT1: CIRCULAR LOOP WITHIN A SOLENOID The graph below shows how the current in a large solenoid of length L, diameter D, and n turns per unit length changes during a certain time interval. Write the specific expression(s) for the induced current in a circular wire loop of resistance R and diameter d, which is inside the solenoid, concentric with it, and at the center of the solenoid. Current (A) I3 I2 I1 0 Time (s) 0 to t1: (assuming that that the entire loop is within the solenoid or d<D) 0 Iloop = t1 t2 t3 Emfinduced I −I d2 ∆i 2 = µ0 ⋅ nsolenoid ⋅ solenoid ⋅π ⋅ rloop ⋅ Rloop == µ0 ⋅ nsolenoid ⋅ 2 1 ⋅π ⋅ ⋅ Rloop Rloop t1 − 0 ∆t 4 t1 to t2: Iloop = Emfinduced = 0 since there is no change in current in the solenoid Rloop t2 to t3: (assuming that that the entire loop is within the solenoid or d<D) Iloop = Emfinduced I −I d2 ∆i 2 = µ0 ⋅ nsolenoid ⋅ solenoid ⋅π ⋅ rloop ⋅ Rloop == µ0 ⋅ nsolenoid ⋅ 3 2 ⋅π ⋅ ⋅ Rloop Rloop t3 − t2 ∆t 4 8 of 11 EMI3c key 6_08_02 EMI3C—PET1: CIRCULAR LOOP WITHIN A SOLENOID A solenoid with a large diameter has a current I1 in it. A circular wire loop of smaller diameter is placed inside the solenoid with the axis of the loop parallel to the axis of the solenoid. The wire loop is connected to an ammeter. The current in the solenoid increases to three times its initial value during a 50-millisecond time interval. What will we observe on the ammeter? Explain. A steady or constant current will be observed since the induced emf will be constant due to the constant rate of change of the current in the solenoid. 9 of 11 EMI3c key 6_08_02 EMI3C—QRT1: CIRCULAR LOOP WITHIN A SOLENOID A wire loop is placed inside and concentric with a large diameter solenoid. The solenoid has an initial current I in it. The initial current decreases to half in a time interval ∆t. This change will produce an induced current IInd in the wire loop. Explain how each of the changes described below would affect the induced current in the wire loop. The basic equation for this situation is Iloop = ∆i Emfinduced = Aloop ⋅ µ0 ⋅ nsolenoid ⋅ solenoid ∆t Rloop Rloop The solenoid current decrease occurs in a longer time than the time above. Smaller since delta t is longer. The diameter of the wire loop is doubled but the loop is still inside the solenoid. Larger by a factor of two since the area increases by a factor of four while the resistance increases by a factor of two so the ratio is the factor of two. The diameter of the solenoid is increased. No difference since the magnetic field is the same. The current decrease in the solenoid occurs in less than the original time interval. Larger since the time interval is shorter. The initial solenoid current is three times the value above but decreases to half in the same time interval. ∆i Emfinduced The basic equation for the situation Iloop = = Aloop ⋅ µ0 ⋅ nsolenoid ⋅ solenoid Rloop ∆t Rloop 3I 3I Everything is the same except ∆isolenoid = 3 I0 − 0 = 0 so the induced current is larger by a 2 2 factor or 3 (the initial situation gave a change of 1/2) 10 of 11 EMI3c key 6_08_02 EMI3C—M/MCT1: CIRCULAR LOOP WITHIN A SOLENOID A student is working with a situation where a solenoid (n = 120 turns / m) with a large radius, r = 12 cm, has a current of 4 A in it. A circular wire loop of smaller radius, r = 5 cm, is placed inside the solenoid with the axis of the loop parallel to the axis of the solenoid. The wire loop is connected to an ammeter. The current in the solenoid increases to three times its initial value during a 50-millisecond time interval. The resistance of the wire loop is 15 ohms. The student performs the calculation below for the induced current in the loop. Is this calculation meaningful or meaningless? I= (µo )(120m-1 )(12 A − 4 A)(π )(.05m) 2 (5 × 10 -3 s)(15Ω) Explain fully. Meaningful since the basic equation is ∆i Emfinduced Iloop = = Aloop ⋅ µ0 ⋅ nsolenoid ⋅ solenoid ∆t Rloop 11 of 11 EMI3c key Rloop 6_08_02