Physics 231 Exam III Dec. 1, 2003

Transcription

1 Physics 231 Exam III Dec. 1, One night while asleep you find you are having a nightmare. You find that you are in the middle of the French revolution. Unfortunately you are being led up to the guillotine. You notice that the blade runs in an east-west direction, parallel to the ground. You remember that the Earth s magnetic field runs south to north. For this scenario take the Earth s magnetic field to be 0.3 Tesla. You notice that there is a current generator nearby that is miraculously hooked to the blade. The guillotine blade has a mass of 300 gms and has a length of 1 meter. What must the magnitude and direction (east or west) of the current be through the blade such that your head remains connected to your shoulders? With a current going through the blade, there will be a magnetic force on the blade, since the blade will act as a current carrying wire. You want the magnetic force to balance the downward acting gravitational force on the blade. F Magnetic = I L B Since L and B are perpendicular to each other F = ILB The gravitational force is acting downward and is given by F Gravity = mg Themagneticforceonthewireshouldbedirectedupwardsandbeequaltothegravitational force downward F Magnetic = F Gravity ILB = mg I = mg / LB = (0.3)(9.8) / (1)(0.3) = 9.8 Amps Now for the direction: If the current goes east to west, <, then the magnetic force will be downwards, the same direction as the gravitational force, facilitating your head coming off. If the current goes west to east, ->, then the magnetic force will be upwards, hindering the blade. So the desired direction for the current is west to east. 1

2 2. A sliding bar, of length 4 meters, rests on conducting rails as shown. The conducting rails have a net resistance of 8 ohms. a. Suppose that B= 4.5 Tesla and directed as shown and that it is constant in time. What is the magnitude and direction of the induced current in the closed loop if the bar moves to the right a constant velocity of 1 9 m/sec? then The induced emf is given by ε = dφ φ = BA ε = ( db A + B da ) Since B is constant, db =0, and it is the area that is changing time, the induced emf is then given by ε = B da So da A = l x = l dx = lv ε = Blv = = 2Volts Since the flux through the loop is increasing, the current that is set up will create a magnetic field opposing this increasing flux. To do this the current will have to be counterclockwise a magnitude of I = ε R = 2 8 =0.25Amps b. Suppose now that the bar is held fixed in the position shown and that the magnetic field now increases time as B =0.5 t 2 Tesla. What is the magnitude and direction of the induced current in the closed loop at t =3sec? Now the area is held fixed and it is the magnetic field that is varying. Again the induced emf is given by ε = dφ 2

3 then Since the area is constant, da φ = BA ε = ( db A + B da ) =0, and the field is varying, the induced emf is given by ε = A db db = d 0.5 t 2 = t So ε = At= 5 4 t = 20 3= 60Volts The current is then given by I = ε R = 60 8 =7.5Amps Since the emf is negative the flux through the loop is increasing, therefore the current will be set up so as to oppose this increasing magnetic flux and it will be counterclockwise c. Suppose now that the field is increasing time as B =0.5t 2 Tesla and also that the bar is moving to the left a velocity of 1 m/sec. What is the 9 magnitude and direction of the current in the loop at t =27sec? We now have both the area and the magnetic field changing, so we use both of the previous results ε = ( db A + B da ) and db = d 0.5 t 2 = t da dx = l = lv (Bar is moving to the left) We also need to know what the area is at t =27sec. Remember that the bar is now moving to the left. If at t =0x =5meters, att =27xisgivenby x = 5 vt = = 2meters We therefore have ε = t A +0.5t 2 l v = ( ) = ( ) = 54volts AGain the current is given by I = ε R = 54 8 =6.75Amps Since the emf is negative this means that the flux is still increasing through the loop, the current that will be set up will be so as to oppose this increase in the magnetic flux and therefore will be in a counterclockwise direction. 3

4 3. At the instant when the current in an inductor is increasing at a rate of Amp/sec, the magnitude of the self-induced emf is V a. What is the inductance of the inductor? The self induced emf in an inductor is given by or ε = L di L = ε di = =0.25Henry b. If the inductor is a solenoid 400 turns, what is the average magnetic fluxthrougheachturnwhenthecurrentis0.720amp? For a solenoid, the inductance is given by L = NΦ B i where N is the number of turns in the solenoid, Φ B is the flux through one turn, and i is the current in the solenoid. Then we have Φ B = Li N = 400 = Weber 4

5 4. A particle carries a charge of 2 coulombs. When it moves a velocity of 3 m/s at an angle of 30 above the x-axis in the xy plane, a uniform magnetic field exerts a force on the particle in the negative y direction of magnitude 6 Newtons Whenthesameparticlemovesavelocityof2m/salongthepositivez-axis, thereisaforceof4newtonsexertedontheparticleinthepositivex-direction. What are the magnitude and direction of the magnetic field? We remember that F = q v B We write this as bi bj b k F = q v x v y v z B x B y B z h F = q b i (v y B z v z B y )+bj (v z B x v x B z )+ b i k (v x B y v y B x ) (1) We start the first situation. Since the force is only in the negative y-direction only the second term of equation 1 comes into play. F y = q (v z B x v x B z ) We have that v z =0while v x = v cos θ =3cos30=2.60m/s. So we therefore have B z = F = ( 6) =1.154 Tesla qv x For the second situation, we have a force on the x direction, so only the first term of equation 1comesintoplay F x = q (v y B z v z B y ) We have that v y =0and that v z =2m/s. We then have B y = F x qv z = 4 = 1 Tesla 2 2 We can not say anything about B x. Soatbestwehave ³ B = 1bj b k Tesla 5

6 MaybeUsefulformulaeandconstants: F = q v B F = I L B I µ = NIA τ = µ B db = µ 0 Id l br B 4π r d l = 2 µ0 I V = dφ B Φ = B A L = NΦ B i ε = L di k =8.988x10 9 Nm 2 /C 2 ; 0 =8.854x10 12 C 2 /N m 2 ; q electron = 1.6x10 19 C µ 0 =4π 10 7 N/A 2 6