Couse Updates http://www.phys.hawaii.edu/~vane/phys272-sp10/physics272.html Remindes: 1) Assignment #8 will be able to do afte today 2) Finish Chapte 28 today 3) Quiz next Fiday 4) Review of 3 ight-hand ules
B Foce (single chage): F = qv B Lots of q: F = q v i F = qnv i avg B B F = IL B F z I B y x 06
Staight wie I Cuent in Wie Resulting B field :08
Magnetic Dipole Moment Aea vecto Magnitude = Aea Diection uses R.H.R. Magnetic Dipole moment μ NIA 19
Consistent? Yes! I Resulting B field Cuent in Wie
Today is Ampee s Law Day "High symmety" v B dl = μ0i Integal aound a path hopefully a simple one Cuent enclosed by that path I
Calculation of Electic Field Two Ways to calculate Coulomb s Law dq de = k ˆ 2 "Bute foce" Gauss Law ε 0 E ds = q "High symmety" What ae the analogous equations fo the Magnetic Field?
Calculation of Magnetic Field Two Ways to calculate Biot-Savat Law ( Bute foce ) db = μ ˆ 0I dl 2 4π Ampee s Law ( High symmety ) v B dl = μ0i I AMPERIAN SURFACE/LOOP These ae the analogous equations
B-field of Staight Wie, evisited Calculate field at distance R fom wie using Ampee's Law: Choose loop to be cicle of adius R centeed on the wie in a plane to Why? wie. Magnitude of B is constant (function of R only) Diection of B is paallel to the path. Evaluate line integal in Ampee s Law: B dl = B( 2πR) Cuent enclosed by path = I Apply Ampee s Law: 2 πrb= μ0i Ampee's Law simplifies the calculation thanks to symmety of the cuent! (axial/cylindical) B = μ 0 I 2πR I X dl R
Abitay closed path about a staight cuent wie will satisfy B dl = μ0i enclosed Abitay closed path outside a cuent wie will satisfy B dl = 0
Question 1: Two identical loops ae placed in poximity to two identical cuent caying wies. Fo which loop is B dl the geatest? A B A) B) C) A B Same
Question 1: Two identical loops ae placed in poximity to two identical cuent caying wies. Fo which loop is B dl the geatest? A B A) B) C) A B Same
Question 2: B C Now compae loops B and C. Fo which loop is B dl the geatest? A) B) C) B C Same
Question 2: B C Now compae loops B and C. Fo which loop is B dl the geatest? A) B) C) B C Same
B Field Inside a Long Wie Suppose a total cuent I flows x x x x x though the wie of adius a into x x x x x x x x the sceen as shown. dl x x x x x x x x x Calculate B field as a function of x x x x x x x x, the distance fom the cente of B field diection tangent to cicles. a the wie. x x x x x B field is only a function of take path to be cicle of adius : B dl = B(2 π ) Cuent passing though cicle: I B dl = μ o I Ampee's Law: enclosed a 2 I = enclosed 2 B μ I 2 π a 0 B = 2
B Field of a Long Wie Inside the wie: ( < a) B = μ0i 2π a 2 B a Outside the wie: ( > a ) μ0i B = 2π
Question 3 Two cylindical conductos each cay cuent I into the sceen as shown. The conducto on the left is solid and has adius R = 3a. The conducto on the ight has a hole in the middle and caies cuent only between R = a and R = 3a. What is the elation between the magnetic field at R = 6a fo the (a) Btwo L (6a)< cases B(L R (6a) = left, (b) R = Bight)? L (6a)= B R (6a) (c) B L (6a)> B R (6a) 3a I 3a I 2a a
Question 3 Two cylindical conductos each cay cuent I into the sceen as shown. The conducto on the left is solid and has adius R=3a. The conducto on the ight has a hole in the middle and caies cuent only between R=a and R=3a. What is the elation between the (a) Bmagnetic field at R = 6a fo the two L (6a)< B cases (L=left, R (6a) (b) B R=ight)? L (6a)= B R (6a) (c) B L (6a)> B R (6a) Use Ampee s Law in both cases by dawing a loop in the plane of the sceen at R=6a Both fields have cylindical symmety, so they ae tangent to the loop at all points, thus the field at R=6a only depends on cuent enclosed I enclosed = I in both cases 3a I 3a I 2a a
Question 4 Two cylindical conductos each cay cuent I into the sceen as shown. The conducto on the left is solid and has adius R = 3a. The conducto on I I the ight has a hole in the middle and caies cuent only between R = a and R = 3a. 3a 3a 2a a What is the elation between the magnetic field at R = 2a fo the two cases (L = left, R = ight)? (a) B L (2a)< B R (2a) (b) B L (2a)= B R (2a) (c) B L (2a)> B R (2a)
Question 4 Two cylindical conductos each cay cuent I into the sceen as shown. The conducto on the left is solid and has adius R=3a. The conducto on the ight has a hole in the middle and caies cuent only between R=a and R=3a. What is the elation between the magnetic field at R = 2a fo the (a) B L (2a)< B R (2a) (b) B L (2a)= B R (2a) (c) B two cases (L=left, R=ight)? L (2a)> B R (2a) Again, field only depends upon cuent enclosed... LEFT cylinde: 2 I (2 a) 4 I = I L 2 (3 a) 9 3a I RIGHT cylinde: π ( 2 2 π (2a) a ) = = π π ( 2 2 (3a ) a ) 3a I 3 I = I R 8 I 2a a
B Field of Cuent Sheet Conside an sheet of cuent descibed by n wies/length each caying cuent i into the sceen as shown. Calculate the B field. What is the diection of the field? Symmety vetical diection Calculate using Ampee's law fo a squae of side w: B dl = Bw + 0 + Bw + 0 = 2Bw I = nwi theefoe, B dl = μ0i B = μ ni 0 2 constant w x x x x x constant
B Field of an ideal Solenoid A constant magnetic field can (in pinciple) be poduced by an sheet of cuent. In pactice, howeve, a constant magnetic field is often poduced by a solenoid. A solenoid is defined by a cuent i flowing though a wie that is wapped n tuns pe unit length on a cylinde of adius a and length L. To coectly calculate the B-field, we should use Biot-Savat, and add up the field fom the diffeent loops. If a << L, the B field is to fist ode contained within the solenoid, in the axial diection, and of constant magnitude. In this limit, we can calculate the field using Ampee's Law. Ideal Solenoid L a
B Field of an Solenoid To calculate the B field of the solenoid using Ampee's Law, we need to justify the claim that the B field is nealy 0 outside the solenoid (fo an solenoid the B field To do this, exactly view the 0 outside). solenoid fom the x x x x x x x x x x x side as 2 cuent sheets. The fields ae in the same diection in the egion between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid). B is unifom inside solenoid and zeo outside. Daw squae path of side w: B dl = Bw B = μ0ni I = nwi x x x x x x x x Note: B Amp Length
Question 5: A cuent caying wie is wapped aound an ion coe, foming an electo-magnet. Which diection does the magnetic field point inside the ion coe? a) left b) ight c) up d) down e) out of the sceen
Question 5: A cuent caying wie is wapped aound an ion coe, foming an electo-magnet. Which diection does the magnetic field point inside the ion coe? a) left b) ight c) up d) down e) out of the sceen
Question 6: A cuent caying wie is wapped aound an ion coe, foming an electo-magnet. Which side of the solenoid should be labeled as the magnetic noth pole? a) ight b) left
Question 6: A cuent caying wie is wapped aound an ion coe, foming an electo-magnet. Which side of the solenoid should be labeled as the magnetic noth pole? a) ight b) left
Use the wap ule to find the B-field: wap you finges in the diection of the cuent, the B field points in the diection of the thumb (to the left). Since the field lines leave the left end of solenoid, the left end is the noth pole.
Solenoids The magnetic field of a solenoid is essentially identical to that of a ba magnet. The big diffeence is that we can tun the solenoid on and off! It attacts/epels othe pemanent magnets; it attacts feomagnets, etc.
Digital [on/off]: Doobells Solenoid Applications Magnet off plunge held in place by sping Magnet on plunge expelled stikes bell Powe doo locks Magnetic canes Electonic Switch elay Close switch cuent magnetic field pulls in plunge closes lage cicuit Advantage: A small cuent can be used to switch a much lage one State in washe/dye, ca ignition,
Analog (deflection I ): Vaiable A/C valves Speakes Solenoid Applications Solenoids ae eveywhee! In fact, a typical ca has ove 20 solenoids!
Tooid Tooid defined by N total tuns with cuent i. B=0 outside tooid! (Conside integating B on cicle outside tooid) Diection? tangent to cicle. Magnitude depends on? (not theta To o z). find B inside, conside cicle of adius, centeed at the cente of the tooid. B dl = B(2 π ) Apply Ampee s Law: I = Ni B dl = μ0i x x x x x x x x x x x x x x x x B B = μ0ni 2π
ITER Tokamak; giant tooid fo fusion A joint US-Euope- Japan poject to be built in southen Fance by 2016. A tooidal field contains the hot plasma. Fusion should povide clean powe. ITER is pototype fo futue machines and is suppose to poduce 500MW of powe. peson
Summay Example B-field Calculations: Inside a Long Staight Wie B = μ 0 I 2 π a 2 Infinite Cuent Sheet Solenoid B = μ0ni B = μ 0 ni 2 Tooid μ0ni B = 2π Cicula Loop B z μ 0 ir 2 2z 3
28.30 II dl I R III dl P dl db = Idl μo 2 4π ˆ On segments I and III, dl x ^ is zeo Is the field fom a staight wie always zeo? No!!! See 28.76 Make sue you undestand why db II = μ dl = μ π R = μo o o 2 4π 2 4π 4 R R R What is the diection of the B field? By the ight hand ule, it points into the pape
28.37, 28.38 Simplified coaxial cable a b c Cuent I flowing out of the page though inne conducto Cuent I flowing into the page though oute conducto B dl =μ I B π μ I o C 2 o c μ I o = Fo a<<b B = Fo >c, I 2π C =+I-I =0 hence B=0 fo >c Fo a<<b
28.77 J 2I0 2 1 ( ) kˆ 2 = π a a Find B fo >a B dl =μ I o c J is cuent density, so integate da to obtain cuent passing though a I = JdA () = J ()2πd I o a 2I0 = 2 π a o a o a 2 1 ( ) 2 π d
a 2 1 2 1 4 2 I (4 Io / a ) = / a = Io 2 4 o B(2 π ) = μoic= μoi 0 B μ I /2π = Outside the cylinde o How should this calculation be modified inside the cylinde <a?
Moe weekend fun? HW #8 get cacking due Monday Office Hous immediately afte this class (9:30 10:00) in WAT214 (1:30-2/1-1:30 M/WF) Next week: Chap 29 [Midtem 2 is Chap 25 29] Quiz next Fiday