MAGNETIC FIELD INTRODUCTION
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1 MAGNETIC FIELD INTRODUCTION It was found when a magnet suspended fom its cente, it tends to line itself up in a noth-south diection (the compass needle). The noth end is called the Noth Pole (N-pole), and the south end is called the South Pole (S-pole). These two poles cannot be isolated, that is, poles always appea in pais. Opposite poles attact each othe, while like poles epel each othe. As we did fo electic field, we epesent the magnetic field by field lines called the magnetic field lines. These lines can be dawn such that (1) they point out fom the noth pole towad the south pole of a magnet, (2) the diection of the magnetic field at any point is tangent to these lines at that point, and (3) the numbe of lines pe unit aea is popotional to the stength of the magnetic field. Notice that the magnetic field lines continue inside the magnet foming closed loops, unlike electic filed lines that begin on positive chages and teminate on negative chages. The eath is consideed as a magnet with its geogaphic south pole is the N- pole. This explains the diection of a suspended magnet. (a) Figue 13.1 (a) Magnetic field lines fo a ba magnet. (b) Magnetic filed patten fo a ba magnet as displayed by ion filings. (b)
2 MAGNETIC FIELD AND FORCES We can define the magnetic inteaction between two moving chages as follow: 1- A moving chage, o a cuent, ceates a magnetic field in the space aound it. 2 Any moving chage q exists in the egion of will be affected by a magnetic foce accoding to F = qv = qvsinθ 29.1 whee v is the velocity of the chage q and θ is the angle between v and. Note that F is zeo if v and ae paallel and maximum if v and ae pependicula. In contay to the electostatic foce which is paallel to E, the diection of F is always nomal to both v and. It is given by the ight hand ule: Let the outstetched finges of you ight hand to point along the diection of v and then bend them towad the diection of. The thumb then gives the diection of the magnetic foce. This is tue if q is positive only, and if q is negative the diection must be evesed. The SI unit of is N.s/C.m, which is called Tesla (T). The cgs unit of is the Gauss (G), which is elated to the Tesla though 1 T = 10 4 G Since F and v ae pependicula, then ds F v = F dt = 0
3 This means that the magnetic foce is always pependicula to the displacement ds, and hence it does not do any wok in a displaced paticle. The wok-enegy theoem states that the net wok exeted on a paticle must equal the change in its kinetic enegy. Theefoe, the magnetic foce cannot change the speed of the paticle. Example 29.1 An electon moves with a speed of m/s along the x- axis in a egion of unifom m. field of 0.025T, diected at an angle 60 o to the x- axis. Calculate the magnitude and the diection of the magnetic foce acting on the electon as it entes the egion of. Solution Expessing v and in vecto notations we have v = iˆ m/s And o ˆ o = cos60 i sin 60 ˆj = ( 0.013ˆ i ˆj )T 19 6 F = qv = iˆ 0.013ˆ i ˆj = = ( kˆ ) N ( ) ( ) 6 ( )( 0.022)( iˆ ˆj ) y v 60 o x 29.4 MOTION OF A CHARGED PARTICLE IN MAGNETIC FIELD In the pevious section we said that the static magnetic foce could not alte the speed of the paticle. Hence, the only thing the magnetic foce can do is to change the diection of the velocity. The acceleation, theefoe, is centipetal, and so the foce. This means that a moving paticle in a magnetic field follows a cicula path. If the magnetic is unifom and the initial velocity of the paticle is nomal to the diection of, the paticle will move in a cicle. Since the magnetic foce is always pependicula to both v and, the cicle in a plane nomal to the diection of the magnetic field. Since v and ae pependicula we have fom Equation v qv = m 29.2
4 Fom which we get mv = 29.3 q The angula fequency of the motion is v q ω = = 29.4 m and the peiod is v +q F Figue 13.4 A positively chaged paticle moves in a unifom magnetic field diected into the page. The path of the motion is a cicle nomal to the magnetic field and the sense of otation is counteclockwise. 2 π 2πm T = = 29.5 ω q The sense of otation can be detemined again fom the ight hand ule. The thumb must now point towad the cente, the diection of the foce. If is diected into the page. A positive chage will otate counteclockwise, while a negative chage will otate clockwise. If the velocity of the chaged paticle and the magnetic field ae not pependicula the path is a helix whose axis is paallel to.
5 Example 29.7 An electon is acceleated fom est by a potential diffeence of 350 V. It then entes a egion of unifom magnetic field and follows a cicula path of adius 7.5 cm. If its velocity is pependicula to. Calculate a) The magnitude of, b) the angula speed of the electon. Solution a) To find the speed of the electon we use the consevation of enegy pinciple, that is eV 2( )( 350) 7 mv = ev o v = = = Now fom the Equation mv = q = m 31 7 ( )( ) 19 ( )( 0.075) = T m/s b) ω = v = 7 ( ) 1 8 ( 0.075) = ad / s 29.2 MAGNETIC FORCE ON A CURRENT-CARRYING WIRE It is known that cuent is a collective of moving chages of the same sign. This means that any cuent-caying wie expeiences a magnetic foce when placed in a magnetic field. This foce is the vecto sum of the individual foces on the chaged paticles. Let us conside a staight potion of wie of length l and cosssectional aea A. The wie is assumed in to cay a cuent I and to be placed vd A in a unifom magnetic field as shown in the Figue. The magnetic L foce on a single chage dq moving with dift velocity v d is
6 df = dqv 29.6 ut dq = Idt df = ( Idt) v Knowing that vdt = dl, the dift velocity is constant (because the wie is staight), we can integate the last Equation to obtain the total foce on the wie, that is, F = I L (fo staight wies) 29.7 Note that the diection of L is detemined by the diection of the cuent. If the wie is not staight we have to find the foce on small segment and then integate, i.e., F = I f dl i If is unifom f F = I dl = il i With L is the staight line fom i tp f. Not that fo closed loop L is zeo The m. foce on any closed cuent-loop in a unifom m. field must be zeo. The diection of the magnetic foce on a wie is detemine by the ight hand ule descibed befoe with the fou finges now point towad the diection of the cuent instead of the velocity. Example 29.2 A wie bent into a semicicle of adius R and caies a cuent I. The wie lies in the x-y plane, and a unifom m.f. is diected along the +ve y-axis. Find F m on the staight potion and on the cuved potion of the wie. R Solution Fo the staight potion we have I
7 F = I L = I ( 2 Riˆ ) ( j ˆ) = 2IRkˆ And fo the cuved potion we have F = I L = I ( 2Riˆ ) ( j ˆ) = 2IRkˆ Not that the net foce fo the whole wie is zeo as expected TORQUE ON A CURRENT LOOP Conside a ectangula loop of sides a and b caying a cuent I and immesed in a unifom magnetic field. Fo the hoizontal sides of the loop the diection of cuent is paallel to, and hence the foces on the two sides is zeo. Howeve, the magnitude of the foces on the vetical sides of length b is Ib, and the two foces point in opposite diections. If the loop is assumed to otates about its cente, these two foces will poduce a toque that tends to otate the loop clockwise. The magnitude of this toque is F L a Figue 13.5 A ectangula loop of cuent I in a unifom magnetic field. Two equal but opposite foces acting on the vetical sides of the loop that make the loop to otate about its cente. I F R b a a τ = Ib + Ib whee (a/2) is the moment am fo each foce. Knowing the A=ab is the aea of the loop, Equation 29.8 can be expessed as τ = IA 29.9 If is not paallel to the plane of the loop, Equation 13.9 can be genealized as τ = IAsin θ 29.10
8 whee θ is the angle between and a unit vecto n pependicula to the plane of the loop. Equation shows that the toque has a maximum value IA when is paallel to the plane of the loop (θ =90 o ), and zeo when is pependicula to the loop s plane (θ =0). Now, let us define the magnetic dipole moment m as µ = I A The diection of m is pependicula to the plane of the loop and its sense is detemined by the ight hand ule: Culing the fou finges of the ight hand in the diection of the cuent, the thumb points in the diection of m. The SI unit of m is ampee-mete squae (A.m 2 ). The toque tends to otate the loop such that m and be in paallel (that is, to the smallest value of θ ). Example 29.3 A ectangula loop of 25 tuns has dimensions 5.4 cm 8.5 cm and caies a cuent of 15 ma. The loop exists in a unifom magnetic field of 0.35 T and initially paallel to the plane of the loop. What is the magnitude of the initial toque acting on the loop? Solution Since the coil has 25 tuns, we have ( ) = A.m µ = NIA= Now noting that m and ae pependicula we have τ = m = ( )(0.35) = N.m
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