# Samples of conceptual and analytical/numerical questions from chap 21, C&J, 7E

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1 CHAPTER 1 Magnetism CONCEPTUAL QUESTIONS Cutnell & Johnson 7E 3. ssm A chaged paticle, passing though a cetain egion of space, has a velocity whose magnitude and diection emain constant, (a) If it is known that the extenal magnetic field is zeo evey whee in this egion, can you conclude that the extenal electic field is also zeo? Explain. (b) If it is known that the extenal electic field is zeo eveywhee, can you conclude that the extenal magnetic field is also zeo? Explain. 3. SSM REASONING AND SOLUTION A chaged paticle, passing though a cetain egion of space, has a velocity whose magnitude and diection emain constant. a. If it is known that the extenal magnetic field is zeo eveywhee in the egion, we can conclude that the electic field is also zeo. Any chaged paticle placed in an electic field will expeience a foce given by = qe, whee q is the chage and E is the electic field. If the magnitude and diection of the velocity of the paticle ae constant, then the paticle has zeo acceleation. om Newton's second law, we know that the net foce on the paticle is zeo. But thee is no magnetic field and, hence, no magnetic foce. Theefoe, the net foce is the electic foce. Since the electic foce is zeo, the electic field must be zeo. b. If it is known that the extenal electic field is zeo eveywhee, we cannot conclude that the extenal magnetic field is also zeo. In ode fo a moving chaged paticle to expeience a magnetic foce when it is placed in a magnetic field, the velocity of the moving chage must have a component that is pependicula to the diection of the magnetic field. If the moving chaged paticle entes the egion such that its velocity is paallel o antipaallel to the magnetic field, it will expeience no magnetic foce, even though a magnetic field is pesent. In the absence of an extenal electic field, thee is no electic foce eithe. Thus, thee is no net foce, and the velocity vecto will not change in any way. 7. The dawing shows a top view of fou inteconnected chambes. A negative chage is fied into chambe 1. By tuning on sepaate magnetic fields in each chambe, the chage can be made to exit fom chambe 4, as shown. (a) Descibe how the magnetic field in each chambe should be diected. (b) If the speed of the chage is v when it entes chambe 1, what is the speed of the chage when it exits chambe 4? Why? Compiled by DJJ Page 1 of 8 9/13/6

2 7. REASONING AND SOLUTION The dawing shows a top view of fou inteconnected chambes. A negative chage is fied into chambe 1. By tuning on sepaate magnetic fields in each chambe, the chage is made to exit fom chambe 4. a. In each chambe the path of the paticle is onequate of a cicle. The dawing at the ight also shows the diection of the centipetal foce that must act on the paticle in each chambe in ode fo the paticle to tavese the path. The chaged 1 paticle can be made to move in a cicula path by launching it into a egion in which thee exists a magnetic field that is pependicula to the velocity of the paticle. Using RHR-1, we see that if the palm of the ight hand wee facing in the diection of in chambe 1 so that the thumb points along the path of the paticle, the finges of the ight hand must point out of the page. This is the diection that the magnetic field must have to make a positive chage move along the path shown in chambe 1. Since the paticle is negatively chaged, the field must point opposite to that diection o into the page. Simila easoning using RHR-1, and emembeing that the paticle is negatively chaged, leads to the following conclusions: in egion the field must point out of the page, in egion 3 the field must point out of the page, and in egion 4 the field must point into the page. b. If the speed of the paticle is v when it entes chambe 1, it will emege fom chambe 4 with the same speed v. The magnetic foce is always pependicula to the velocity of the paticle; theefoe, it cannot do wok on the paticle and cannot change the kinetic enegy of the paticle, accoding to the wok-enegy theoem. Since the kinetic enegy is unchanged, the speed emains constant. 15. o each electomagnet at the left of the dawing, explain whethe it will be attacted to o epelled fom the pemanent magnet at the ight. 4 v q 3 Compiled by DJJ Page of 8 9/13/6

3 15. REASONING AND SOLUTION The figue below shows the aangements of electomagnets and magnets. (a) S N N S (b) S N N S We can detemine the polaity of the electomagnets by using RHR-. Imagine holding the cuent-caying wie of the electomagnet in the ight hand as the wie begins to coil aound the ion coe. The thumb points in the diection of the cuent. o the electomagnet in figue (a), the finges of the ight hand wap aound the wie on the left end so that they point, inside the coil, towad the ight end. Thus, the ight end of the coil must be a noth pole. Simila easoning can be used to identify the noth and south poles of the electomagnet in figue (b). The esults ae shown in the figue above. Since the like poles of two diffeent magnets epel each othe and the dissimila poles of two diffeent magnets attact each othe, we can conclude that in both aangements, the electomagnet is epelled fom the pemanent magnet at the ight. CHAPTER 1 MAGNETIC ORCES AND MAGNETIC IELDS Samples of solutions to Poblems fom chapte 1Cutnell & Johnson 7E 4. When a chaged paticle moves at an angle of 5 with espect to a magnetic field, it expeiences a magnetic foce of magnitude. At what angle (less than 9 ) with espect to this field will this paticle, moving at the same speed, expeience a magnetic foce of magnitude? 4. REASONING Accoding to Equation 1.1, the magnetic foce has a magnitude of = q vb sin θ, whee q is the magnitude of the chage, B is the magnitude of the magnetic field, v is the speed, and θ is the angle of the velocity with espect to the field. As θ inceases fom to 9, the foce inceases. Theefoe, the angle we seek must lie between 5 and 9. Compiled by DJJ Page 3 of 8 9/13/6

4 SOLUTION Letting θ 1 = 5 and θ be the desied angle, we jcan apply Equation 1.1 to both situations as follows: = q vbsin θ and = q vbsinθ 1 Situation 1 Situation Dividing the equation fo situation by the equation fo situation 1 gives qvbsinθ = o sin θ=sin θ1 =sin 5 =.85 q vbsinθ 1 1 ( ) θ = sin.85 = A chaged paticle entes a unifom magnetic field and follows the cicula path shown in the dawing. (a) Is the paticle positively o negatively chaged? Why? (b) The paticle s speed is 14 m/s, the magnitude of the magnetic field is.48 T, and the adius of the path is 96 m. Detemine the mass of the paticle, given that its chage has a magnitude of. 11. REASONING a. The dawing shows the velocity v of the paticle at the top of its path. The magnetic foce, which povides the centipetal foce, must be diected towad the cente of the cicula path. Since the diections of v,, and B ae known, we can use Right-Hand Rule No. 1 (RHR-1) to detemine if the chage is positive o negative. v B (out of pape) b. The adius of the cicula path followed by a chaged paticle is given by Equation 1. as = mv/ q B. The mass m of the paticle can be obtained diectly fom this elation, since all othe vaiables ae known. SOLUTION a. If the paticle wee positively chaged, an application of RHR-1 would show that the foce would be diected staight up, opposite to that shown in the dawing. Thus, the chage on the paticle must be negative. b. Solving Equation 1. fo the mass of the paticle gives ( C)(.48 T)( 96 m) qb 3 m = = =.7 1 kg v 14 m/s Compiled by DJJ Page 4 of 8 9/13/6

5 19. ssm Review Conceptual Example as an aid in undestanding this poblem. The dawing shows a positively chaged paticle enteing a.5-t magnetic field (diected out of the pape). The paticle has a speed of 7 m/s and moves pependicula to the magnetic field. Just as the paticle entes the magnetic field, an electic field is tuned on. What must be the magnitude and diection of the electic field such that the net foce on the paticle is twice the magnetic foce? 19. SSM REASONING AND SOLUTION Accoding to Right-Hand Rule No. 1, the magnetic foce on the positively chaged paticle is towad the bottom of the page in the dawing in the text. If the pesence of the electic field is to double the magnitude of the net foce on the chage, the electic field must also be diected towad the bottom of the page. Note that this esults in the electic field being pependicula to the magnetic field, even though the electic foce and the magnetic foce ae in the same diection. uthemoe, if the magnitude of the net foce on the paticle is twice the magnetic foce, the electic foce must be equal in magnitude to the magnetic foce. In othe wods, combining Equations 18. and 1.1, we find qe= qvbsinθ, with sinθ = sin 9. =1.. Then, solving fo E E = vbsin θ = ( 7 m / s)(.5 T)(1.) = 14 V / m 9. A squae coil of wie containing a single tun is placed in a unifom.5-t magnetic field, as the dawing shows. Each side has a length of.3 m, and the cuent in the coil is 1 A. Detemine the magnitude of the magnetic foce on each of the fou sides. 9. REASONING AND SOLUTION The foce on each side can be found fom = ILB sin θ. o the top side, θ = 9., so = (1 A)(.3 m)(.5 T) sin 9. =.96 N The foce on the bottom side (θ = 9. ) is the same as that on the top side, =.96 N. o each of the othe two sides θ =, so that the foce is = N. Compiled by DJJ Page 5 of 8 9/13/6

6 41. ssm www The ectangula loop in the dawing consists of 75 tuns and caies a cuent of I = 4.4 A. A 1.8-T magnetic field is diected along the +y axis. The loop is fee to otate about the z axis. (a) Detemine the magnitude of the net toque exeted on the loop and (b) state whethe the 35 angle will incease o decease. 41. SSM WWW REASONING The toque on the loop is given by Equation 1.4, τ = NIABsinφ. om the dawing in the text, we see that the angle φ between the nomal to the plane of the loop and the magnetic field is 9 35 = 55. The aea of the loop is.7 m.5 m =.35 m. SOLUTION a. The magnitude of the net toque exeted on the loop is τ = NIAB sin φ = (75)(4.4 A)(.35 m )(1.8 T) sin 55 = 17 N m b. As discussed in the text, when a cuent-caying loop is placed in a magnetic field, the loop tends to otate such that its nomal becomes aligned with the magnetic field. The nomal to the loop makes an angle of 55 with espect to the magnetic field. Since this angle deceases as the loop otates, the 35 angle inceases. 59. The dawing shows an end-on view of thee wies. They ae long, staight, and pependicula to the plane of the pape. Thei coss sections lie at the cones of a squae. The cuents in wies 1 and ae I 1 = I = I and ae diected into the pape. What is the diection of the cuent in wie 3, and what is the atio I 3 /I, such that the net magnetic field at the empty cone is zeo? 59. REASONING AND SOLUTION The cuents in wies 1 and poduce the magnetic fields B 1 and B at the empty cone, as shown in the following dawing. The diections of these fields can be obtained using RHR-. Since thee ae equal cuents in wies 1 and and since these wies ae each the same distance fom the empty cone, B 1 and B have equal magnitudes. Using Equation 1.5, we can wite the field magnitude as B = B = µ I 1 /b g. Since the fields B 1 and B ae pependicula, it follows fom the Pythagoean theoem that they combine to poduce a net magnetic field that has the diection shown in the dawing at the ight and has a magnitude B 1+ given by Compiled by DJJ Page 6 of 8 9/13/6

7 B B B = + = H G I + KJ H G I K J = µ I µ I µ I The cuent in wie 3 poduces a field B 3 at the empty cone. Since B 3 and B 1+ combine to give a zeo net field, B 3 must have a diection opposite to that of B 1+. Thus, B 3 must point upwad and to the left, and RHR- indicates that B 3 Wie 1 X B 1 X Wie 3 Wie B B 1+ B the cuent in wie 3 must be diected out of the plane of the pape. Moeove, the magnitudes of B 3 and B 1+ must be the same. Recognizing that wie 3 is a distance of d = + = fom the empty cone, we have µ I µ I I 3 3 = B o = so that = + I 3 1 d i 6. The wie in igue 1.4 caies a cuent of 1 A. Suppose that a second long, staight wie is placed ight next to this wie. The cuent in the second wie is 8 A. Use Ampèe s law to find the magnitude of the magnetic field at a distance of =.7 m fom the wies when the cuents ae (a) in the same diection and (b) in opposite diections. 6. REASONING Since the two wies ae next to each othe, the net magnetic field is eveywhee paallel to in igue 1.4. Moeove, the net magnetic field B has the same magnitude B at each point along the cicula path, because each point is at the same distance fom the wies. Thus, in Ampèe's law (Equation 1.8), B = B, I = I1+ I, and we have ( ) µ ( ) ΣB = B Σ = I + I 1 But Σ is just the cicumfeence () of the cicle, so Ampèe's law becomes This expession can be solved fo B. ( ) = µ ( + ) B I I 1 SOLUTION a. When the cuents ae in the same diection, we find that Compiled by DJJ Page 7 of 8 9/13/6

8 7 ( ) ( )( ) I1+ I 4π 1 T m/a 8A+ 1A 5 µ B = = = T.7 m ( ) b. When the cuents have opposite diections, a simila calculation shows that 7 ( ) ( )( ) I1 I 4π 1 T m/a 8 A 1 A 6 µ B = = = T.7 m ( ) 65. ssm A long solenoid has 14 tuns pe mete of length, and it caies a cuent of 3.5 A. A small cicula coil of wie is placed in side the solenoid with the nomal to the coil oiented at an angle of 9. with espect to the axis of the solenoid. The coil consists of 5 tuns, has an aea of m, and caies a cuent of.5 A. ind the toque exeted on the coil. 65. SSM REASONING The coil caies a cuent and expeiences a toque when it is placed in an extenal magnetic field. Thus, when the coil is placed in the magnetic field due to the solenoid, it will expeience a toque given by Equation 1.4: τ = NIABsinφ, whee N is the numbe of tuns in the coil, A is the aea of the coil, B is the magnetic field inside the solenoid, and φ is the angle between the nomal to the plane of the coil and the magnetic field. The magnetic field in the solenoid can be found fom Equation 1.7: B = µ ni, whee n is the numbe of tuns pe unit length of the solenoid and I is the cuent. SOLUTION The magnetic field inside the solenoid is ( ) 7 3 B= µ ni = (4π 1 T m/a) 14 tuns/m (3.5 A)=6. 1 T The toque exeted on the coil is τ = NIABsin φ = (5)(.5 A)(1. 1 m )(6. 1 T)(sin 9. )= N m Compiled by DJJ Page 8 of 8 9/13/6

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