Homework (lecture ): 3, 5, 9, 3,, 5, 9, 3, 4, 45, 49, 5, 57, 6
3. An electron that has velocity: moves through the uniform magnetic field (a) Find the force on the electron. (b) Repeat your calculation for a proton having the same velocity. Recall: (a) The force acts on the electron: j s m i s m v ˆ ) / (3. )ˆ / (. 6 6 j T i T ˆ ) (.5 )ˆ.3 ( qv F k v v q F x y y x ˆ ) ( k F ˆ.3)] (3.5) ( )[.6 ( 6 6 9 k F ˆ 6.4 4 b a c 3 3 3 3 3 b a b a b a b a b a b a c c c
(b) The proton has a positive charge: F p 6.4 4 So, the force acting on the proton has the same magnitude but opposite in direction to the force acting on the electron kˆ
5. An electron moves through a uniform magnetic field given by iˆ x (3.x ) ˆj. At a particular instant, the electron has velocity and the magnetic force acting on it is (6.4 v (.ˆ i 4. ˆ) j m/ 9 N) k ˆ F Find x. s q( v v ) kˆ x y y x F z kˆ x q(3v F x z v y ) ( T)
9. In Fig. 8-33, an electron accelerated from rest through potential difference V =. kv enters the gap between two parallel plates having separation d =. mm and potential difference V = V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap? Animation The electric force: E F e qe V ev F e d d The magnetic field force acts on the electron with an external : F mv qv F ev ev v F e E ev m F +
To travel in a straight line: F e F ev V ev d vd d V ev m 3.6 9 9. 3 4.67 ( T ) 4 (.67 T) kˆ
3. A strip of copper 5 m thick and 4.5 mm wide is placed in a uniform magnetic field of magnitude.65 T, with perpendicular to the strip. A current i = 3 A is then sent through the strip such that a Hall potential difference V appears across the width of the strip. Calculate V. (The number of charge carriers per unit volume for copper is 8.47 x 8 electron/m 3 ). n l: thickness of the strip i Vle V V 8.47 7.4 6 8 5 ( V ).65 3 6.6 9
. An electron of kinetic energy. kev circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 5. cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion. (a) The speed: K mv v K m (b) The centripetal force here is the magnetic force: (c) The frequency: (d) The period: F ma m f T v R T qv v R ( s) f mv qr ( Hz)
5. (a) Find the frequency of revolution of an electron with an energy of ev in a uniform magnetic field of magnitude 35. T. (b) Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field. (a) the frequency: For the electron case: f f q f m e f.6 9 35. 3.49. 9.8 5 ( Hz ) (b) the radius: r mv q m K q / m e 6 3 m Km e e.96( m)
9. An electron follows a helical path in a uniform magnetic field of magnitude.3 T. The pitch of the path is 6. m, and the magnitude of the magnetic force on the electron is. x -5 N. What is the electron's speed? F v qv The period: v v v v F e v v cos and sin 5 4.7 4.6 9.3 T qp m r v 5.3 v ep m e 4 v ( m/ s) ( m / s) m p ; T q v.6 9.3 6 6 3 3.49. v 6.53 4 ( m / s)
3. A particular type of fundamental particle decays by transforming into an electron e - and a positron e +. Suppose the decaying particle is at rest in a uniform magnetic field of magnitude 3.53 mt and the e - and e + move away from the decay point in paths lying in a plane perpendicular to. How long after the decay do the e - and e + collide? When the electron and the positron travel half of the circular path, they will collide: t m T e 5. 9 ( s ) e ray e e
4. A wire.3 m long carries a current of 3. A and makes an angle of 35. with a uniform magnetic field of magnitude =.5 T. Calculate the magnetic force on the wire. F il F ilsin F 3.3.5 sin35 5.7( N)
45. A wire 5. cm long carries a.5 A current in the positive direction of an x axis through a magnetic field: ( 3.mT ) ˆj (.mt ) kˆ. In unit-vector notation, what is the magnetic force on the wire? F L ˆ i F (3. i[(.5 il (.5m)ˆ i ˆj 3 T) ˆj ) ˆj kˆ ( T) kˆ (.53. 3 ) kˆ] F 3 ) ˆ 3 (.5 j (.75 ) kˆ
49. Figure 8-4 shows a rectangular -turn coil of wire, of dimensions cm by 5. cm. It carries a current of. A and is hinged along one long side. It is mounted in the xy plane, at angle = 3 to the direction of a uniform magnetic field of magnitude.5 T. In unit-vector notation, what is the torque acting on the coil about the hinge line. n is the angle between and. (5 ( NiA) n sin ' ' 4 ).5sin 4.33 Using the right hand rule to determine the torque direction: ( 4.33 3 N. m) ˆj 3 ( N. m)
5. Figure 8-44 shows a wood cylinder of mass m =.5 kg and length L =. m, with N =. turns of wire wrapped around it longitudinally, so that the plane of the wire coil contains the long central axis of the cylinder. The cylinder is released on a plane inclined at an angle to the horizontal, with the plane of the coil parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude.5 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane? F 4 F f s F F 3 F g
To keep the cylinder from rolling down: () The net force is zero: mg sin f ma f s : frictional force AND () The net torque acting on the cylinder about the cylinder axis must be zero: s f NiAsin sin f s r I ; mgr mgr mg sinr i NA NLr NiA mg NL
57. A circular coil of 6 turns has a radius of.9 cm. (a) Calculate the current that results in a magnetic dipole moment of magnitude.3 A.m. (b) Find the maximum magnitude of the torque that the coil, carrying this current, can experience in a uniform 35. mt magnetic field. (a) The magnetic dipole moment: i NA (b) The torque acts on the coil: NiA.3 63.4.9.7( A) is maximum if = 9 : sin max.335. 3.8( N. m)
6. In Fig. 8-47a, two concentric coils, lying in the same plane, carry currents in opposite directions. The current in the larger coil i is fixed. Current i in coil can be varied. Figure 8-47b gives the net magnetic moment of the two-coil system as a function of i. If the current in coil is then reversed, what is the magnitude of the net magnetic moment of the two-coil system when i = 7. ma? i and i are in opposite directions: At i = : net At i = 5 ma: if i is inversed: net net 5 ( A. m net NiA NA 5 3 NA 4. 5 NiA 4. 7 4.8 ( A. m ) 5 3 3 3 5 )
Homework (lecture ): 4, 7,, 6, 8,, 35, 38, 43, 46, 49, 5, 57, 6
4. A straight conductor carrying current i = 5. A splits into identical semicircular arcs as shown in Fig. 9-34. What is the magnetic field at the center C of the resulting circular loop? i 4R We use the right hand rule to determine the field direction of the two arcs: upper lower
7. In Fig. 9-36, two circular arcs have radii a = 3.5 cm and b =.7 cm, subtend angle = 74., carry current i =.4 A, and share the same center of curvature P. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at P? We choose the positive direction if is directed into the page The upper arc produces directed into the page and the lower arc produces out of the page, so: net i i i net upper lower 4a 4b 4 a b 4 7.4(74 /8) 4 3.5.7.3 ( ) net 7 T the page and the net field direction is out of
. In Fig. 9-38, two long straight wires at separation d = 6. cm carry currents i = 3.6 ma and i = 3.i out of the page. (a) At what point on the x axis shown is the net magnetic field due to the currents equal to zero? (b) If the two currents are doubled, is the point of zero magnetic field shifted toward wire, shifted toward wire,or unchanged? i : is up (positive) i : is down (negative) (a) net net d i i d d i d 4cm; d 3 d cm i d d 3d (b) unchanged
6. In Fig. 9-47, two concentric circular loops of wire carrying current in the same direction lie in the same plane. Loop L has radius.5 cm and carries 4. ma. Loop has radius.5 cm and carries 6. ma. Loop is to be rotated about a diameter while the net magnetic field set up by the two loops at their common center is measured. Through what angle must loop be rotated so that the magnitude of that net field is nt? The net field: i R arccos 44 cos(8 ) net
8. A current is set up in a wire loop consisting of a semi-circle of radius 4.5 cm, a smaller concentric semi-circle and two radial straight lengths, all in the same plane. Figure 9-48a shows the arrangement but is not drawn to scale. The magnitude of the magnetic field produced at the center of curvature is 47.5 T. The smaller semicircle is then flipped over (rotated) until the loop is again entirely in the same plane (Fig. 9-48b). The magnetic field produced at the (same) center of curvature now has magnitude 5.75 T and its direction is reversed. What is the radius of the smaller semicircle? For a semi-circle: Figure a: Figure b: i 4R i i i a 4R 4r 4 R r i i i b 4R 4r 4 R r R r r R a b 47.5 5.75 3 r R.5( cm)
. Figure 9-4a shows, in cross section, two long, parallel wires carrying current and separated by distance L. The ratio i /i of their currents is 4.; the directions of the currents are not indicated. Figure 9-4b shows the y component y of their net magnetic field along the x axis to the right of wire. (a) At what value of x > is y maximum? (b) If i = 3 ma, what is the value of that maximum? What is the direction (into or out of the page) of (c) i and (d) i? = at x = cm, so i and i are in opposite directions: i i i 4 net ( L x) x L x x
is maximum when x matches the following equation: x Now, we need to calculate L. At x = cm, y =, so: (b) if i = 3 ma: f '( x) 3x L 4 (c), (d): for x > cm, net >, and > : so > and <, thus i is directed out and i into the page x x Lx L L; x L/3( discarded as x ) 3x L L 3x 3( cm) i net, max net 7 L 4 4 3 9, max ( T).3 L ' L 3 i L
35. Figure 9-55 shows wire in cross section; the wire is long and straight, carries a current of 4. ma out of the page, and is at distance d =.4 cm from a surface. Wire, which is parallel to wire and also long, is at horizontal distance d = 5. cm from wire and carries a current of 6.8 ma into the page. What is the x component of the magnetic force per unit length on wire due to wire? F, x F, x F F cos Li d i Li d i cos Lii d F, x i i d d L d d d F
F x /L (N/m) 38. Figure 9-57a shows, in cross section, three current carrying wires that are long, straight, and parallel to one another. Wires and are fixed in place on an x axis, with separation d. Wire has a current of.75 A, but the direction of the current is not given. Wire 3, with a current of.5 A out of the page, can be moved along the x axis to the right of wire. As wire 3 is moved, the magnitude of the net magnetic force F on wire due to the currents in wires and 3 changes. The x component of that force is F, and the value per unit length of wire is F x /L. Figure 9-57b gives F x /L versus the position x of wire 3. The plot has an asymptote F x /L = -.67 N/m as x. What are the (a) size and (b) direction (into or out of the page) of the current in wire? At x = 4 cm, F x =, so i is in the same direction as i 3. When i 3 approaches i, F is positive and dominated by the force from i 3, so i is out of the page. 3
At x = 4 cm: F, net L d When x : i i F, net L 3 x i x F L i 3.75.5 4 d i i i d i ( cm) i.5( A).67 F F 3 6 3
43. Figure 9-6 shows a cross section across a diameter of a long cylindrical conductor of radius a =. cm carrying uniform current 7 A. What is the magnitude of the current's magnetic field at radial distance (a), (b). cm, (c). cm (wire's surface), and (d) 4. cm? The magnetic field is inside a wire with current: (a) = ; (b) (c) (d) i a r 8.5 i r r i R 4 7 4 8.5 r 7 4 r. 4 7. ( T ) 8.5 4 ( T) 8.5 r 8.5 4 ( T)
i r.5 i R r
46. Eight wires cut the page perpendicularly at the points shown in the figure below. A wire labeled with the integer k (k =,,,8) carries the current ki, where i = 6. ma. For those wires with odd k, the current is out of the page; for those with even k, it is into the page. Evaluate ds. along the closed path in the direction shown. Using Ampere s law: ds enc i ( i i3 i6i7 ) Using the curled-straight right hand rule: i, i 3, i 7 are positive and i 6 is negative ds ds ( i 3i 5 6. 6i 3 7i) 5i 4 7 3.77 8 ( T. m)
49. A toroid having a square cross section, 5. cm on a side, and an inner radius of 5. cm has 5 turns and carries a current of.8 A. (It is made up of a square solenoid instead of a round one as in Fig. 9-7 -bent into a doughnut shape.) What is the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius? (a) at the inner radius, r = 5 cm: 7 in (b) at the outer radius, r = 5 + 5 = cm: 4.85 5.33 4.5 r ( T) 7 4.85 4. 4. ( T )
5. A solenoid that is 95. cm long has a radius of. cm and a winding of 5 turns; it carries a current of 3.6 A. Calculate the magnitude of the magnetic field inside the solenoid. The magnetic field inside a solenoid is computed by: in where n is the number of turns per unit length So: n L N 4 7 5.95 3.6579 579 (turns/m) 7.4 3 ( T )
57. A student makes a short electromagnet by winding 3 turns of wire around a wooden cylinder of diameter d = 5. cm. The coil is connected to a battery producing a current of 4. A in the wire. (a) What is the magnitude of the magnetic dipole moment of this device? (b) At what axial distance z >> d will the magnetic field have the magnitude 5. T (approximately one-tenth that of Earth's magnetic field)? (a) The magnetic dipole moment: 3 (b) if z >> d: 4. ( z) (.5 NiA z 3 NiA ) z 3.36( A.m z 3 ) z 3 4 7 5.36 6.46( m)
6. In Fig. 9-66, current i = 56. ma is set up in a loop having two radial lengths and two semicircles of radii a = 5.7 cm and b = 8.57 cm with a common center P. What are the (a) magnitude and (b) direction (into or out of the page) of the magnetic field at P and the (c) magnitude and (d) direction of the loop's magnetic dipole moment? (a) The field due to a circular arc at its center: For the two semicirles: a b i 4R (b) Using the right hand rule, the direction points into the page (c) The magnetic dipole moment: (d) points into the page (the same direction as ) i i i 4a 4b 4 a NiA ia i ( a b b ) n