m e = m/s. x = vt = t = x v = m

Transcription

1 5. (a) The textbook uses geomagnetic north to refer to Earth s magnetic pole lying in the northern hemisphere. Thus, the electrons are traveling northward. The vertical component of the magnetic field is downward. The right-hand rule indicates that v B is to the west, but since the electron is negatively charged (and F = q v B), the magnetic force on it is to the east. (b) We combine F = m e a with F = evb sin φ. Here, B sin φ represents the downward component of Earth sfield(givenintheproblem). Thus,a = evb/m e. Now, the electron speed can be found from its kinetic energy. Since K = 1 2 mv2, Therefore, v = 2K 2( = 3 ev)( J/eV) m e = m/s. kg a = evb = ( C)( m/s)( T) m e = m/s 2. kg (c) We ignore any vertical deflection of the beam which might arise due to the horizontal component of Earth s field. Technically, then, the electron should followa circular arc. However, the deflection is so small that many of the technicalities of circular geometry may be ignored, and a calculation along the lines of projectile motion analysis (see Chapter 4) provides an adequate approximation: x = vt = t = x v = m m/s which yields a time of t = s. Then, with our y axis oriented eastward, y = 1 2 at2 = 1 ( )( ) 2 = m. 2

2 8. (a) Letting F = q( E + v B) = 0, we get vb sin φ = E. We note that (for given values of the fields) this gives a minimum value for speed whenever the sin φ factor is at its maximum value (which is 1, corresponding to φ =90 ). So v min = E/B =( V/m)/(0.400 T) = m/s. (b) Having noted already that v B,wenowpointoutthat v B (which direction is given by the right-hand rule) must be in the direction opposite to E. Thus, we can use the left hand to indicate the arrangement of vectors: if one points the thumb, index finger, and middle finger on the left hand so that all three are mutually perpendicular, then the thumb represents v, the index finger indicates B, and the middle finger represents E.

3 12. We use Eq to solve for V : V = ib nle = (23 A)(0.65 T) ( /m 3 )(150 µm)( C) = V.

4 26. The equation of motion for the proton is Thus, F = q v B = q(v x î+v y ĵ+v zˆk) Bî =qb(vz ĵ v yˆk) [( ) ( ) ( ) ] dvx dvy dvz = m p a = m p î+ ĵ+ ˆk. dt dt dt dv x dt dv y dt dv z dt = 0 = ωv z = ωv y, where ω = eb/m p. The solution is v x = v 0x, v y = v 0y cos ωt and v z = v 0y sin ωt. In summary, we have v(t) =v 0x î+v 0y cos(ωt)ĵ v 0y (sin ωt)ˆk.

5 33. The magnitude of the magnetic force on the wire is given by F B = ilb sin φ, wherei is the current in the wire, L is the length of the wire, B is the magnitude of the magnetic field, and φ is the angle between the current and the field. In this case φ =70.Thus, F B = (5000 A)(100 m)( T) sin 70 =28.2 N. We apply the right-hand rule to the vector product F B = i L B to show that the force is to the west.

6 47. We use Eq where µ is the magnetic dipole moment of the wire loop and B is the magnetic field, as well as Newton s second law. Since the plane of the loop is parallel to the incline the dipole moment is normal to the incline. The forces acting on the cylinder are the force of gravity mg, actingdownward from the center of mass, the normal force of the incline N, acting perpendicularly to the incline through the center of mass, and the force of friction f, acting up the incline at the point of contact. We take the x axis to be positive down the incline. Then the x component of Newton s second law for the center of mass yields mg sin θ f = ma. For purposes of calculating the torque, we take the axis of the cylinder to be the axis of rotation. The magnetic field produces a torque with magnitude µb sin θ, and the force of friction produces a torque with magnitude fr,wherer is the radius of the cylinder. The first tends to produce an angular acceleration in the counterclockwise direction, and the second tends to produce an angular acceleration in the clockwise direction. Newton s second law for rotation about the center of the cylinder, τ = Iα,gives fr µb sin θ = Iα. Since we want the current that holds the cylinder in place, we set a =0andα = 0, and use one equation to eliminate f from the other. The result is mgr = µb. The loop is rectangular with two sides of length L and two of length 2r, soitsareaisa =2rL and the dipole moment is µ = NiA =2NirL. Thus, mgr =2NirLB and i = mg 2NLB = (0.250 kg)(9.8m/s 2 ) 2(10.0)(0.100 m)(0.500 T) =2.45 A.

7 54. Let a =30.0cm, b =20.0cm, and c =10.0cm.Fromthegivenhint,wewrite Thus, using the Pythagorean theorem, µ = µ 1 + µ 2 = iab( ˆk) + iac(ĵ) = ia(cĵ bˆk) = (5.00 A)(0.300 m)[(0.100 m)ĵ (0.200 m)ˆk] = (0.150ĵ 0.300ˆk) A m 2. µ = (0.150) 2 +(0.300) 2 =0.335 A m 2, and µ is in the yz plane at angle θ to the +y direction, where ( ) ( ) θ =tan 1 µy =tan 1 = µ x 0.150

8 8. Recalling the straight sections discussion in Sample Problem 30-1, we see that the current in segments AH and JD do not contribute to the field at point C. Using Eq (with φ = π) andtheright-hand rule, we find that the current in the semicircular arc HJ contributes µ 0 i/4r 1 (into the page) to the field at C. Also, arc DA contributes µ 0 i/4r 2 (out of the page) to the field there. Thus, the net field at C is B = µ 0i 4 ( 1 R 1 1 R 2 ) into the page.

9 11. Our x axis is along the wire with the origin at the midpoint. The current flows in the positive x direction. All segments of the wire produce magnetic fields at P 1 that are out of the page. According to the Biot-Savart law, the magnitude of the field any(infinitesimal) segment produces at P 1 is given by db = µ 0i sin θ 4π r 2 dx where θ (the angle between the segment and a line drawn from the segment to P 1 )andr (the length of that line) are functions of x. Replacing r with x 2 + R 2 and sin θ with R/r = R/ x 2 + R 2,we integrate from x = L/2 tox = L/2. The total field is B = µ 0iR 4π L/2 L/2 dx (x 2 + R 2 ) = µ 0iR 3/2 4π 1 R 2 If L R, thenr 2 in the denominator can be ignored and B = µ 0i 2πR x (x 2 + R 2 ) 1/2 L/2 L/2 = µ 0i 2πR L L2 +4R 2. is obtained. This is the field of a long straight wire. For points veryclose to a finite wire, the field is quite similar to that of an infinitelylong wire.

10 12. The center of a square is a distance R = a/2 from the nearest side (each side being of length L = a). There are four sides contributing to the field at the center, so the result of problem 11 leads to ( ) ( ) µ0 i a B center =4 = 2 2 µ 0 i. 2π(a/2) a2 +4(a/2) 2 πa

11 28. (a) Consider a segment of the projectile between y and y + dy. We use Eq to find the magnetic force on the segment, and Eq for the magnetic field of each semi-infinite wire (the top rail referred to as wire 1 and the bottom as wire 2). The current in rail 1 is in the +î direction, and the current in the rail 2 is in the î direction. The field (in the region between the wires) set up by wire 1 is into the paper (the ˆk direction) and that set up by wire 2 is also into the paper. The force element (a function of y) acting on the segment of the projectile (in which the current flows in the ĵ direction) is given below. The coordinate origin is at the bottom of the projectile. Thus, the force on the projectile is F = df = i2 µ 0 4π d F = d F 1 + d F 2 = idy( ĵ) B 1 + dy( ĵ) B 2 = i[b 1 + B 2 ] î dy [ µ 0 i = i 4π(2R + w y) + µ ] 0i î dy. 4πy R+w R ( 1 2R + w y + 1 ) dy î = µ 0i 2 (1+ y 2π ln w ) î. R (b) Using the work-energy theorem, we have K = 1 2 mv2 f = W ext = F d s = FL. Thus, the final speed of the projectile is v f = = ( ) 1/2 [ 2Wext 2 µ 0 i 2 = (1+ m m 2π ln w ) ] 1/2 L R [ 2(4π 10 7 T m/a)( A) 2 ln( cm/6.7cm)(4.0m) = m/s. 2π( kg) ] 1/2

12 37. (a) The the magnetic field at a point within the hole is the sum of the fields due to two current distributions. The first is that of the solid cylinder obtained by filling the hole and has a current density that is the same as that in the original cylinder (with the hole). The second is the solid cylinder that fills the hole. It has a current density with the same magnitude as that of the original cylinder but is in the opposite direction. If these two situations are superposed the total current in the region of the hole is zero. Now, a solid cylinder carrying current i, uniformly distributed over a cross section, produces a magnetic field with magnitude B = µ 0ir 2πR 2 a distance r from its axis, inside the cylinder. Here R is the radius of the cylinder. For the cylinder of this problem the current density is J = i A = i π(a 2 b 2 ), where A = π(a 2 b 2 ) is the cross-sectional area of the cylinder with the hole. The current in the cylinder without the hole is I 1 = JA = πja 2 = ia2 a 2 b 2 and the magnetic field it produces at a point inside, a distance r 1 from its axis, has magnitude Thecurrentinthecylinderthatfillstheholeis B 1 = µ 0I 1 r 1 µ 0 ir 1 a 2 2πa 2 = 2πa 2 (a 2 b 2 ) = µ 0 ir 1 2π(a 2 b 2 ). I 2 = πjb 2 = ib2 a 2 b 2 and the field it produces at a point inside, a distance r 2 from the its axis, has magnitude B 2 = µ 0I 2 r 2 µ 0 ir 2 b 2 2πb 2 = 2πb 2 (a 2 b 2 ) = µ 0 ir 2 2π(a 2 b 2 ). At the center of the hole, this field is zero and the field there is exactly the same as it would be if theholewerefilled.placer 1 = d in the expression for B 1 and obtain B = µ 0 id 2π(a 2 b 2 ) for the field at the center of the hole. The field points upward in the diagram if the current is out of the page. (b) If b = 0 the formula for the field becomes B = µ 0id 2πa 2. This correctly gives the field of a solid cylinder carrying a uniform current i, at a point inside the cylinder a distance d from the axis. If d = 0 the formula gives B = 0. This is correct for the field on the axis of a cylindrical shell carrying a uniform current. (c) Consider a rectangular path with two long sides (side 1 and 2, each with length L) and two short sides (each of length less than b). If side 1 is directly along the axis of the hole, then side 2 would be also parallel to it and also in the hole. To ensure that the short sides do not contribute significantly to the integral in Ampere s law, we might wish to make L very long (perhaps longer than the length

13 44. (a) The ideal solenoid is long enough (and we are evaluating the field at a point far enough inside) such that the open ends of the solenoid are out of sight and the situation displays a horizontaltranslational symmetry (assuming the axis of the cylindrical shape of the solenoid is horizontal). A view of a slice of, say, the bottom of the solenoid would therefore appear similar to that shown in Fig , where point P is in the interior of the solenoid and point P is outside the coil. Now, Fig differs in at least one respect from our slice view of the solenoid in that the field at P would be zero instead of what is shown in that figure. The field vanishes there because the top of the solenoid (similar to that shown in Fig , in slice view, but with the currents and field directions reversed) would contribute an equal and opposite field to any exterior point, thus canceling it. For interior points, the top and bottom slices each contribute 1 2 µ 0λ (in the same direction) [this is shown in the solution to problem 39] and thus produce an interior field equal to B = µ 0 λ. (b) Applying Ampere s law to a rectangular path which passes through points P (interior) and P (exterior) similar to that described in the solution to part (b) of problem 39, we are not surprised to find B d s = ( BP B P ) î x = µ 0 λ x just as we found in part (b) of problem 39 (except that we are now taking the +x directioninthe same direction as the field ( at P, to avoid confusion with signs). The difference with the previous solution is that in 39, BP B P ) î wasequaltob ( B) =2B, whereas in this case we have B( 0=B. Although the value of B is different in the two problems, we see that the change BP B P ) î isthesame:µ 0 λ.

14 51. (a) The magnitude of the magnetic dipole moment is given by µ = NiA,whereN is the number of turns, i is the current, and A is the area. We use A = πr 2,whereR is the radius. Thus, µ = NiπR 2 = (300)(4.0A)π(0.025 m) 2 =2.4 A m 2. (b) The magnetic field on the axis of a magnetic dipole, a distance z away, is given by Eq : B = µ 0 µ 2π z 3. We solve for z: z = ( µ0 2π µ B ) ( 1/3 (4π 10 7 T m/a)(2.36 A m 2 ) 1/3 ) = 2π( =46cm. T)