http://iml.umkc.edu/physics/wrobel/phy250/homework.htm Homework 9 chapter 29: 23, 29, 35, 61 Problem 29.23 The picture in a television uses magnetic deflection coils rather than electric deflection plates. Suppose an electron beam is accelerated through a 50 kV potential difference and then through a region of uniform magnetic field 1 cm wide. The screen is located 10 cm from the center of the coils and is 50 cm wide. When the field is turned off, the electron beam hits the center of the screen. What field magnitude is necessary to deflect the beam to the side of the screen? L r α y V D The electrons leave the gun with a speed dependent on the potential difference. From the work-energy theorem the speed of the electrons is v= 2 Wel 2eV = m m where m is the electron mass, e the elementary charge and V the accelerating potential difference. When the electrons enter the magnetic field, they move along a circular path due to the magnetic interaction. The radius of the path can be determined from Newton’s second law – the centripetal acceleration is proportional to the magnetic force r= mv eB sin 90° 2 The radius of the circular path (and the width of the magnetic field) affects the deflections angle sin α = D r That angle also determines where the electron beam hits the screen y ≈ L ⋅ tan α Hence B= mv = er e ⋅ m⋅ 2eV m D sin arctan = 2meV y ⋅ sin arctan L eD y L For the beam to reach a side of the screen, the magnetic field strength must be Bmax = ( )( ) ⋅ sin arctan 25cm = 70.4mT 10cm C ) ⋅ 0.01m 2 ⋅ 9.1 ⋅ 10 − 31 kg ⋅ 50 ⋅ 103 V (1.6 ⋅10 −19 3 Problem 29.29 A rod of mass 0.720 kg and radius 6 cm rests on two parallel rails that are 12 cm apart and 45 cm long. The rod carries a current of 48 A (in the direction shown in the figure) and rolls along the rails without slipping. A uniform magnetic field of magnitude 0.240 T is directed perpendicular the rod and the rails. If it starts from rest, what is the speed of the road as it leaves the rails? B d I F L The simplest solution is based on the work energy theorem. Three objects interact with the rod – the earth (gravity), the rails (normal and static friction) and the (source of) magnetic field (magnetic). Assuming that the rails are horizontal, only the magnetic interaction results in non-zero total work Wnet = WB = ∫ FB ⋅ ds = I ⋅ d ⋅ B ⋅ sin 90° ⋅ L ⋅ cos 0° = IBLd path This work must be equal to the change in the total kinetic energy of the rod. K tot = K rot 2 1 ⎛1 Mv 2 3Mv 2 2⎞v + K trans. = ⋅ ⎜ Mr ⎟ 2 + = 2 ⎝2 2 4 ⎠r Therefore v= 4 Wnet 4IBLd 4 ⋅ 48A ⋅ 0.24T ⋅ 0.45m ⋅ 0.12m m = = = 1.07 3M 3M 3 ⋅ 0.72kg s (Note that the radius of the rod has no effect on the speed.) 4 Problem 29.35 A rectangular loop consists of N = 100 closely wrapped turns and has dimensions a = 0.40 m by b = 0.30 m. The loop is hinged along the y axis, and the plane of the coil makes an angle of θ = 30° with the x axis (Fig. P29.25). What is the magnitude of the torque exerted on the loop by a uniform magnetic field of B = 0.8 T directed along the x axis when the current has a value of I = 1.2 A in the direction shown? What is the expected direction of rotation of the loop? y The magnetic torque exerted on an object with magnetic moment μ is related to the magnetic field vector (definition of the magnetic moment): (1) τ = μ×B I=1.2A μ a =0.4m B τ x α =30 b =0.3m z From the text of the problem we know that the magnetic field is in the x-direction: B = B, 0, 0 (2) We have to find the magnetic moment of the loop. The direction of the loop's magnetic moment is perpendicular to the surface of the loop following the right-hand rule. I marked the direction in the figure. Its magnitude depends on the number of turns, the electric current in the wire and the area of the loop. μ = nIab Consistent with the choice of the coordinate system, the magnetic moment is therefore: μ = nIabsin α , 0, − nIabcos α (3) Using (2) and (3) in (1) we find the magnetic torque: 5 τ = μ yBz − μ z By , μ z Bx − μ x Bz , μ x By − μ yBx = 0, − nIBabcos α , 0 = = 0, − 100 ⋅1. 2 A ⋅ 0.8T ⋅ 0. 4 m ⋅ 0.3m ⋅ cos 30o , 0 ≈ 0,10, 0 Nm Recalling Newton's second law, we can relate the net torque with the rate of change in the angular momentum. dL = τ net dt The hinges and the gravity cause a horizontal torque, which cannot affect the vertical component of the angular momentum. The hinges constrain the loop to rotation about the y-axis. For a rectangular loop, the direction of angular momentum L coincides with the direction of angular velocity ω. Considering the y-components we obtain: dω Ia =τ dt where Ia is the moment of inertia of the loop about the axis of rotation. If initially the loop did not rotate, the magnetic moment will cause it to rotate with angular velocity down. 6 Problem 29.61 A heart surgeon monitors the flow rate of blood through an artery using an electromagnetic flowmeter. Electrodes A and B make contact with the outer surface of the blood vessel, which has interior diameter 3 mm. (a) For a magnetic field magnitude of 0.04 T, an emf of 160 μV appears between the electrodes. Calculate the speed of the blood. (b) Verify that electrode A has higher potential, as shown. Does the sign of the emf depend on whether the mobile ions in the blood are predominantly positively or negatively charged? Explain. a) The ions move along the vessel while the magnetic force is transverse to the vessel. The redistritbution of ions results in a transverse electric field resulting in a force on the ions in the opposite direction Artery + A F B N v - B to voltmeter S Blood flow qE = − qv × B or in terms of magnitudes qE = qvB sin 90° Hence the speed of the ions must be v= 160μV m E V = = = 1.33 s B Bd 0.04T ⋅ 3mm b) Consistent with the right hand rule, the direction of the magnetic force exerted on the ions with a positive charge is upward while the force exerted on the negatively charged ions is downward. In both cases, the polarity of the electric field is consistent with the indicated potential of the electrodes. Both types of ions contribute to the electromotive force. 7