Problem 27.15 An electron at point A has a speed of 1.41 x 10 6 m/s. Find (a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B, and (b) the time required for the electron to move from A to B. Fig. 27.46 PHYS 153 08W 1
Problem 27.31 Determining the mass of an isotope. The electric field between the plates of the velocity selector in the Bainbridge mass spectrometer (Fig. 27.24) is 1.12 x 10 5 V/m, and the magnetic field is 0.540 T. A stream of singly charged selenium ions moves in a circular path with a radius of 31.0 cm in the magnetic field. Determine the mass of one selenium ion and the mass number of the selenium isotope. (The mass number is equal to the mass of the isotope in atomic mass units, rounded to the nearest integer. One atomic mass unit = 1u = 1.66 x 10-27 kg). PHYS 153 08W Fig. 27.24 2
The force exerted by a magnetic field. The force on a single charge is given by F = qvxb x x x x x x x x x x x x x x x x x x x x x positive charge entering a magnetic field When a wire carries a current, in a magnetic field, there is a force on the wire that is equal to the sum of the magnetic forces on the charged particles whose motion produces the current. The number of charges in length l is nal, where n is the density and Al is the volume. = F ( qvd xb) nal But the current I = nqv d A PHYS 153 08W 3 Fig. 27.25
Hence the force can be written F = I l xb Note: Current is not a vector. is. The direction in which current flows is given by, or. D.C. motor I l d l l In everyday life, there are countless examples of motors. It is important to understand how they work. First though, we have to learn about - Forces and torques on a current loop. Recall for a current carrying wire in a magnetic field, F = IlxB= F = IlB when IlB sinφ o φ =90 The diagrams illustrate these two cases. To understand this is important when studying the case of torques on a current carrying loop of wire in a B field. I (valid for both + and charges) B PHYS 153 08W 4 l φ l B I I
This figure shows a current carrying loop of wire in a magnetic field. A normal to the plane of the loop makes an angle φ to the magnetic field, B. Consider the sides of length b. The force on each side is ' F = IbBsin(90 φ ) = IbB cosφ They are designated and F ' j ' F( j) Fig. Fig. 27.31 27.31 and are equal and opposite. There is no net force in the direction. Consider the sides of length a. The force on each side is F =IaB These forces are designated and i F F i ) ( and are equal and opposite. There is no net force in the direction. ± PHYS 153 08W 5 y ±x
But there is a torque acting on the sides of length a. τ = 2F ( b )sinφ = ( IaB)( bsinφ) = IABsinφ 2 o o Torque is a maximum when φ =90 (Fig. 27.31b), and zero when φ =0, or (Fig. 27.31c). o 180 We can write torque as τ µ Bsinφ µ =IA NIA = where (or where is the number of turns of wire) is the magnetic dipole moment of the loop. A µ I Since is a vector, is a vector and the direction is given by the R.H. rule: Place the fingers of the R.H. in the direction of and the thumb points in µ the direction of (when it is perpendicular to the plane of the loop). τ = µxb (recall the similarity to the torque exerted by an electric field on an electric dipole moment, τ = pxe ). N PHYS 153 08W 6
Using this symmetry and recalling that the potential energy of an electric dipole is, can make a reasonable guess that the potential U = p E energy for a magnetic dipole moment is U = µ B U Is greatest when µ and B are antiparallel, and least when they are parallel. when and are perpendicular to each other. U =0 µ B Note that the rectangular current loop is not a special case. The relationships above are valid for a plane loop of any shape. The irregular shape to the right can be approximated by an infinite number of rectangles. The currents in adjacent sides cancel and that leaves just the current in the outer boundary. The results above are valid for each of the rectangles, and hence valid for the irregular shape. Fig. 27.33 PHYS 153 08W 7
Question: What is the torque on a solenoid situated in a magnetic field as shown? For a single loop, τ =IAB sinφ = µ Bsinφ A solenoid is equivalent to N circular loops, side by side. So in this case, µ = NIA The torque as usual is given by τ = µ Bsinφ, so for the solenoid, τ = NIABsinφ Fig. 27.34 Now we are ready to discuss the D-C motor. The above ideas of current carrying loops and torques are applicable in this case. PHYS 153 08W 8
Fig. 27.39 A current carrying loop lies in a magnetic field generated by a permanent magnet. The battery is connected to the coil through commutators which keep the current flowing in the same direction in the loop as it turns. µ In (a), note the orientation of with respect to. The torque will turn the loop counter-clockwise. In (b), each brush which touches the commutators touches both commutators simultaneously. B PHYS 153 08W 9
I =0 µ = 0 = 0 Hence, through the coil,, and. Momentum keeps the loop turning. In (c), the coil has turned o with respect to (a), but because of the µ 180 commutators, is in the same direction as it was in (a) and the torque will continue to turn the loop counter-clockwise. The coil will turn at top speed when the turning torque is balanced by a resisting torque due to air resistance and friction. For N turns of wire in the loop, the torque will be N times stronger. Question: What improvements could be made to this simple motor set-up to make it more efficient? Back emfs When a wire moves in a magnetic field, a current is induced in it to oppose the motion causing it. In the motor discussed above, the induced current is in the direction opposite to the current shown. This induced emf is called a back emf. We will study this in more detail later. τ PHYS 153 08W 10
Hall Effect The Hall effect is the classic way of determining the sign of the mobile charge carriers in a material, and the density of current carrying charges in the material. A current flows to the right in the +x direction. For a metal conductor, the drift velocity, v d, of the charge carriers (electrons) is to the left (diagram (a)). For a semiconductor, for example, the drift velocity of the charge carriers (holes) is to the right (diagram (b)). In both cases, the magnetic field produces a force on the moving charges in the +z direction. Fig. 27.41 PHYS 153 08W 11
In (a), an excess of negative charges accumulates near the top edge of the conductor. In (b), the charge accumulating along the top is positive. The net result of this is to create an electric field to oppose the motion of the charge, negative charge in (a), and positive charge in (b). The transverse potential difference is called the Hall voltage. Eventually, the electrostatic force balances the magnetic force. qe( k) + qvdb( k) = 0, and E= vdb v (when is negative, is positive, and vice versa). d E J x = nqvd From before, the current density Eliminate v d from these last two equation and obtain, nq = J x E B y (Hall effect) E J and x By are positive. If is positive (diagram (a)), then nq is negative, and vice versa (diagram (b)). PHYS 153 08W 12
Since the magnitude of q is known (equal to the electron charge), then n can be calculated. For a metal conductor, How magnets work n 29 ~10 m 3 What will happen to a bar magnet when it is placed in a magnetic field, and at some angle to it? It will rotate until it is aligned with the field. This means that the bar magnet has a magnetic dipole moment. The origin of this dipole moment is the electron which spins and acts somewhat like a current loop (look back on slide 5). An electron has a magnetic moment. In a permanent magnet, these dipoles are well aligned. The magnetic dipole moment for a S permanent magnet points from S to N. µ B bar magnet B N PHYS 153 08W 13
Fig. 27.37 In an unmagnetized object (e.g. iron in diagram (a)), there is no overall alignment of the magnetic moments, and the vector sum is zero. In the presence of a magnet, the magnetic moments align with the B field and the object acquires a net magnetic moment (diagram (b)). This new magnetic moment of the object (e.g. the magnet above, or an iron nail) is the reason why the object is attracted by the B field of the magnet. Attraction and repulsion µ Fig. 27.36 µ Fig. 27.37 PHYS 153 08W 14
In Fig. 27.36 (a), the current loop has a magnetic dipole moment which is anti-aligned with the magnetic moment of the bar magnet. The force on a section of the loop has a radial component, and a component to the right. The radial components cancel and the net force on the loop is to the right. So, when the magnetic moments are anti-aligned, the loop is repelled from the bar magnet. In (b), the current loop has a magnetic dipole moment which is aligned with that of the bar magnet. The net force on the current loop is now to the left. So, when the magnetic moments are aligned, the loop is attracted to the bar magnet. For a class of materials which are called ferromagnetic (iron, nickel, cobalt and alloys of these) which are unmagnetized, the atomic magnetic moments tend to align with the B field when in the presence of a permanent magnet. The aligned dipole moments are represented by (b) in Fig. 27.36, and the object is attracted to the magnet. Also (b) explains the attraction between two bar magnets end to end, with the S pole of one next to the N pole of the other. Diagram (a) in Fig. 27.36 explains the repulsion between two bar magnets when similar poles are put next to each other. PHYS 153 08W 15
Problem 27.51 The figure shows a portion of a silver ribbon with z 1 =11.8 mm and y 1 = 0.23 mm, carrying a current of 120 A in the +x direction. The ribbon lies in a uniform magnetic field, in the y direction, with magnitude 0.95 T. If there are 5.85 x 10 28 free electrons per cubic metre, use the simplified model of the Hall effect to find (a) the magnitude of the drift velocity of the electrons in the x direction, (b) the magnitude and direction of the electric field in the z direction due to the Hall effect, (c) the Hall emf. PHYS 153 08W 16