Faraday’s Law of Induction

Faraday’s Law of Induction 10.1 Faraday’s Law of Induction The electric fields and magnetic fields considered up to now have been produced by stationary charges and moving charges (currents), respectively. Imposing an electric field on a conductor gives rise to a current which in turn generates a magnetic field. One could then inquire whether or not an electric field could be produced by a magnetic field. In 1831, Michael Faraday discovered that, by varying magnetic field with time, an electric field could be generated. The phenomenon is known as electromagnetic induction. Figure 10.1.1 illustrates one of Faraday’s experiments. Figure 10.1.1 Electromagnetic induction Faraday showed that no current is registered in the galvanometer when bar magnet is stationary with respect to the loop. However, a current is induced in the loop when a relative motion exists between the bar magnet and the loop. In particular, the galvanometer deflects in one direction as the magnet approaches the loop, and the opposite direction as it moves away. Faraday’s experiment demonstrates that an electric current is induced in the loop by changing the magnetic field. The coil behaves as if it were connected to an emf source. Experimentally it is found that the induced emf depends on the rate of change of magnetic flux through the coil. 1 10.1.1 Magnetic Flux Consider a uniform magnetic field passing through a surface S, as shown in Figure 10.1.2 below: Figure 10.1.2 Magnetic flux through a surface G Let the area vector be A = A nˆ , where A is the area of the surface and n̂ its unit normal. The magnetic flux through the surface is given by G G Φ B = B ⋅ A = BA cos θ (10.1.1) G where θ is the angle between B and n̂ . If the field is non-uniform, Φ B then becomes G G Φ B = ∫∫ B ⋅ dA (10.1.2) S The SI unit of magnetic flux is the weber (Wb): 1 Wb = 1 T ⋅ m 2 Faraday’s law of induction may be stated as follows: The induced emf ε in a coil is proportional to the negative of the rate of change of magnetic flux: ε =− dΦB dt (10.1.3) For a coil that consists of N loops, the total induced emf would be N times as large: ε = −N dΦB dt (10.1.4) G Combining Eqs. (10.1.3) and (10.1.1), we obtain, for a spatially uniform field B , 2 ε =− d ⎛ dB ⎞ ⎛ dA ⎞ ⎛ dθ ⎞ ( BA cos θ ) = − ⎜ ⎟ A cos θ − B ⎜ ⎟ cos θ + BA sin θ ⎜ ⎟ dt ⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠ (10.1.5) Thus, we see that an emf may be induced in the following ways: G (i) by varying the magnitude of B with time (illustrated in Figure 10.1.3.) Figure 10.1.3 Inducing emf by varying the magnetic field strength G (ii) by varying the magnitude of A , i.e., the area enclosed by the loop with time (illustrated in Figure 10.1.4.) Figure 10.1.4 Inducing emf by changing the area of the loop G G (iii) varying the angle between B and the area vector A with time (illustrated in Figure 10.1.5.) G G Figure 10.1.5 Inducing emf by varying the angle between B and A . 3 10.1.2 Lenz’s Law The direction of the induced current is determined by Lenz’s law: The induced current produces magnetic fields which tend to oppose the change in magnetic flux that induces such currents. To illustrate how Lenz’s law works, let’s consider a conducting loop placed in a magnetic field. We follow the procedure below: G 1. Define a positive direction for the area vector A . G G G 2. Assuming that B is uniform, take the dot product of B and A . This allows for the determination of the sign of the magnetic flux Φ B . 3. Obtain the rate of flux change d Φ B / dt by differentiation. There are three possibilities: dΦB : dt ⎧> 0 ⇒ induced emf ε < 0 ⎪ ⎨< 0 ⇒ induced emf ε > 0 ⎪= 0 ⇒ induced emf ε = 0 ⎩ 4. Determine the direction of the induced current using the right-hand rule. With your G thumb pointing in the direction of A , curl the fingers around the closed loop. The induced current flows in the same direction as the way your fingers curl if ε > 0 , and the opposite direction if ε < 0 , as shown in Figure 10.1.6. Figure 10.1.6 Determination of the direction of induced current by the right-hand rule In Figure 10.1.7 we illustrate the four possible scenarios of time-varying magnetic flux and show how Lenz’s law is used to determine the direction of the induced current I . 4 (b) (a) (c) (d) Figure 10.1.7 Direction of the induced current using Lenz’s law The above situations can be summarized with the following sign convention: ΦB + − d Φ B / dt ε + − + − − + − + I − + − + The positive and negative signs of I correspond to a counterclockwise and clockwise currents, respectively. As an example to illustrate how Lenz’s law may be applied, consider the situation where a bar magnet is moving toward a conducting loop with its north pole down, as shown in Figure 10.1.8(a). With the magnetic field pointing downward and the area vector G A pointing upward, the magnetic flux is negative, i.e., Φ B = − BA < 0 , where A is the area of the loop. As the magnet moves closer to the loop, the magnetic field at a point on the loop increases ( dB / dt > 0 ), producing more flux through the plane of the loop. Therefore, d Φ B / dt = − A(dB / dt ) < 0 , implying a positive induced emf, ε > 0 , and the induced current flows in the counterclockwise direction. The current then sets up an induced magnetic field and produces a positive flux to counteract the change. The situation described here corresponds to that illustrated in Figure 10.1.7(c). Alternatively, the direction of the induced current can also be determined from the point of view of magnetic force. Lenz’s law states that the induced emf must be in the direction that opposes the change. Therefore, as the bar magnet approaches the loop, it experiences 5 a repulsive force due to the induced emf. Since like poles repel, the loop must behave as if it were a bar magnet with its north pole pointing up. Using the right-hand rule, the direction of the induced current is counterclockwise, as view from above. Figure 10.1.8(b) illustrates how this alternative approach is used. Figure 10.1.8 (a) A bar magnet moving toward a current loop. (b) Determination of the direction of induced current by considering the magnetic force between the bar magnet and the loop 10.2 Motional EMF Consider a conducting bar of length l moving through a uniform magnetic field which points into the page, as shown in Figure 10.2.1. Particles with charge q > 0 inside G G G experience a magnetic force FB = qv × B which tends to push them upward, leaving negative charges on the lower end. Figure 10.2.1 A conducting bar moving through a uniform magnetic field G The separation of charge gives rise to an electric field E inside the bar, which in turn G G produces a downward electric force Fe = qE . At equilibrium where the two forces cancel, 6 we have qvB = qE , or E = vB . Between the two ends of the conductor, there exists a potential difference given by Vab = Va − Vb = ε = El = Blv (10.2.1) Since ε arises from the motion of the conductor, this potential difference is called the motional emf. In general, motional emf around a closed conducting loop can be written as G G G ε = v∫ ( v × B) ⋅ d s (10.2.2) G where d s is a differential length element. Now suppose the conducting bar moves through a region of uniform magnetic field G B = − B kˆ (pointing into the page) by sliding along two frictionless conducting rails that are at a distance l apart and connected together by a resistor with resistance R, as shown in Figure 10.2.2. Figure 10.2.2 A conducting bar sliding along two conducting rails G Let an external force Fext be applied so that the conductor moves to the right with a G constant velocity v = v ˆi . The magnetic flux through the closed loop formed by the bar and the rails is given by Φ B = BA = Blx (10.2.3) Thus, according to Faraday’s law, the induced emf is ε =− dΦB d dx = − ( Blx ) = − Bl = − Blv dt dt dt (10.2.4) where dx / dt = v is simply the speed of the bar. The corresponding induced current is I= | ε | Blv = R R (10.2.5) 7 and its direction is counterclockwise, according to Lenz’s law. The equivalent circuit diagram is shown in Figure 10.2.3: Figure 10.2.3 Equivalent circuit diagram for the moving bar The magnetic force experienced by the bar as it moves to the right is G ⎛ B 2l 2 v ⎞ ˆ FB = I (l ˆj) × (− B kˆ ) = − IlB ˆi = − ⎜ ⎟i ⎝ R ⎠ (10.2.6) G which is in the opposite direction of v . For the bar to move at a constant velocity, the net force acting on it must be zero. This means that the external agent must supply a force G G ⎛ B 2l 2 v ⎞ ˆ Fext = −FB = + ⎜ ⎟i ⎝ R ⎠ (10.2.7) G The power delivered by Fext is equal to the power dissipated in the resistor: G G ⎛ B 2l 2 v ⎞ ( Blv)2 ε 2 = = = I 2R P = Fext ⋅ v = Fext v = ⎜ v ⎟ R R ⎝ R ⎠ (10.2.8) as required by energy conservation. From the analysis above, in order for the bar to move at a constant speed, an external G agent must constantly supply a force Fext . What happens if at t = 0 , the speed of the rod is v0 , and the external agent stops pushing? In this case, the bar will slow down because of the magnetic force directed to the left. From Newton’s second law, we have FB = − B 2l 2 v dv = ma = m R dt (10.2.9) or dv B 2l 2 dt =− dt = − τ v mR (10.2.10) 8 where τ = mR / B 2l 2 . Upon integration, we obtain v (t ) = v0 e − t /τ (10.2.11) Thus, we see that the speed decreases exponentially in the absence of an external agent doing work. In principle, the bar never stops moving. However, one may verify that the total distance traveled is finite. 10.3 Induced Electric Field In Chapter 3, we have seen that the electric potential difference between two points A and G B in an electric field E can be written as B G G ∆V = VB − VA = − ∫ E ⋅ d s A (10.3.1) When the electric field is conservative, as is the case of electrostatics, the line integral of G G G G E ⋅ d s is path-independent, which implies v∫ E ⋅ d s = 0 . Faraday’s law shows that as magnetic flux changes with time, an induced current begins to flow. What causes the charges to move? It is the induced emf which is the work done per unit charge. However, since magnetic field can do not work, as we have shown in Chapter 8, the work done on the mobile charges must be electric, and the electric field in this situation cannot be conservative because the line integral of a conservative field must vanish. Therefore, we conclude that there is a non-conservative electric field G Enc associated with an induced emf: G G ε = v∫ E nc ⋅ d s (10.3.2) Combining with Faraday’s law then yields G G dΦ E v∫ nc ⋅ d s = − dt B (10.3.3) The above expression implies that a changing magnetic flux will induce a nonconservative electric field which can vary with time. It is important to distinguish between the induced, non-conservative electric field and the conservative electric field which arises from electric charges. As an example, let’s consider a uniform magnetic field which points into the page and is confined to a circular region with radius R, as shown in Figure 10.3.1. Suppose the 9 G magnitude of B increases with time, i.e., dB / dt > 0 . Let’s find the induced electric field everywhere due to the changing magnetic field. Since the magnetic field is confined to a circular region, from symmetry arguments we choose the integration path to be a circle of radius r. The magnitude of the induced field G Enc at all points on a circle is the same. According to Lenz’s law, the direction of G Enc must be such that it would drive the induced current to produce a magnetic field G opposing the change in magnetic flux. With the area vector A pointing out of the page, the magnetic flux is negative or inward. With dB / dt > 0 , the inward magnetic flux is increasing. Therefore, to counteract this change the induced current must flow G counterclockwise to produce more outward flux. The direction of Enc is shown in Figure 10.3.1. Figure 10.3.1 Induced electric field due to changing magnetic flux G Let’s proceed to find the magnitude of Enc . In the region r < R, the rate of change of magnetic flux is dΦB d G G d ⎛ dB ⎞ 2 = B ⋅ A = ( − BA ) = − ⎜ ⎟π r dt dt dt ⎝ dt ⎠ ( ) (10.3.4) Using Eq. (10.3.3), we have G G dΦ ⎛ dB ⎞ 2 E v∫ nc ⋅ d s = Enc ( 2π r ) = − dt B = ⎜⎝ dt ⎟⎠ π r (10.3.5) which implies Enc = r dB 2 dt (10.3.6) Similarly, for r > R, the induced electric field may be obtained as Enc ( 2π r ) = − d Φ B ⎛ dB ⎞ 2 =⎜ ⎟π R dt ⎝ dt ⎠ (10.3.7) 10 or R 2 dB Enc = 2r dt (10.3.8) A plot of Enc as a function of r is shown in Figure 10.3.2. Figure 10.3.2 Induced electric field as a function of r 10.4 Generators One of the most important applications of Faraday’s law of induction is to generators and motors. A generator converts mechanical energy into electric energy, while a motor converts electrical energy into mechanical energy. Figure 10.4.1 (a) A simple generator. (b) The rotating loop as seen from above. Figure 10.4.1(a) is a simple illustration of a generator. It consists of an N-turn loop rotating in a magnetic field which is assumed to be uniform. The magnetic flux varies with time, thereby inducing an emf. From Figure 10.4.1(b), we see that the magnetic flux through the loop may be written as G G Φ B = B ⋅ A = BA cos θ = BA cos ωt (10.4.1) The rate of change of magnetic flux is 11 dΦB = − BAω sin ωt dt (10.4.2) Since there are N turns in the loop, the total induced emf across the two ends of the loop is ε = −N dΦB = NBAω sin ωt dt (10.4.3) If we connect the generator to a circuit which has a resistance R, then the current generated in the circuit is given by I= | ε | NBAω = sin ωt R R (10.4.4) The current is an alternating current which oscillates in sign and has an amplitude I 0 = NBAω / R . The power delivered to this circuit is P = I | ε |= ( NBAω ) 2 sin 2 ωt R (10.4.5) On the other hand, the torque exerted on the loop is τ = µ B sin θ = µ B sin ω t (10.4.6) Thus, the mechanical power supplied to rotate the loop is Pm = τω = µ Bω sin ωt (10.4.7) Since the dipole moment for the N-turn current loop is µ = NIA = N 2 A2 Bω sin ωt R (10.4.8) the above expression becomes ⎛ N 2 A2 Bω ⎞ ( NABω ) 2 Pm = ⎜ sin ωt ⎟ Bω sin ωt = sin 2 ωt R R ⎝ ⎠ (10.4.9) As expected, the mechanical power put in is equal to the electrical power output. 12 10.5 Eddy Currents We have seen that when a conducting loop moves through a magnetic field, current is induced as the result of changing magnetic flux. If a solid conductor were used instead of a loop, as shown in Figure 10.5.1, current can also be induced. The induced current appears to be circulating and is called an eddy current. Figure 10.5.1 Appearance of an eddy current when a solid conductor moves through a magnetic field. The induced eddy currents also generate a magnetic force that opposes the motion, making it more difficult to move the conductor across the magnetic field (Figure 10.5.2). Figure 10.5.2 Magnetic force arising from the eddy current that opposes the motion of the conducting slab. Since the conductor has non-vanishing resistance R , Joule heating causes a loss of power by an amount P = ε 2 / R . Therefore, by increasing the value of R , power loss can be reduced. One way to increase R is to laminate the conducting slab, or construct the slab by using gluing together thin strips that are insulated from one another (see Figure 10.5.3a). Another way is to make cuts in the slab, thereby disrupting the conducting path (Figure 10.5.3b). Figure 10.5.3 Eddy currents can be reduced by (a) laminating the slab, or (b) making cuts on the slab. 13 There are important applications of eddy currents. For example, the currents can be used to suppress unwanted mechanical oscillations. Another application is the magnetic braking systems in high-speed transit cars. 10.6 Summary • The magnetic flux through a surface S is given by G G Φ B = ∫∫ B ⋅ dA S • Faraday’s law of induction states that the induced emf ε in a coil is proportional to the negative of the rate of change of magnetic flux: ε = −N dΦB dt • The direction of the induced current is determined by Lenz’s law which states that the induced current produces magnetic fields which tend to oppose the changes in magnetic flux that induces such currents. • A motional emf ε is induced if a conductor moves in a magnetic field. The general expression for ε is G G G ε = v∫ ( v × B) ⋅ d s G In the case of a conducting bar of length l moving with constant velocity v through a G magnetic field which points in the direction perpendicular to the bar and v , the induced emf is ε = − Bvl . • An induced emf in a stationary conductor is associated with a non-conservative G electric field Enc : G G ε = v∫ E nc ⋅ d s = − dΦB dt 10.7 Appendix: Induced Emf and Reference Frames In Section 10.2, we have stated that the general equation of motional emf is given by G G G ε = v∫ ( v × B ) ⋅ d s 14 G G where v is the velocity of the length element d s of the moving conductor. In addition, we have also shown in Section 10.4 that induced emf associated with a stationary conductor may be written as the line integral of the non-conservative electric field: G G ε = v∫ E nc ⋅ d s However, whether an object is moving or stationary actually depends on the reference frame. As an example, let’s examine the situation where a bar magnet is approaching a conducting loop. An observer O in the rest frame of the loop sees the bar magnet moving G toward the loop. An electric field Enc is induced to drive the current around the loop, and G G a charge on the loop experiences an electric force Fe = qEnc . Since the charge is at rest according to observer O, no magnetic force is present. On the other hand, an observer O ' in the rest frame of the bar magnet sees the loop moving toward the magnet. Since the G conducting loop is moving with a velocity v , a motional emf is induced. In this frame, G O ' sees the charge q moving with a velocity v , and concludes that the charge G G G experiences a magnetic force FB = qv × B . Figure 10.7.1 Induction observed in different reference frames. In (a) the bar magnet is moving, while in (b) the conducting loop is moving. Since the event seen by the two observer is the same except the choice of reference G G frames, the force acting on the charge must be the same, Fe = FB , which implies G G G Enc = v × B (10.7.1) In general, as a consequence of relativity, an electric phenomenon observed in a reference frame O may appear to be a magnetic phenomenon in a frame O ' that moves at a speed v relative to O . 10.8 Problem-Solving Tips: Faraday’s Law and Lenz’s Law In this chapter we have seen that a changing magnetic flux induces an emf: 15 ε = −N dΦB dt according to Faraday’s law of induction. For a conductor which forms a closed loop, the emf sets up an induced current I = | ε | / R , where R is the resistance of the loop. To compute the induced current and its direction, we follow the procedure below: G 1. For the closed loop of area A on a plane, define an area vector A and let it point in the direction of your thumb, for the convenience of applying the right-hand rule later. Compute the magnetic flux through the loop using G G ⎧⎪B ⋅ A ΦB = ⎨ G G ⎪⎩ ∫∫ B ⋅ dA G (B is uniform) G (B is non-uniform) Determine the sign of Φ B . 2. Evaluate the rate of change of magnetic flux d Φ B / dt . Keep in mind that the change could be caused by (i) changing the magnetic field dB / dt ≠ 0 , (ii) changing the loop area if the conductor is moving ( dA / dt ≠ 0 ), or (iii) changing the orientation of the loop with respect to the magnetic field ( dθ / dt ≠ 0 ). Determine the sign of d Φ B / dt . 3. The sign of the induced emf is the opposite of that of d Φ B / dt . The direction of the induced current can be found by using Lenz’s law discussed in Section 10.1.2. 10.9 10.9.1 Solved Problems Rectangular Loop Near a Wire An infinite straight wire carries a current I is placed to the left of a rectangular loop of wire with width w and length l, as shown in the Figure 10.9.1. 16 Figure 10.9.1 Rectangular loop near a wire (a) Determine the magnetic flux through the rectangular loop due to the current I. (b) Suppose that the current is a function of time with I (t ) = a + bt , where a and b are positive constants. What is the induced emf in the loop and the direction of the induced current? Solutions: (a) Using Ampere’s law: G G v∫ B ⋅ d s = µ I 0 enc (10.9.1) the magnetic field due to a current-carrying wire at a distance r away is B= µ0 I 2π r (10.9.2) The total magnetic flux Φ B through the loop can be obtained by summing over contributions from all differential area elements dA =l dr: G G µ Il s + w dr µ0 Il ⎛ s + w ⎞ Φ B = ∫ d Φ B = ∫ B ⋅ dA = 0 ∫ = ln ⎜ ⎟ 2π s r 2π ⎝ s ⎠ (10.9.3) Note that we have chosen the area vector to point into the page, so that Φ B > 0 . (b) According to Faraday’s law, the induced emf is ε =− µ0l ⎛ s + w ⎞ dI µ0bl ⎛ s + w ⎞ dΦB d ⎡ µ Il ⎛ s + w ⎞ ⎤ ln ⎜ ln ⎜ = − ⎢ 0 ln ⎜ ⎟⎥ = − ⎟⋅ = − ⎟ 2π ⎝ s ⎠ dt 2π dt dt ⎣ 2π ⎝ s ⎠ ⎦ ⎝ s ⎠ (10.9.4) where we have used dI / dt = b . 17 The straight wire carrying a current I produces a magnetic flux into the page through the rectangular loop. By Lenz’s law, the induced current in the loop must be flowing counterclockwise in order to produce a magnetic field out of the page to counteract the increase in inward flux. 10.9.2 Loop Changing Area A square loop with length l on each side is placed in a uniform magnetic field pointing into the page. During a time interval ∆t , the loop is pulled from its two edges and turned into a rhombus, as shown in the Figure 10.9.2. Assuming that the total resistance of the loop is R, find the average induced current in the loop and its direction. Figure 10.9.2 Conducting loop changing area Solution: Using Faraday’s law, we have ε =− ∆Φ B ⎛ ∆A ⎞ = −B ⎜ ⎟ ∆t ⎝ ∆t ⎠ (10.9.5) Since the initial and the final areas of the loop are Ai = l 2 and Af = l 2 sin θ , respectively G G (recall that the area of a parallelogram defined by two vectors l1 and l2 is G G A =| l1 × l2 |= l1l2 sin θ ), the average rate of change of area is l 2 (1 − sin θ ) ∆A Af − Ai = =− <0 ∆t ∆t ∆t (10.9.6) which gives ε= Bl 2 (1 − sin θ ) >0 ∆t (10.9.7) Thus, the average induced current is 18 I= ε R = Bl 2 (1 − sin θ ) ∆tR (10.9.8) Since ( ∆ A / ∆ t ) < 0 , the magnetic flux into the page decreases. Hence, the current flows in the clockwise direction to compensate the loss of flux. 10.9.3 Sliding Rod A conducting rod of length l is free to slide on two parallel conducting bars as in Figure 10.9.3. Figure 10.9.3 Sliding rod In addition, two resistors R1 and R2 are connected across the ends of the bars. There is a uniform magnetic field pointing into the page. Suppose an external agent pulls the bar to the left at a constant speed v . Evaluate the following quantities: (a) The currents through both resistors; (b) The total power delivered to the resistors; (c) The applied force needed for the rod to maintain a constant velocity. Solutions: (a) The emf induced between the ends of the moving rod is dΦB = − Blv dt (10.9.9) |ε | |ε | , I2 = R1 R2 (10.9.10) ε =− The currents through the resistors are I1 = Since the flux into the page for the left loop is decreasing, I1 flows clockwise to produce a magnetic field pointing into the page. On the other hand, the flux into the page for the right loop is increasing. To compensate the change, according to Lenz’s law, I2 must flow counterclockwise to produce a magnetic field pointing out of the page. (b) The total power dissipated in the two resistors is 19 ⎛1 1 ⎞ ⎛1 1 ⎞ PR = I1 | ε | + I 2 | ε |= ( I1 + I 2 ) | ε |= ε 2 ⎜ + ⎟ = B 2l 2v 2 ⎜ + ⎟ ⎝ R1 R2 ⎠ ⎝ R1 R2 ⎠ (10.9.11) (c) The total current flowing through the rod is I = I1 + I 2 . Thus, the magnetic force acting on the rod is ⎛ 1 1 ⎞ 1 ⎞ 2 2 ⎛ 1 FB = IlB =| ε | lB ⎜ + ⎟=B l v⎜ + ⎟ ⎝ R1 R2 ⎠ ⎝ R1 R2 ⎠ (10.9.12) and the direction is to the right. Thus, an external agent must apply an equal but opposite G G force Fext = −FB to the left in order to maintain a constant speed. Alternatively, we note that since the power dissipated in the resistors must be equal to Pext , the mechanical power supplied by the external agent. The same result is obtained since G G Pext = Fext ⋅ v = Fext v 10.9.4 (10.9.13) Moving Bar G A conducting rod of length l moves with a constant velocity v perpendicular to an infinitely long, straight wire carrying a current I, as shown in the Figure 10.9.4. What is the emf generated between the ends of the rod? Figure 10.9.4 A bar moving away from a current-carrying wire Solution: From Faraday’s law, the motional emf is | ε |= Blv (10.9.14) 20 where v is the speed of the rod. However, the magnetic field due to the straight currentcarrying wire at a distance r away is, using Ampere’s law: B= µ0 I 2π r (10.9.15) Thus, the emf between the ends of the rod is given by ⎛µ I⎞ ε = ⎜ 0 ⎟ lv ⎝ 2π r ⎠ 10.9.5 (10.9.16) Time-Varying Magnetic Field A circular loop of wire of radius a is placed in a uniform magnetic field, with the plane of the loop perpendicular to the direction of the field, as shown in Figure 10.9.5. Figure 10.9.5 Circular loop in a time-varying magnetic field The magnetic field varies with time according to B ( t ) = B0 + bt , where B0 and b are positive constants. (a) Calculate the magnetic flux through the loop at t = 0 . (b) Calculate the induced emf in the loop. (c) What is the induced current and its direction of flow if the overall resistance of the loop is R? (d) Find the power dissipated due to the resistance of the loop. Solution: (a) The magnetic flux at time t is given by 21 Φ B = BA = ( B0 + bt ) (π a 2 ) = π ( B0 + bt ) a 2 (10.9.17) where we have chosen the area vector to point into the page, so that Φ B > 0 . At t = 0 , we have Φ B = π B0 a 2 (10.9.18) (b) Using Faraday’s Law, the induced emf is ε =− d ( B0 + bt ) dΦB dB = −A = −(π a 2 ) = −π ba 2 dt dt dt (10.9.19) (c) The induced current is I= | ε | π ba 2 = R R (10.9.20) and its direction is counterclockwise by Lenz’s law. (d) The power dissipated due to the resistance R is 2 ⎛ π ba 2 ⎞ (π ba 2 ) 2 = P= I R=⎜ R ⎟ R ⎝ R ⎠ 2 10.9.6 (10.9.21) Moving Loop G A rectangular loop of dimensions l and w moves with a constant velocity v away from an infinitely long straight wire carrying a current I in the plane of the loop, as shown in Figure 10.9.6. Let the total resistance of the loop be R. What is the current in the loop at the instant the near side is a distance r from the wire? Figure 10.9.6 A rectangular loop moving away from a current-carrying wire Solution: The magnetic field at a distance s from the straight wire is, using Ampere’s law: 22 B= µ0 I 2π s (10.9.22) The magnetic flux through a differential area element dA = lds of the loop is G G µ I d Φ B = B ⋅ dA = 0 l ds 2π s (10.9.23) where we have chosen the area vector to point into the page, so that Φ B > 0 . Integrating over the entire area of the loop, the total flux is ΦB = µ 0 Il r + w ds µ 0 Il ⎛ r + w ⎞ = ln ⎜ ⎟ 2π ∫r s 2π ⎝ r ⎠ (10.9.24) Differentiating with respect to t, we obtain the induced emf as ε =− µ Il d ⎛ r + w ⎞ µ0 Il ⎛ 1 dΦB 1 ⎞ dr µ Il wv =− 0 − ⎟ = 0 ⎜ ln ⎟=− ⎜ 2π dt ⎝ 2π ⎝ r + w r ⎠ dt 2π r (r + w) dt r ⎠ (10.9.25) where v = dr / dt . Notice that the induced emf can also be obtained by using Eq. (10.2.2): G G ⎡ µ0 I µ0 I ⎤ − ⎥ ⎣ 2π r 2π (r + w) ⎦ G ε = v∫ ( v × B) ⋅ d s = vl [ B(r ) − B(r + w) ] = vl ⎢ µ Il vw = 0 2π r (r + w) (10.9.26) The induced current is I= | ε | µ0 Il vw = R 2π R r ( r + w ) (10.9.27) 10.10 Conceptual Questions 1. A bar magnet falls through a circular loop, as shown in Figure 10.10.1 23 Figure 10.10.1 (a) Describe qualitatively the change in magnetic flux through the loop when the bar magnet is above and below the loop. (b) Make a qualitative sketch of the graph of the induced current in the loop as a function of time, choosing I to be positive when its direction is counterclockwise as viewed from above. 2. Two circular loops A and B have their planes parallel to each other, as shown in Figure 10.10.2. Figure 10.10.2 Loop A has a current moving in the counterclockwise direction, viewed from above. (a) If the current in loop A decreases with time, what is the direction of the induced current in loop B? Will the two loops attract or repel each other? (b) If the current in loop A increases with time, what is the direction of the induced current in loop B? Will the two loops attract or repel each other? 3. A spherical conducting shell is placed in a time-varying magnetic field. Is there an induced current along the equator? 4. A rectangular loop moves across a uniform magnetic field but the induced current is zero. How is this possible? 24 10.11 Additional Problems 10.11.1 Sliding Bar A conducting bar of mass m and resistance R slides on two frictionless parallel rails that are separated by a distance A and connected by a battery which maintains a constant emf ε , as shown in Figure 10.11.1. Figure 10.11.1 Sliding bar G A uniform magnetic field B is directed out of the page. The bar is initially at rest. Show that at a later time t, the speed of the bar is v= where τ = mR / B 2 A 2 . ε BA (1 − e − t /τ ) 10.11.2 Sliding Bar on Wedges A conducting bar of mass m and resistance R slides down two frictionless conducting rails which make an angle θ with the horizontal, and are separated by a distance A, as G shown in Figure 10.11.2. In addition, a uniform magnetic field B is applied vertically downward. The bar is released from rest and slides down. Figure 10.11.2 Sliding bar on wedges (a) Find the induced current in the bar. Which way does the current flow, from a to b or b to a? (b) Find the terminal speed vt of the bar. After the terminal speed has been reached, (c) What is the induced current in the bar? 25 (d) What is the rate at which electrical energy has been dissipated through the resistor? (e) What is the rate of work done by gravity on the bar? 10.11.3 RC Circuit in a Magnetic Field Consider a circular loop of wire of radius r lying in the xy plane, as shown in Figure 10.11.3. The loop contains a resistor R and a capacitor C, and is placed in a uniform magnetic field which points into the page and decreases at a rate dB / dt = −α , with α > 0 . Figure 10.11.3 RC circuit in a magnetic field (a) Find the maximum amount of charge on the capacitor. (b) Which plate, a or b, has a higher potential? What causes charges to separate? 10.11.4 Sliding Bar A conducting bar of mass m and resistance R is pulled in the horizontal direction across two frictionless parallel rails a distance A apart by a massless string which passes over a frictionless pulley and is connected to a block of mass M, as shown in Figure 10.11.4. A uniform magnetic field is applied vertically upward. The bar is released from rest. Figure 10.11.4 Sliding bar (a) Let the speed of the bar at some instant be v. Find an expression for the induced current. Which direction does it flow, from a to b or b to a? You may ignore the friction between the bar and the rails. 26 (b) Solve the differential equation and find the speed of the bar as a function of time. 10.11.5 Rotating Bar A conducting bar of length l with one end fixed rotates at a constant angular speed ω , in a plane perpendicular to a uniform magnetic field, as shown in Figure 10.11.5. Figure 10.11.5 Rotating bar (a) A small element carrying charge q is located at a distance r away from the pivot point O. Show that the magnetic force on the element is FB = qBrω . (b) Show that the potential difference between the two ends of the bar is ∆V = 1 2 Bω l 2 . 10.11.6 Rectangular Loop Moving Through Magnetic Field A small rectangular loop of length l = 10 cm and width w = 8.0 cm with resistance R = 2.0 Ω is pulled at a constant speed v = 2.0 cm/s through a region of uniform magnetic field B = 2.0 T , pointing into the page, as shown in Figure 10.11.6. Figure 10.11.6 At t = 0 , the front of the rectangular loop enters the region of the magnetic field. (a) Find the magnetic flux and plot it as a function of time (from t = 0 till the loop leaves the region of magnetic field.) (b) Find the emf and plot it as a function of time. (c) Which way does the induced current flow? 27 10.11.7 Magnet Moving Through a Coil of Wire Suppose a bar magnet is pulled through a stationary conducting loop of wire at constant speed, as shown in Figure 10.11.7. Figure 10.11.7 Assume that the north pole of the magnet enters the loop of wire first, and that the center of the magnet is at the center of the loop at time t = 0. (a) Sketch qualitatively a graph of the magnetic flux Φ B through the loop as a function of time. (b) Sketch qualitatively a graph of the current I in the loop as a function of time. Take the direction of positive current to be clockwise in the loop as viewed from the left. (c) What is the direction of the force on the permanent magnet due to the current in the coil of wire just before the magnet enters the loop? (d) What is the direction of the force on the magnet just after it has exited the loop? (e) Do your answers in (c) and (d) agree with Lenz's law? (f) Where does the energy come from that is dissipated in ohmic heating in the wire? 10.11.8 Alternating-Current Generator An N-turn rectangular loop of length a and width b is rotated at a frequency f in a G uniform magnetic field B which points into the page, as shown in Figure 10.11.8 At time t = 0, the loop is vertical as shown in the sketch, and it rotates counterclockwise when viewed along the axis of rotation from the left. 28 Figure 10.11.8 (a) Make a sketch depicting this “generator” as viewed from the left along the axis of rotation at a time ∆t shortly after t = 0, when it has rotated an angle θ from the vertical. G Show clearly the vector B , the plane of the loop, and the direction of the induced current. (b) Write an expression for the magnetic flux Φ B passing through the loop as a function of time for the given parameters. (c) Show that an induced emf ε appears in the loop, given by ε = 2π fNbaB sin(2π ft ) = ε 0 sin(2π ft ) (d) Design a loop that will produce an emf with ε 0 = 120 V when rotated at 60 revolutions/sec in a magnetic field of 0.40 T. 10.11.9 EMF Due to a Time-Varying Magnetic Field G A uniform magnetic field B is perpendicular to a one-turn circular loop of wire of negligible resistance, as shown in Figure 10.11.9. The field changes with time as shown (the z direction is out of the page). The loop is of radius r = 50 cm and is connected in series with a resistor of resistance R = 20 Ω. The "+" direction around the circuit is indicated in the figure. Figure 10.11.9 (a) What is the expression for EMF in this circuit in terms of Bz (t ) for this arrangement? 29 (b) Plot the EMF in the circuit as a function of time. Label the axes quantitatively (numbers and units). Watch the signs. Note that we have labeled the positive direction of G the emf in the left sketch consistent with the assumption that positive B is out of the paper. [Partial Ans: values of EMF are 1.96 V, 0.0 V, 0.98 V]. (c) Plot the current I through the resistor R. Label the axes quantitatively (numbers and units). Indicate with arrows on the sketch the direction of the current through R during each time interval. [Partial Ans: values of current are 98 mA, 0.0 mA, 49 mA] (d) Plot the rate of thermal energy production in the resistor. [Partial Ans: values are 192 mW, 0.0 mW, 48 mW]. 10.11.10 Square Loop Moving Through Magnetic Field An external force is applied to move a square loop of dimension l × l and resistance R at a constant speed across a region of uniform magnetic field. The sides of the square loop make an angle θ = 45° with the boundary of the field region, as shown in Figure 10.11.10. At t = 0 , the loop is completely inside the field region, with its right edge at the boundary. Calculate the power delivered by the external force as a function of time. Figure 10.11.10 10.11.11 Falling Loop A rectangular loop of wire with mass m, width w, vertical length l, and resistance R falls out of a magnetic field under the influence of gravity, as shown in Figure 10.11.11. The G magnetic field is uniform and out of the paper ( B = B ˆi ) within the area shown and zero outside of that area. At the time shown in the sketch, the loop is exiting the magnetic G field at speed v = −v kˆ . 30 Figure 10.11.11 (a) What is the direction of the current flowing in the circuit at the time shown, clockwise or counterclockwise? Why did you pick this direction? (b) Using Faraday's law, find an expression for the magnitude of the emf in this circuit in terms of the quantities given. What is the magnitude of the current flowing in the circuit at the time shown? (c) Besides gravity, what other force acts on the loop in the ±kˆ direction? Give its magnitude and direction in terms of the quantities given. (d) Assume that the loop has reached a “terminal velocity” and is no longer accelerating. What is the magnitude of that terminal velocity in terms of given quantities? (e) Show that at terminal velocity, the rate at which gravity is doing work on the loop is equal to the rate at which energy is being dissipated in the loop through Joule heating. 31