Chapter 9 FARADAY'S LAW Recommended Problems:

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1 Chapter 9 FARADAY'S LAW Recommended Problems: 5,7,9,10,11,13,15,17,20,21,28,29,31,32,33,34,49,50,52,58,63,64.

2 Faraday's Law of Induction We learned that e. current produces magnetic field. Now we want to verify that m. field produces, in some conditions, e. current. Consider a loop of wire that is connected to a galvanometer and a magnet is moved in the vicinity of the loop. It is observed experimentally that

3 (a) If both the loop and the magnet held stationary relative to each other there will be no deflection in the galvanometer. N S (b) If the loop is held stationary while the magnet is moved toward the loop, the galvanometer will deflect in one direction. (a) N S (c) If the magnet is moved away from the loop, the galvanometer will deflect in the opposite direction. The same observations is occurred when the loop is moved while the magnet is held stationary. (b) (c) N S

4 From these observations one concludes that a current is produced in the loop as long as there is relative motion between the loop and the magnet. Such a current is called the induced current, and its source is called the induced emf. The phenomenon itself (the production of electric current from changing magnetic flux) is called the electromagnetic induction. Michael Faraday studied these observations quantitatively and put them in a mathematical formula considered as one of the fundamental laws in electromagnetic theory. He found that the induced emf is proportional to the rate of change of the magnetic flux, i.e., N d dt m N is the number of turns in the loop. For a uniform magnetic field the magnetic flux becomes m BAcos is the angle between the magnetic field and the area

5 N d BAcos dt From this expression we conclude that an induced emf can be created if either B, A,, or a combination of them vary with time. The minus sign in Faraday's law is a consequence of the law of conservation of energy. Example 31.1 A coil consists of 200 turns of wire. Each turn is a square of side 18 cm, and a uniform m. field perpendicular to the plane of the coil is turned on. If B=0.5 T in 0.8 s, what is the induced emf in the coil. Solution: Knowing that, and the area A is constant we have db db NAcos NA dt dt 2 200(0.18) V

6 Example 31.3 Two bulbs are connected to opposite sides of a circular loop of wire, as shown. A changing magnetic field (confined to the smaller circular area shown in the figure) induces an emf in the loop that causes the two bulbs to light. When the switch is closed, the resistance-free wires connected to the switch short out bulb 2 and it goes out. What happens if the wires containing the closed switch are lifted up and moved to the other side of the field, as in Figure 3.18b? The wire is still connected to bulb 2 as it was before, so does it continue to stay dark?

7 Test Your Understanding (1) A circular loop of wire is held in a uniform magnetic field, with the plane of the loop perpendicular to the field lines. Which of the following will not cause a current to be induced in the loop? a) crushing the loop; b) rotating the loop about an axis perpendicular to the field lines; c) keeping the orientation of the loop fixed and moving it along the field lines; (d) pulling the loop out of the field.

8 MOTIONAL EMF To understand how the induced emf is originated we now study in details the nature of the induced emf. Consider a rod of length l moving with speed v in a uniform magnetic field B directed into the page The free electrons inside the rod will experience a magnetic force according to B l v E F m evi ˆ Bkˆ evb ˆj The electrons then will accumulate at the lower end of the rod leaving a net positive at its upper end. As a result of this charge separation, a net electric field E will be set up inside the rod. Therefore, free electrons now will be affected by an upward electric force F e =ee in addition to the magnetic field.

9 Charges continue to build up at the ends of the rod until the two forces balanced. At this point motion of charges ceases leaving the rod with two opposite polarities at its end, that is an emf is produced across the rod. To calculate this emf we have, from the equilibrium condition F m F e evb ee E vb Since the electric field in uniform inside the rod, the potential difference across the rod is related to this electric field according to V El vlb This potential difference is maintained across the ends of the rod as long as it is moving in the field and is called the motional emf.

10 Let us now prove the last equation using Faraday's law. Consider now the rod is sliding along conducting rails in the magnetic field such that it forms a closed loop. As the area of the loop at some instant is A=lx, the magnetic flux through the loop is m Blx l B I x l (b) v Where x, the width of the loop, is changing with time as the rod moves. Using Faraday's law, we find that the induced emf in the loop is d m d dx dt dt dt Blx Bl Blv Which is the same result except of the minus sign.

11 Example 31.2 A conducting bar of length l rotates with a constant angular speed about one end. A uniform m. B is perpendicular to the plane of the rotation. What is the potential difference induced between the ends of the bar. Solution v It is clear that v is not constant along the length of the bar We have to divide the bar into small elements each of length dr. Now the emf across one of these e elements is r l dv vbdr Brdr Integrate to find the emf across the bar V B L 0 rdr 1 2 Bl 2

12 Lenz s Law It tells us that the induced current must be in a direction such that it produces a magnetic field to oppose the change in the magnetic flux. Lenz's law can be explained by the following two rules: (i) If the magnetic flux through the loop is increasing, the direction of the induced current is such that it produces a magnetic field opposite to the source magnetic field. (ii) If the magnetic flux through the loop is decreasing, the direction of the induced current is such that it produces a magnetic filed in the same direction as that of the source magnetic field.

14 Test Your Understanding (2) The figure shows a circular loop of wire being dropped toward a wire carrying a current to the left. The direction of the induced current in the loop of wire is a) clockwise (b) counterclockwise c) zero d) impossible to determine.

15 Example 31.2 A square loop of side L and resistance R moves with constant speed v through a region of width 4L in which a uniform magnetic field B directed out of page, as shown. Plot the flux, the induced emf, and the force on the loop as a function of x, the position of the right side of the loop. 4L x Solution

16 4L It is clear that is zero as long as the loop is totally outside the field. It increases as it is entering and decreasing as it is leaving the field. It is constant =Bl 2 when it is totally inside the field. is zero as long as is constant. Its direction while is increasing is opposite to its direction while it is decreasing. The force is zero as long as =0. As the loop is entering, the force is on its right side while the force is on the left side as the loop is leaving the field. In both cases the m.force is in the same direction. F x

17 Test Your Understanding (3) The figure shows a rectangular loop moves to the right at constant velocity through a region of uniform magnetic field directed out of the page. Which of the following is correct concerning the loop at its middle position? a) Both the magnetic flux and the induced emf is zero b) The magnetic flux is maximum while the induced emf is zero c) Both the magnetic flux and the magnetic force is zero d) Both the magnetic flux and the induced emf is maximum.

18 INDUCED ELECTRIC FIELD A changing magnetic flux creates an induced emf and thus an induced current in a conducting loop. Therefore, an electric field must be present along the loop. This field, which is created by changing magnetic flux, is called induced electric field and given by. in E ds Using Faraday s law we get E ds d dt m It should be noted that this result is also valid for any hypothetical closed path.

19 The induced electric field is quite different from the electrostatic field (produced by static charges). The induced e.f is a nonconservative field produced by a changing magnetic flux. Hence no electric potential can be associated by the induced electric field. The potential difference between two points i and f is V f V i f E ds i which would be zero for a closed path, contrary to the induced e.f. The direction of the induced e.f. is determined by Lenz's rule

20 Example 31.8 A long solenoid of radius R has n turns of wire per unit length and carries a current given by, with B o and are constant and the time t is in seconds. Calculate the induced electric field inside and outside the Solenoid. Solution Let first draw an imaginary closed loop of radius r < R The m. flux through the area enclosed by this closed loop is r n r I cost m BAcos ni E ds d dt o m 2 n r o o I 2 max sin t By symmetry, the magnitude of E is constant around the path and tangent to it 2 max r R

21 2 ( max E 2r ) on r I sin t To calculate the electric field outside the sphere, the closed path has now a radius r R. Since the magnetic field is confined only to the region r R, the magnetic flux through the path is m R 2 2 n R I cos t oni o The electric field is now max n I E o max r sin t 2 d d m 2 E s E2r on R Imax sin t dt n I E o 2 max R sin t 2r R r

22 Test Your Understanding (4) In a region of space, the magnetic field increases at a constant rate. This changing magnetic field induces an electric field that a) increases in time b) is conservative c) is in the direction of the magnetic field d) has a constant magnitude.

Chapter 27, 28 & 29: Magnetism & Electromagnetic Induction. Magnetic flux Faraday s and Lenz s law Electromagnetic Induction Ampere s law

Chapter 27, 28 & 29: Magnetism & Electromagnetic Induction Magnetic flux Faraday s and Lenz s law Electromagnetic Induction Ampere s law 1 Magnetic Flux and Faraday s Law of Electromagnetic Induction We