Chapter 31 Faraday’s Law 31.1 Faraday’s Law of Induction 31.2 Motional emf Norah Ali Al- moneef

Physics C 4/16/2017 Example: Sketch the path and calculate the acceleration. E=2000 N/C, and v = 2,000 m/s (figure on the right) Norah Ali Al- moneef Bertrand

Example A wire carries a current of 25 A. What is the magnetic field 10 cm from this wire? B = moI / 2p r B = (4p X 10-7 T m/A)(25A) (2p)(0.10 m) B = 5.0 X 10-5 T Norah Ali Al- moneef

Ampere’s law: Closed Loops i1 i2 i3 ds i4 Thumb rule for sign; ignore i4 Norah Ali Al- moneef

Norah Ali Al- moneef

Example A 10 cm long solenoid has a total of 400 turns of wire and carries a current of 2.0 A. Calculate the magnetic field inside the solenoid. n = 400 turns/0.10 m = 4000 m-1 B = monI B = (4p X 10-7 T m/A)(4000m-1)(2.0 A) B = 0.01 T Norah Ali Al- moneef

Sample Problem: Sketch the path and calculate the acceleration. B = 2 Sample Problem: Sketch the path and calculate the acceleration. B = 2.0 T, and v = 2,000 m/s (for figure on the right). Physics C 4/16/2017 Norah Ali Al- moneef Bertrand

Magnetic Force What can be concluded about the sign of the charge of each particle? positive neutral 1 3 negative 2 Norah Ali Al- moneef

Example An electron travels at 2.0 X 107 m/s in a plane perpendicular to a 0.010-T magnetic field. Describe its path. Path is circular (right-hand rule, palm positive) F = m v2/r qvB = mv2/r r = mv /qB r = mv / qB r = (9.1 X 10-31 kg)(2.0 X 107 m/s ) (1.6 X 10-19 C)(0.010 T) r = 0.011 m Norah Ali Al- moneef

Sample Problem: A wire carries a current of 2 Sample Problem: A wire carries a current of 2.40 Amperes through a uniform magnetic field B=1.6 k Tesla. What is the force on a 0.75 meter long section of the wire if the current moves in the +x direction? Physics C 4/16/2017 Norah Ali Al- moneef Bertrand

Physics C Sample problem: How much current must be flowing through a horizontal 20.0 meter long wire to suspended in a 5.0 T magnetic field if the wire has a mass of 0.20 kg? 4/16/2017 If the current is flowing north, in what direction must be the magnetic field? Norah Ali Al- moneef Bertrand

Example What is the force on a wire carrying 30 A through a length of 12 cm? The magnetic field is 0.90 T and the angle is 60o. F = IlBsinq F = (30A)(0.12 m)(0.90 T)sin60o F = 2.8 N into the page Norah Ali Al- moneef

Physics C 4/16/2017 Example Two parallel wires have currents of I1 and I2 in exactly the same direction. Define the magnitude and direction of the force they exert on each other. Norah Ali Al- moneef Bertrand

Example A loop of wire carries 0.245 A and is placed in a magnetic field. The loop is 10.0 cm wide and experiences a force of 3.48 X 10-2 N downward (on top of gravity). What is the strength of the magnetic field? F = IlBsinq F = IlBsin90o F = IlB(1) F = IlB B = F Il B = 3.48 X 10-2 N = 1.42 T (0.245 A)(0.100m) Norah Ali Al- moneef

Example A magnetic pole face has a rectangular section having dimensions 200 mm by 100 mm. If the total flux emerging from the pole is 150 μ Wb. Calculate the B. Solution: Φ = 150 μ Wb = 150 x 10-6 Wb c.s.a = 200 x 100 = 20000 mm2 = 20000 x 10-6 m2 B = Φ/A = 150 x 10-6/20000 x 10-6 = 0.0075 T or 7.5mT Norah Ali Al- moneef

Force Between Two Parallel Currents Force on I2 from I1 RHR  Force towards I1 Force on I1 from I2 RHR  Force towards I2 Magnetic forces attract two parallel currents I1 I2 I2 I1 Norah Ali Al- moneef

Force Between Two Anti-Parallel Currents Force on I2 from I1 RHR  Force away from I1 Force on I1 from I2 RHR  Force away from I2 Magnetic forces repel two antiparallel currents I1 I2 I1 I2 Norah Ali Al- moneef

Parallel Currents (cont.) Look at them edge on to see B fields more clearly B B Antiparallel: repel 2 1 2 1 F F B B 2 1 Parallel: attract 2 1 F F Norah Ali Al- moneef

Example Two wires in a 2.0 m long cord are 3.0 mm apart. If they carry a dc current of 8.0 A, calculate the force between the wires. F = mo I1I2 l 2p L F = (4p X 10-7 T m/A)(8.0A)(8.0A)(2.0m) (2p)(3.0 X 10-3 m) F = 8.5 X 10-3 N Norah Ali Al- moneef

Example The top wire carries a current of 80 A. How much current must the lower wire carry in order to leviate if it is 20 cm below the first and has a mass of 0.12 g/m? F = mo l I1I2 2p L mg = mo l I1I2 Solve for I2 I2 = 15 A Norah Ali Al- moneef

Physics C 4/16/2017 Example: Use Ampere’s Law to derive the magnetic field a distance r from the center of a wire of radius R carrying current Io, where r<R. Io r B R Norah Ali Al- moneef Bertrand

Formula found from Ampere’s law i = current n = turns / meter B ~ constant inside solenoid B ~ zero outside solenoid Most accurate when L>>R Example: i = 100A, n = 10 turns/cm n = 1000 turns / m Norah Ali Al- moneef

Sample problem: Using Ampere’s Law, show that the magnetic field inside a solenoid has magnitude B = moIon Physics C 4/16/2017 Norah Ali Al- moneef Bertrand

example: Use Ampere’s Law to derive the magnetic field at various spots around the Toroid. Physics C 4/16/2017 Norah Ali Al- moneef Bertrand

Example Find the magnetic field in which the flux in 0.1m2 is 800μWb. Solution B = Φ/A = 800μWb/0.1m2 = 8000μT Norah Ali Al- moneef

Example B = Φ / A = (6 x 1) / 2 = 3 T = 1 x 10-4 Wb In an area of 2 m2 there are 6 field lines each representing a flux of 1 Wb. What is the flux density? B = Φ / A = (6 x 1) / 2 = 3 T A magnetic field has a magnitude of 0.0078 T and is uniform over a circular surface that has radius of 0.10 m. The field is orientated at an angle of of θ= 25o with respect to the surface. What is the flux through the surface? Φ = BA cos θ= 0.0078 x (π x 0.12) cos 75 = 1 x 10-4 Wb Norah Ali Al- moneef

Example In a coil of 3 m2 and 4 turns there are 12 field lines each representing a flux of 2 Wb. What is the flux density? B = Φ / AN = (12 X 2) / (3 X 4) = 2 T Norah Ali Al- moneef

31.1 Faraday’s Law of Induction We have observed that force is exerted on a charge by either and E field or a B field (when charge is moving): Consequences of the Lorentz Force: A B field can exert a force on an electric current (moving charge) A changing B-field (such as a moving magnet) will exert a magnetic force on a static charge, producing an electric current → this is called electromagnetic induction Faraday’s contribution to this observation: For a closed loop, a current is induced when: The B-field through the loop changes The area (A) of the loop changes The orientation of B and A changes q N S q N S A current is induced ONLY when any or all of the above are changing The magnitude of the induced current depends on the rate of change of 1-3 Moving charge Moving magnet

Magnetic Flux Faraday referred to changes in B field, area and orientation as changes in magnetic flux inside the closed loop The formal definition of magnetic flux (FB) (analogous to electric flux): When B is uniform over A, this becomes: Magnetic flux is a measure of the # of B field lines within a closed area (or in this case a loop or coil of wire) Changes in B, A and/or f change the magnetic flux Faraday’s Law: changing magnetic flux induces electromotive force (& thus current) in a closed wire loop f Norah Ali Al- moneef

Norah Ali Al- moneef

loop 1 loop 2 Norah Ali Al- moneef

Norah Ali Al- moneef

Norah Ali Al- moneef

N S magnet motion Norah Ali Al- moneef

N S magnet motion Norah Ali Al- moneef

S N magnet motion Norah Ali Al- moneef

S N magnet motion Norah Ali Al- moneef

Electromagnetism & EM Waves 05/19/08 Induced Current The next part of the story is that a changing magnetic field produces an electric current in a loop surrounding the field called electromagnetic induction, or Faraday’s Law Norah Ali Al- moneef Lecture 18

(a) A bar magnet is moved to the right toward a stationary loop of wire. As the magnet moves, the magnetic flux increases with time (b) The induced current produces a flux to the left to counteract the increasing external flux to the right Norah Ali Al- moneef

Faraday’s Experiment - 1: G G N S N S N S G G Norah Ali Al- moneef

i) the relative motion between the coil and the magnet Observe: i) the relative motion between the coil and the magnet ii) the induced polarities of magnetism in the coil iii) the direction of current through the galvanometer and hence the deflection in the galvanometer iv) that the induced current (e.m.f) is available only as long as there is relative motion between the coil and the magnet Note: i) coil can be moved by fixing the magnet ii) both the coil and magnet can be moved (towards each other or away from each other) i.e. there must be a relative velocity between them iii) magnetic flux linked with the coil changes relative to the positions of the coil and the magnet iv) current and hence the deflection is large if the relative velocity between the coil and the magnet and hence the rate of change of flux across the coil is more Norah Ali Al- moneef

Faraday’s Experiment - 2: When the primary circuit is closed current grows from zero to maximum value. During this period, changing current induces changing magnetic flux across the primary coil. This changing magnetic flux is linked across the secondary coil and induces e.m.f (current) in the secondary coil. Induced e.m.f (current) and hence deflection in galvanometer lasts only as long as the current in the primary coil and hence the magnetic flux in the secondary coil change. N S N S P S K G E N S N S P E S K G Norah Ali Al- moneef

When the primary circuit is open current decreases from maximum value to zero. During this period changing current induces changing magnetic flux across the primary coil. This changing magnetic flux is linked across the secondary coil and induces current (e.m.f) in the secondary coil. However, note that the direction of current in the secondary coil is reversed and hence the deflection in the galvanometer is opposite to the previous case. Faraday’s Laws of Electromagnetic Induction: I Law: Whenever there is a change in the magnetic flux linked with a circuit, an emf and hence a current is induced in the circuit. However, it lasts only so long as the magnetic flux is changing. II Law: The magnitude of the induced emf is directly proportional to the rate of change of magnetic flux linked with a circuit. ε α dΦ / dt ε = (Φ2 – Φ1) / t Norah Ali Al- moneef

Norah Ali Al- moneef

Lenz’s Law: The induced emf or current always tends to oppose or cancel the change that caused it. Norah Ali Al- moneef

Currents (I) induced in a wire loop. Norah Ali Al- moneef

Faraday’s Law and Electromagnetic Induction The instantaneous emf induced in a circuit equals the time rate of change of magnetic flux through the circuit If a circuit contains N tightly wound loops and the flux through each loop changes by ΔΦ during an interval Δt, the average emf induced is given by Faraday’s Law: Norah Ali Al- moneef

Faraday’s Law and Lenz’ Law The minus sign is included because of the polarity of the emf. The induced emf in the coil gives rise to a current whose magnetic field OPPOSES ( Lenz’s law) the change in magnetic flux that produced it Norah Ali Al- moneef

Faraday’s Law Induced Potential The number of loops (N) in the coil Physics C 4/16/2017 Calculates induced emf due to changing magnetic flux. Unit of Flux: Weber (Wb = Tm2) The number of loops (N) in the coil Note: magnetic flux changes when either the magnetic field (B), the area (A) or the orientation (cos f) of the loop changes Induced Potential Changing the flux B = BAcos() a) Change B b) Change A c) Change  Norah Ali Al- moneef Bertrand

How can we change the flux? Change flux by: Change area Change angle Change field Norah Ali Al- moneef

Changing Area What is the induced e in the loop? A loop of wire (N=10) contracts from 0.03 m2 to 0.01 m2 in 0.5 s, where B is 0.5 T and f is 0o (Rloop is 1 W). What is the induced e in the loop? What is the induced current in the loop? Norah Ali Al- moneef

Changing Orientation What is the induced e in the loop? A loop of wire (N=10) rotates from 0o to 90o in 1.5 s, B is 0.5 T and A is 0.02 m2 (Rloop is 2 W). What is the induced e in the loop? What is the induced current in the loop? Norah Ali Al- moneef

Faraday’s Law of Induction Faraday’s Law: The instantaneous voltage in a circuit (w/ N loops) equals the rate of change of magnetic flux through the circuit: The minus sign indicates the direction of the induced voltage. To calculate the magnitude: Norah Ali Al- moneef

Lenz’s Law When the magnetic flux changes within a loop of wire, the induced current resists the changing flux The direction of the induced current always produces a magnetic field that resists the change in magnetic flux (blue arrows) Review the previous examples and determine the direction of the current Magnetic flux, FB Increasing FB i Increasing FB i Norah Ali Al- moneef

There are three possibilities to produce an emf 1) Time-varying magnetic field e=-N[(A cos)(DB/Dt)+ 2) Time-varying loop area   +(B cos )(DA/Dt)+  3) Turning of the loop (generator)  +BA(D[cos]/Dt)] Norah Ali Al- moneef

Example A UHF television loop antenna has a diameter of 11 cm. The magnetic field of a TV signal is normal to the plane of the loop and, at one instant of time, its magnitude is changing at the rate 0.16 T/s. The magnetic field is uniform. What emf is induced in the antenna? Norah Ali Al- moneef

example The magnetic flux through the loop shown in Fig increases according to the relation FB = 6.0t 2 + 7.0t, where FB is in milliwebers and t is in seconds. (a) What is the magnitude of the emf induced in the loop when t = 2.0 s? (b) What is the direction of the current through R? Norah Ali Al- moneef

example The magnetic field through a single loop of wire, 12 cm in radius and of 8.5 W resistance, changes with time as shown in the figure. Calculate the emf in the loop as a function of time. Consider the time intervals (a) t = 0 to t = 2.0 s, (b) t = 2.0 s to t = 4.0 s, (c) t = 4.0 s to t = 6.0 s. The (uniform) magnetic field is perpendicular to the plane of the loop. Norah Ali Al- moneef

Example A uniform magnetic field is normal to the plane of a circular loop 10 cm in diameter and made of copper wire (of diameter 2.5 mm). (a) Calculate the resistance of the wire. (see copper resistivity in table) (b) At what rate must the magnetic field change with time if an induced current of 10 A is to appear in the loop? Norah Ali Al- moneef

example : In the figure a 120-turn coil of radius 1.8 cm and resistance 5.3 W is placed outside a solenoid like that in the figure . The solenoid current drops from 1.5 A to zero in time interval Δt = 25 ms, what current appears in the coil while the solenoid current is being changed? Norah Ali Al- moneef

Physics C example: What is the current produced in a loop of radius 16 cm and resistance 8.5 W at different times shown below? Assume the loop has an orientation in which it receives maximum magnetic flux. 4/16/2017 2 4 6 8 t (s) 0.5 1.0 B (T) Norah Ali Al- moneef Bertrand

Example Find the emf in a coil of 200 turns when there is a change of flux of 30 mWb linking it in 40 ms. Solution Δϕ = 30 x 10-3 Wb Δt = 40 x 10-3 s Induced emf, E Norah Ali Al- moneef

Physics C 4/16/2017 example: A 50 turn rectangular coil of dimensions 5.0 cm x 10.0 cm is allowed to fall from a position where B = 0 to a new position where B = 0.500 T and is directed perpendicular to the plane of the coil. Calculate the magnitude of the average emf if this occurs in 0.250 seconds. Norah Ali Al- moneef Bertrand

Physics C 4/16/2017 example : The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction of the current in the wire loop? Norah Ali Al- moneef Bertrand

Physics C 4/16/2017 example The magnetic field is increasing at a rate of 4.0 mT/s. What is the direction of the current in the wire loop? Norah Ali Al- moneef Bertrand

Physics C 4/16/2017 example The magnetic field is decreasing at a rate of 4.0 mT/s. The radius of the loop is 3.0 m, and the resistance is 4 W. What is the magnitude and direction of the current? Norah Ali Al- moneef Bertrand

example B decreases by 3.00 mT/s in a circular region of radius r =5.00 cm. What is the EMF around the loop? x x x x x x

Example A flat circular coil of 600 turns and with a mean radius of 1.5 cm is connected between the terminals of an ammeter. The total resistance of the coil and the ammeter is 0.5 Ω. The plane of the coil is placed at right angles to a uniform magnetic field of magnetic flux density 0.34 T. The coil is removed from the magnetic field in a time of 60 ms. Calculate the magnitude of : a) the average induced emf across the ends of the coil magnetic flux linkage = NΦ = BAN Initial magnetic flux linkage = 600 x 0.34 x (π x 0.0152) = 0.144 Wb After the coil is removed from the magnetic field, B = 0 therefore final magnetic flux linkage = 0. ε = - (d N Φ / dt) = (0 – 0.144) / 60 x 10-3 = 2.4 V b) the average induced current in the coil I = ε / R = 2.4 / 0.5 = 4.8 A Norah Ali Al- moneef

Example A flat coil of 680 turns and mean cross-sectional area 4.5 x 10-4 m2 is placed in a region of uniform magnetic field of flux density 580 μT. Initially, the plane of the coil is at right angles to the magnetic field. Calculate the mean induced ε across the ends of the coil when: a) the coil is withdrawn from the region of field in a time of 60 ms. Initial magnetic flux linkage = NΦ = BAN = 680 x 580 μ x (4.5 x 10-4) = 1.77 x 10-4 Wb Final magnetic flux linkage = 0. ε = - (d N Φ / dt ) = (0 – 1.77 x 10-4) / 60 x 10-3 = 3.0 mV b) the coil is rotated through 90o in a time of 60 ms. Final magnetic flux linkage = 0, therefore ε = 3.0 mV c) the direction of the field is reversed in a time of 30 ms. Initial flux = 1.77 x 10-4 Wb Final flux = -1.77 x 10-4 Wb ε = - (d N Φ / dt) = (-1.77 x 10-4 -1.77 x 10-4 ) / 30 x 10-3 = 12 mV Norah Ali Al- moneef

example A 0.25 T magnetic field is perpendicular to a circular loop of wire with 50 turns and a radius 15 cm. The magnetic field is reduced to zero in 0.12 s. What is the magnitude of the induced EMF? Flux = BA = (B)(pr2) = (0.25 T)p(0.15m)2 = 0.0176 T·m2 Rate of change of flux = DF/Dt DF = Ff-F i= 0-BA = - 0.0176 T·m2 EMF = - N DF/Dt = (-50)(-0.0176 T·m2 )/(0.12 s) EMF = 7.35 T·m2 /s T·m2 /s = [N / (A·m)] (m2/s) =(N·m)/(A·s) = J/C =V EMF = 7.35 V Norah Ali Al- moneef

31.2 Motional emf A straight conductor of length ℓ moves perpendicularly with constant velocity through a uniform field The electrons in the conductor experience a magnetic force F = q v B The electrons tend to move to the lower end of the conductor ℓ As the negative charges accumulate at the base, a net positive charge exists at the upper end of the conductor As a result of this charge separation, an electric field is produced in the conductor Charges build up at the ends of the conductor until the downward magnetic force is balanced by the upward electric force There is a potential difference between the upper and lower ends of the conductor Norah Ali Al- moneef

Charges stop moving when: Norah Ali Al- moneef

As long as the wire keeps moving, there will be a charge separation. Electron holes continue to move up, but only until FE down equals FB up, at which charge separation ceases. As long as the wire keeps moving, there will be a charge separation. The magnetic force is doing work to maintain that charge separation. For a wire of length L, moving in a direction perpendicular to an external B field: Charges stop moving when: Norah Ali Al- moneef

V=Bℓv, voltage across the conductor Motional emf, cont. V =Eℓ F=qvB   V=Bℓv, voltage across the conductor   If the motion is reversed, the polarity of the potential difference is also reversed F=qE=q (V/ℓ ) =qvB Norah Ali Al- moneef

Magnitude of the Motional emf A conducting bar sliding with v along two conducting rails under the action of an applied force Fapp. The magnetic force Fm opposes the motion, and a counterclockwise current is induced. Magnitude of the Motional emf Norah Ali Al- moneef

Yet another magnetic force generated, this time to the left. Once there is a current flowing through the wire, charges move in 2 directions. They all move right with the moving wire. Current moves up. Yet another magnetic force generated, this time to the left. Norah Ali Al- moneef

Motional emf The emf due to the charge separation exists whether or not the loop is closed (battery analogy). If there is a closed loop: I = vLB/R The induced current is due to magnetic forces on moving charges. Norah Ali Al- moneef

As the bar moves to the right, the magnetic flux through the circuit increases with time because the area of the loop increases The induced current must be in a direction such that it opposes the change in the external magnetic flux Norah Ali Al- moneef

Norah Ali Al- moneef

Magnetic force (F = IL x B) due to the induced current is toward the left, opposite to velocity v. Norah Ali Al- moneef

This magnetic force opposes the original velocity of the moving wire. The moving wire will slow down and stop. Need a constant Fpull to the right to make the contraption work. To keep the wire at constant speed, and continue the emf and current Fpull = Fmag F = ILB Fpull = Fmag = ILB I = vLB/R F = vL2B2/R Norah Ali Al- moneef

Fpull = Fmag Wpull = Wmag The rate at which work is done on the circuit equals the power dissipated by the circuit: P = I2R = v2L2B2/R Norah Ali Al- moneef

Linear Generator with Faraday’s Law By Lenz’s Law, what is the direction of current? Norah Ali Al- moneef

Motional emf in a Circuit, cont. The changing magnetic flux through the loop and the corresponding induced emf in the bar result from the change in area of the loop The induced, motional emf, acts like a battery in the circuit Norah Ali Al- moneef

Lenz’ Law, Bar Example, cont The flux due to the external field is increasing into the page The flux due to the induced current must be out of the page Therefore the current must be counterclockwise when the bar moves to the right Norah Ali Al- moneef

Lenz’ Law, Bar Example, final The bar is moving toward the left The magnetic flux through the loop is decreasing with time The induced current must be clockwise to to produce its own flux into the page Norah Ali Al- moneef

Force & Magnetic Induction What about the force applied by the hand to keep the rail moving? The moving rail induces an electric current and also produces power to drive the current: P = e.i = (5 V)(2.5 A) = 12.5 W The power (rate of work performed) comes from the effort of the hand to push the rail Since v is constant, the magnetic field exerts a resistive force on the rail: The force of the hand can be determined from the power: Verify FB direction with RHR Norah Ali Al- moneef

Power moving the bar Same result! Norah Ali Al- moneef

Example: (a) e=vBℓ e =(5.0 m/s)(0.80 T)(1.6 m)=6.4 V (b) I=e/R Rod and rail have negligible resistance but the bulb has a resistance of 96 W, B=0.80 T, v=5.0 m/s and ℓ =1.6 m. Calculate (a) emf in the rod, (b) induced current (c) power delivered to the bulb and (d) the energy used by the bulb in 60 s. (a) e=vBℓ e =(5.0 m/s)(0.80 T)(1.6 m)=6.4 V (b) I=e/R I=(6.4V)/(96 W)=0.067 A (c) P=eI P=eI=(6.4 V)(0.067 A)=0.43 W (d) E=Pt E=(0.43 W)(60 s)=26 J (=26 Ws) Norah Ali Al- moneef

Operating a light bulb with motional EMF Consider a rectangular loop placed within a magnetic field, with a moveable rail (Rloop= 2 W). B = 0.5 T v = 10 m/s L = 1.0 m Questions: 1) What is the induced e in the loop? 2) What is the induced current in the loop? 3) What is the direction of the current? Norah Ali Al- moneef

(a) ccw (b) cw (c) no induced current (a) ccw (b) cw x y A conducting rectangular loop moves with constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown. What is the direction of the induced current in the loop? (a) ccw (b) cw (c) no induced current 1B x y A conducting rectangular loop moves with constant velocity v in the -y direction and a constant current I flows in the +x direction as shown. What is the direction of the induced current in the loop? (a) ccw (b) cw (c) no induced current Norah Ali Al- moneef

(a) ccw (b) cw (c) no induced current x y A conducting rectangular loop moves with constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown. What is the direction of the induced current in the loop? (c) no induced current There is a non-zero flux FB passing through the loop since B is perpendicular to the area of the loop. Since the velocity of the loop and the magnetic field are CONSTANT, however, this flux DOES NOT CHANGE IN TIME. Therefore, there is NO emf induced in the loop; NO current will flow!! Norah Ali Al- moneef

(a) ccw (b) cw (c) no induced current (a) ccw (b) cw y A conducting rectangular loop moves with constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown. What is the direction of the induced current in the loop? x (a) ccw (b) cw (c) no induced current A conducting rectangular loop moves with constant velocity v in the -y direction and a constant current I flows in the +x direction as shown. What is the direction of the induced current in the loop? y i x (a) ccw (b) cw (c) no induced current The flux through this loop DOES change in time since the loop is moving from a region of higher magnetic field to a region of lower field. Therefore, by Lenz’ Law, an emf will be induced which will oppose the change of flux. The current i is induced in the clockwise direction to restore the flux. Norah Ali Al- moneef

Example: A coil is wrapped with 200 turns of wire on the perimeter of a square frame of side 18 cm. The total resistance of the coil is 2Ω . B is  the plane of the coil and changes linearly from 0 to 0.5 T in 0.80 seconds. Find the emf in the coil while the field is changing. What is the induced current? Ans. 4.05 V, 2.03 A. Norah Ali Al- moneef

Another example: a bar magnet is moved to the left/ right toward a stationary loop of wire. Norah Ali Al- moneef

Example Induced EMF, Induced Current. A solenoid (similar to one used for a class demonstration) has a diameter of 10 cm, a length of 10 cm, and contains 3500 windings with a total resistance of 60 Ohm. The solenoid is connected in a simple loop, modeled above. Initially, the solenoid is embedded in a magnetic field of 0.100 T, parallel to the axis of the solenoid, as shown. This external field is reduced to zero in 0.10 sec. During this 0.1 sec, what is the EMF in the coil, what is the current in the circuit, and what is the direction and magnitude of the magnetic field in the solenoid generated by this current? Norah Ali Al- moneef

Flux & induced EMF First, what is the initial flux? Let us chose current flowing around the circuit in the clockwise direction to be positive. Such a current would generate a magnetic field pointing up in the solenoid (and pointing down outside the solenoid). Thus the initial flux is positive. Flux in one winding of solenoid F = (Area)(Magnetic Field) = p r2 B = 3.14 (0.05m)2 (0.100 T). F = 7.85e-4 T m2 DF = Ff - Fi = 0 - 7.85e-4 Tm2 Dt = 0.100 sec EMF = - N (DF/Dt) = -(3500)(-7.85e-3 T m2/s) = 27.5 V Norah Ali Al- moneef

Induced Current EMF in loop = 27.5 V EMF – IR = 0 Changing flux acts just like a battery EMF – IR = 0 I = EMF/R = (27.5 V) / 60 W =0.458 A Norah Ali Al- moneef

example: In the figure, a long rectangular conducting loop, of width L, resistance R, and mass m, is hung in a horizontal, uniform magnetic field B that is directed into the page and that exists only above line aa. The loop is then dropped; during its fall, it accelerates until it reaches a certain terminal speed vt. Ignoring air drag, find that terminal speed. Norah Ali Al- moneef

example: A metal rod is forced to move with constant velocity along two parallel metal rails, connected with a strip of metal at one end, as shown in the figure. A magnetic field B = 0.350 T points out of the page. (a) If the rails are separated by 25.0 cm and the speed of the rod is 55.0 cm/s, what emf is generated? (b) If the rod has a resistance of 18.0 W and the rails and connector have negligible resistance, what is the current in the rod? (c) At what rate is energy being transferred to thermal energy? Norah Ali Al- moneef

Physics C 4/16/2017 example How much current flows through the resistor? How much power is dissipated by the resistor?  50 cm B = 0.15 T 3 W v = 2 m/s Counter clock wise ( direction) Bertrand

Norah Ali Al- moneef

Norah Ali Al- moneef

Norah Ali Al- moneef

Norah Ali Al- moneef

Example An airplane travels at 1000 km/hr in a region where the earth’s magnetic field is 5 X 10-5T (vertical). What is the potential difference between the wing tips if they are 70 m apart? 1000 km/hr = 280 m/s ε = Blv ε = (5 X 10-5T )(70 m)(280 m/s) = 1.0 V

Example Blood contains charged ions. A blood vessel is 2.0 mm in diameter, the magnetic field is 0.080 T, and the blood meter registers a voltage of 0.10 mV. What is the flow velocity of the blood? E = Blv v = E /Bl v = (1.0 X 10-4 V) (0.080 T)(0.0020m) v = 0.63 m/s