Physics Solution Manual for 1100 and 2101

39

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: in question, the flux through a single turn changes by ∆Φ = BA cos 45° – BA cos 90° = BA cos 45° during the interval of ∆ t = 0.010 s . magnitude of the emf as Therefore, for N turns, Faraday's law gives the ξ = −N BA cos 45° ∆t Since the loops are circular, the area A of each loop is equal to π r2. Solving for B, we have ξ ∆t (0.065 V)(0.010 s) = 8.6 ×10 –5 T 2 N π r cos 45° (950)π (0.060 m) cos 45° ______________________________________________________________________________ B= 2 = Chapter 22 Problems 1203 22. REASONING According to Equation 22.3, Faraday’s law specifies the emf induced in a coil of N loops as ∆Φ ξ = −N ∆t where ∆Φ / ∆t is the rate of change of the magnetic flux in a single loop. Recognizing that ∆Φ / ∆t is the same for each of the coils, we will apply Faraday’s law to each coil to obtain our solution. SOLUTION Applying Faraday’s law to each coil gives ξ1 = − N1 ∆Φ ∆t and ξ2 = −N2 ∆Φ ∆t Dividing these equations and remembering that the rate of change of the flux is the same for each coil, we find that ∆Φ ξ 2 − N 2 ∆t N 2 = = ξ1 − N ∆Φ N1 1 or ∆t N 2 = N1 ξ2 4.23 V = 184 = 276 ξ1 2.82 V 23. REASONING We will use Faraday’s law of electromagnetic induction, Equation 22.3, to find the emf induced in the loop. Once this value has been determined, we can employ Equation 22.3 again to find the rate at which the area changes. SOLUTION a. The magnitude ξ of the induced emf is given by Equation 22.3 as Φ−Φ BA cos φ − B0 A cos φ B − B0 = − NA cos φ t − t0 t − t0 0 = −N ξ = −N t −t 0 2.1 T − 0 T = (1)( 0.35 m × 0.55 m ) cos 65° = 0.38 V 0.45 s − 0 s b. When the magnetic field is constant and the area is changing in time, Faraday’s law can be written as 1204 ELECTROMAGNETIC INDUCTION Φ−Φ BA cos φ − BA0 cos φ t − t0 0 ξ = −N = −N t −t 0 A − A0 ∆A = − N B cos φ = − N B cos φ t −t ∆t 0 Solving this equation for ∆A and substituting in the value of 0.38 V for the magnitude of ∆t the emf, we find that ξ ∆A 0.38 V = = = 0.43 m 2 / s ∆t N B cos φ (1)( 2.1 T ) cos 65° ______________________________________________________________________________ 24. REASONING According to Faraday’s law the emf induced in either single-turn coil is given by Equation 22.3 as ∆Φ ∆Φ ξ = −N =− ∆t ∆t since the number of turns is N = 1. The flux is given by Equation 22.2 as Φ = BA cos φ = BA cos 0° = BA where the angle between the field and the normal to the plane of the coil is φ = 0° , because the field is perpendicular to the plane of the coil and, hence, parallel to the normal. With this expression for the flux, Faraday’s law becomes ξ =− ∆ ( BA ) ∆Φ ∆B =− = −A ∆t ∆t ∆t (1) In this expression we have recognized that the area A does not change in time and have separated it from the magnitude B of the magnetic field. We will apply this form of Faraday’s law to each coil. The current induced in either coil depends on the resistance R of the coil, as well as the emf. According to Ohm’s law, the current I induced in either coil is given by ξ (2) I= R SOLUTION Applying Faraday’s law in the form of Equation 1 to both coils, we have ξ square = − Asquare ∆B ∆B = − L2 ∆t ∆t and ξ circle = − Acircle ∆B ∆B = −π r 2 ∆t ∆t Chapter 22 Problems 1205 The area of a square of side L is L2, and the area of a circle of radius r is πr2. The rate of change ∆B/∆t of the field magnitude is the same for both coils. Dividing these two expressions gives 2 ∆B ξsquare − L ∆t L2 (3) = =2 ξ circle −π r 2 ∆B π r ∆t The same wire is used for both coils, so we know that the perimeter of the square equals the circumference of the circle L 2π π = = 4 L = 2π r or r 4 2 Substituting this result into Equation (3) gives ξsquare ξ circle = L2 π2 π = 2= 4 π r2 π 2 or ξsquare = π 4 ξcircle = π 4 ( 0.80 V ) = 0.63 V (4) Using Ohm’s law as given in Equation (2), we find for the induced currents that I square I circle ξ square = R ξ circle = ξ square ξ circle R Here we have used the fact that the same wire has been used for each coil, so that the resistance R is the same for each coil. Using the result in Equation (4) gives I square I circle = ξ square ξ circle = π 4 or I square = π π I circle = ( 3.2 A ) = 2.5 A 4 4 25. REASONING Using Equation 22.3 (Faraday’s law) and recognizing that N = 1, we can write the magnitude of the emfs for parts a and b as follows: ∆Φ ∆t a ξa = − (1) and ∆Φ ∆t b ξb = − ( 2) To solve this problem, we need to consider the change in flux ∆Φ and the time interval ∆t for both parts of the drawing in the text. 1206 ELECTROMAGNETIC INDUCTION SOLUTION The change in flux is the same for both parts of the drawing and is given by ( ∆Φ )a = ( ∆Φ )b = Φinside – Φ outside = Φinside = BA (3) In Equation (3) we have used the fact that initially the coil is outside the field regio...
View Full Document

Ask a homework question - tutors are online