Physics Solution Manual for 1100 and 2101

# The current is decreasing so the magnetic field is

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: change in flux ∆Φ is given by ∆Φ = Β ∆Α cos φ, where ∆Α is the area of the loop that leaves the region of the magnetic field in a time ∆t. This area is the product of the width of the rectangle (0.080 m) and the length v ∆t of the side that leaves the magnetic field, ∆A = (0.080 m) v ∆t. ∆Φ = B ∆A cos φ = (2.4 T) (0.080 m)(0.020 m/s)(2.0 s) cos 0.0° = 7.7 × 10 –3 Wb ______________________________________________________________________________ 1200 ELECTROMAGNETIC INDUCTION 17. SSM WWW REASONING The general expression for the magnetic flux through an area A is given by Equation 22.2: Φ = BA cos φ , where B is the magnitude of the magnetic field and φ is the angle of inclination of the magnetic field B with respect to the normal to the surface. SOLUTION Since the magnetic field B is parallel to the surface for the triangular ends and the bottom surface, the flux through each of these three surfaces is 0 Wb . The flux through the 1.2 m by 0.30 m face is Φ = BA cos φ = ( 0.25 T ) (1.2 m ) ( 0.30 m ) cos 0.0 ° = 0.090 Wb For the 1.2 m by 0.50 m side, the area makes an angle φ with the magnetic field B, where 0.30 m = 53° 0.40 m φ = 90° – tan −1 Therefore, Φ = BA cos φ = ( 0.25 T ) (1.2 m ) ( 0.50 m ) cos 53 ° = 0.090 Wb ______________________________________________________________________________ 18. REASONING An emf is induced during the first and third intervals, because the magnetic field is changing in time. The time interval is the same (3.0 s) for the two cases. However, the magnitude of the field changes more during the first interval. Therefore, the magnetic flux is changing at a greater rate in that interval, which means that the magnitude of the induced emf is greatest during the first interval. The induced emf is zero during the second interval, 3.0 – 6.0 s. According to Faraday’s law of electromagnetic induction, Equation 22.3, an induced emf arises only when the magnetic flux changes. During this interval, the magnetic field, the area of the loop, and the orientation of the field relative to the loop are constant. Thus, the magnetic flux does not change, so there is no induced emf. During the first interval the magnetic field in increasing with time. During the third interval, the field is decreasing with time. As a result, the induced emfs will have opposite polarities during these intervals. If the direction of the induced current is clockwise during the first interval, it will be counterclockwise during the third interval. SOLUTION a. The induced emf is given by Equations 22.3 and 22.3: Chapter 22 Problems 1201 0–3.0 s: ξ =−N BA cos φ − B0 A cos φ ∆Φ =−N ∆t t − t0 B − B0 = − NA cos φ t −t 0 0.40 T − 0 T 2 = − ( 50 ) 0.15 m ( cos 0° ) = −1.0 V 3.0 s − 0 s ( ) 3.0–6.0 s: B − B0 0.40 T − 0.40 T 2 = − ( 50 ) 0.15 m ( cos 0° ) = 0V 6.0 s − 3.0 s t − t0 ( ξ = − NA cos φ ) 6.0–9.0 s: B − B0 0.20 T − 0.40 T 2 = − ( 50 ) 0.15 m ( cos 0° ) = +0.50 V t − t0 9.0 s − 6.0 s ( ξ = − NA cos φ ) b. The induced current is given by Equation 20.2 as I = ξ/R. 0–3.0 s: I= 6.0–9.0 s: I= ξ R ξ R = −1.0 V = −2.0 A 0.50 Ω = +0.50 V = +1.0 A 0.50 Ω As expected, the currents are in opposite directions. ______________________________________________________________________________ 19. REASONING According to Faraday’s law, as given in Equation 22.3, the magnitude of the ∆Φ emf is ξ = − (1) , where we have set N = 1 for a single turn. Since the normal is ∆t parallel to the magnetic field, the angle φ between the normal and the field is φ = 0° when calculating the flux Φ from Equation 22.2: Φ = BA cos 0° = BA. We will use this expression for the flux in Faraday’s law. SOLUTION Representing the flux as Φ = BA, we find that the magnitude of the induced emf is ∆ ( BA) ∆Φ B ∆A ξ=− =− =− ∆t ∆t ∆t In this result we have used the fact that the field magnitude B is constant. Rearranging this equation gives 1202 ELECTROMAGNETIC INDUCTION ∆ A ξ 2.6 V = = = 1.5 m 2 /s ∆t B 1.7 T ______________________________________________________________________________ 20. REASONING An emf is induced in the body because the magnetic flux is changing in time. According to Faraday’s law, as given by Equation 22.3, the magnitude of the emf is ∆Φ . This expression can be used to determine the time interval ∆t during which ξ = −N ∆t the magnetic field goes from its initial value to zero. The magnetic flux Φ is obtained from Equation 22.2 as Φ = BA cos φ, where φ = 0° in this problem. SOLUTION The magnitude ξ of the induced emf is BA cos φ − B0 A cos φ ∆t ∆Φ ξ = −N = −N ∆t Solving this relation for ∆t gives ∆t = = − NA cos φ ( B − B0 ) ( ) = 4.8 s 0.010 V ______________________________________________________________________________ 21. ξ = − (1) 0.032 m 2 cos 0° ( 0 − 1.5 T ) SSM REASONING According to Equation 22.3, the average emf induced in a coil of N loops is ξ = –N ∆Φ / ∆ t . SOLUTION For the circular coil...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern