HW8%20solutions

# HW8%20solutions - Homework 8 solutions Physics 204, morning...

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Unformatted text preview: Homework 8 solutions Physics 204, morning summer 2008 Chapter 22 Problems 10,30,24,27 10. REASONING According to Equation 22.2, the magnetic flux is the product of the magnitude B of the magnetic field, the area A of the surface, and the cosine of the angle between the direction of the magnetic field and the normal to the surface. The area of a circular surface is A = π r 2 , where r is the radius. SOLUTION The magnetic flux through the surface is Φ = BA cos ϕ = B (π r 2 ) cos ϕ = ( 0.078 T ) π ( 0.10 m ) cos 25° = 2.2 ×10−3 Wb 2 30. REASONING According to Lenz’s law, the induced current in the triangular loop flows in such a direction so as to create an induced magnetic field that opposes the original flux change. SOLUTION a. As the triangle is crossing the +y axis, the magnetic flux down into the plane of the paper is increasing, since the field now begins to penetrate the loop. To offset this increase, an induced magnetic field directed up and out of the plane of the paper is needed. By applying RHR‐2 it can be seen that such an induced magnetic field will be created within the loop by a counterclockwise induced current . b. As the triangle is crossing the −x axis, there is no flux change, since all parts of the triangle remain in the magnetic field, which remains constant. Therefore, there is no induced magnetic field, and no induced current appears . c. As the triangle is crossing the −y axis, the magnetic flux down into the plane of the paper is decreasing, since the loop now begins to leave the field region. To offset this decrease, an induced magnetic field directed down and into the plane of the paper is needed. By applying RHR‐2 it can be seen that such an induced magnetic field will be created within the loop by a clockwise induced current . d. As the triangle is crossing the +x axis, there is no flux change, since all parts of the triangle remain in the field‐free region. Therefore, there is no induced magnetic field, and no induced current appears . 24. REASONING The energy dissipated in the resistance is given by Equation 6.10b as the power P dissipated multiplied by the time t, Energy = Pt. The power, according to Equation 20.6c, is the square of the induced emf ξ 2 divided by the resistance R, P = ξ /R. The induced emf can be determined from Faraday’s law of electromagnetic induction, Equation 22.3. 2 SOLUTION Expressing the energy consumed as Energy = Pt, and substituting in P = ξ /R, we find Energy = P t = ξ 2t R The induced emf is given by Faraday’s law as ξ = − N ( ΔΦ/Δt ) , and the resistance R is equal to the resistance per unit length (3.3 × 10 –2 /m) times the length of the circumference of the loop, 2 r. Thus, the energy dissipated is 2 2 ⎡ ⎛ Φ − Φ0 ⎞⎤ ΔΦ ⎞ ⎛ ⎢− N ⎜ −N t ⎜ t − t ⎟⎥ t ⎟ ⎜ ⎟ ⎢ Δt ⎠ 0 ⎠⎥ ⎝ ⎝ ⎦ Energy = = ⎣ −2 −2 3.3 × 10 Ω / m 2π r 3.3 × 10 Ω / m 2π r ( ) ( ) 2 2 ⎡ ⎡ ⎛ BA cos φ − B0 A cos φ ⎞ ⎤ ⎛ B − B0 ⎞ ⎤ ⎢− N ⎜ ⎢ − NA cos φ ⎜ ⎟⎥ t ⎜ ⎟ ⎜ t − t ⎟⎥ t ⎟ t − t0 ⎢ 0 ⎠⎥ ⎝ ⎠⎥ ⎝ ⎣ ⎦ = ⎢ ⎣ ⎦ = −2 −2 3.3 × 10 Ω / m 2π r 3.3 × 10 Ω / m 2π r ( ) ( ) 2 ⎡ ⎛ 0.60 T − 0 T ⎞ ⎤ 2 ⎢ − (1) π ( 0.12 m ) ( cos 0° ) ⎜ ⎟ ⎥ ( 0.45 s ) ⎝ 0.45 s − 0 s ⎠ ⎦ −2 ⎣ = 6.6 × 10 J = −2 3.3 × 10 Ω / m 2π ( 0.12 m ) ( ) 27. SSM REASONING According to Ohm’s law, the average current I induced in the coil is given by I = ξ /R, where ξ is the magnitude of the induced emf and R is the resistance of the coil. To find the induced emf, we use Faraday's law of electromagnetic induction SOLUTION The magnitude of the induced emf can be found from Faraday’s law of electromagnetic induction and is given by Equation 22.3: ξ = −N Δ ( BA cos 0° ) ΔΦ ΔB = −N = NA Δt Δt Δt We have used the fact that the field within a long solenoid is perpendicular to the cross‐sectional area A of the solenoid and makes an angle of 0° with respect to the normal to the area. The field is given by Equation 21.7 as B = μ0 n I , so the change in the field is ΔB = μ0 nΔI , where is the change in the current. The induced current is, then, μ n ΔI ΔB NA 0 Δt Δt = R R NA I= (10)(6.0 × 10−4 m 2 ) = (4π × 10 −7 T ⋅ m/A)(400 turns/m)(0.40 A) (0.050 s) = 1.6 × 10−5 A 1.5 Ω ...
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## This note was uploaded on 10/04/2009 for the course PHYS 750:204 taught by Professor Rollino during the Spring '09 term at Rutgers.

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