Physics Solution Manual for 1100 and 2101

# The time interval t is 1 t as discussed in the

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Unformatted text preview: ude B of the magnetic field, the area A of the surface, and the cosine of the angle φ 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 Φ 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 12. REASONING a. The magnetic flux Φ though a surface of area A due to a uniform magnetic field of magnitude B is given by Φ = BA cos φ (Equation 22.2) where φ is the angle between the direction of the magnetic field and the normal to the surface (see the drawing, which shows an edge-on view of the situation). The magnetic field B is horizontal, and the surface makes an angle of 12° with the horizontal, so the normal to the surface is φ = 90.0° − 12° = 78°. We will use Equation 22.2 to determine the surface area A. Surface 12° φ B Normal to surface Edge-on view of first surface b. When exposed to a uniform magnetic field, the magnetic flux through a flat surface is greatest when the surface is perpendicular to the direction of the magnetic field. When this occurs, the normal to the surface is parallel to the direction of the magnetic field, and the angle φ is zero. Therefore, to find the smallest surface area that has same amount of magnetic flux passing through it as the surface in part (a), we will take φ = 0.0° in Equation 22.2. SOLUTION a. Solving Φ = BA cos φ (Equation 22.2) for A yields A = find that Φ . Substituting φ = 78°, we B cos φ 1198 ELECTROMAGNETIC INDUCTION A= b. From A = Φ 8.4 ×10−3 Wb = = 0.086 m2 o B cos φ ( 0.47 T ) cos 78 Φ , with φ = 0.0°, the minimum possible area of the second surface is B cos φ A= 13. 8.4 ×10−3 Wb = 0.018 m2 0.47 T ) cos 0.0o ( SSM 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 area. The magnetic flux through the door is a maximum when the magnetic field lines are perpendicular to the door and φ1 = 0.0° so that Φ1 = Φ max = BA(cos 0.0°) = BA . SOLUTION When the door rotates through an angle φ 2 , the magnetic flux that passes through the door decreases from its maximum value to one-third of its maximum value. Therefore, Φ 2 = 1 Φ max , and we have 3 Φ 2 = BA cos φ2 = 1 BA 3 or cos φ2 = 1 3 or φ2 = cos –1 ( 1 ) = 70.5° 3 ______________________________________________________________________________ 14. REASONING At any given moment, the flux Φ that passes through the loop is given by Φ = BA cos φ (Equation 22.2), where B is the magnitude of the magnetic field, A is the area of the loop, and φ = 0° is the angle between the normal to the loop and the direction of the magnetic field (both directed into the page). As the handle turns, the area A of the loop changes, causing a change ∆Φ in the flux passing through the loop. We can think of the loop as being divided into a rectangular portion and a semicircular portion. Initially, the area A0 of the loop is equal to the rectangular area Arec plus the area Asemi = 1 π r 2 of the semicircle, 2 where r is the radius of the semicircle: A0 = Arec + 1 π r 2 . After half a revolution, the 2 semicircle is once again within the plane of the loop, but now as a reduction of the area of the rectangular portion. Therefore, the final area A of the loop is equal to the area of the rectangular portion minus the area of the semicircle: A = Arec − 1 π r 2 . 2 Chapter 22 Problems 1199 SOLUTION The change ∆Φ in the flux that passes through the loop is the difference between the final flux Φ = BA cos φ (Equation 22.2) and the initial flux Φ 0 = BA0 cos φ : ∆Φ = Φ − Φ0 = BA cos φ − BA0 cos φ = B cos φ ( A − A0 ) (1) Substituting A0 = Arec + 1 π r 2 and A = Arec − 1 π r 2 into Equation (1) yields 2 2 ( )( ) ∆Φ = B cos φ ( A − A0 ) = B cos φ Arec − 1 π r 2 − Arec + 1 π r 2 = −π r 2 B cos φ 2 2 Therefore the change in the flux passing through the loop during half a revolution of the semicircle is 2 ∆Φ = −π r 2 B cos φ = −π ( 0.20 m) ( 0.75 T ) cos 0o = −0.094 Wb 15. REASONING The definition of magnetic flux Ф is Φ = BA cos φ (Equation 22.2), where B is the magnitude of the magnetic field, A is the area of the surface, and φ is the angle between the magnetic field vector and the normal to the surface. The values of B and A are the same for each of the surfaces, while the values for the angle φ are different. The z axis is the normal to the surface lying in the x, y plane, so that φ x y = 35° . The y axis is the normal to the surface lying in the x, z plane, so that φ x z = 55° . We can apply the definition of the flux to obtain the desired ratio directly. SOLUTION Using Equation 22.2, we find that Φxz Φxy = BA cos φ x z BA cos φ x y = cos 55° = 0.70 cos 35° 16. REASONING AND SOLUTION The...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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