Physics Solution Manual for 1100 and 2101

10 the peak 2 voltage v0 across the bulb is equal to

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: Φ ∆B =N = NA ∆t ∆t ∆t (1) Substituting Equation (1) into Equation 20.2, we obtain ∆B NA ∆t NA ∆B = I= = R R R ∆t ξ (2) Substituting Equation (2) into Equation (21.6) yields BI = N µ0 I 2r =N µ0 NA ∆B 2 µ0 A ∆B =N 2r R ∆t 2rR ∆t (3) Lastly, we substitute A = π r 2 into Equation (3), because the coil is circular. This yields BI = N 2 µ0π r 2 ∆B 2 µ0π r ∆B =N 2 R ∆t 2 r R ∆t = (105) 2 ( )( 4π ×10−7 T ⋅ m/A π 4.00 ×10−2 m 2 ( 0.480 Ω ) ) ( 0.783 T/s ) = 1.42 ×10−3 T (4) 29. REASONING AND SOLUTION In time ∆t the flux change through the loop ABC is ∆Φ = B∆A, where ∆A 1 2 ∆A = 1 R 2 ( ω ∆ t ) s o = Rω 2 ∆t 2 Faraday's law gives the magnitude ξ of the induced emf to be ξ=− ∆Φ ∆A = −B =B ∆t ∆t ( 1 R2ω ) = ( 3.8 ×10−3 T ) 1 ( 0.50 m )2 (15 rad/s ) = 7.1×10−3 V 2 2 1210 ELECTROMAGNETIC INDUCTION The current through the resistor is 7.1×10−3 V = 2.4 × 10 –3 A R 3.0 Ω ______________________________________________________________________________ I= ξ = 30. REASONING A greater magnetic flux passes through the coil in part b of the drawing. The flux is given as Φ = BA cos φ (Equation 22.2), where B is the magnitude of the magnetic field, A is the area of the surface through which the field passes, and φ is the angle between the normal to the plane of the surface and the field. In part b the coil area has two parts, and so does the flux. There is the larger semicircular area and the smaller semicircular 2 ( 12 π rlarger ) on the horizontal surface 2 ( 12 π rsmaller ) area perpendicular to the horizontal surface. The field is parallel to the normal for the larger area and perpendicular to the normal for the smaller area. ω ω (b) (a) Hence, the flux is Φ b = BAlarger cos φlarger + BAsmaller cos φsmaller =B 2 2 2 ( 12 π rlarger ) cos 0° + B ( 12 π rsmaller ) cos 90° = B ( 12 π rlarger ) (1) In part a the entire area of the coil lies on the horizontal surface and is the area between the two semicircles. The plane of the coil is perpendicular to the magnetic field. The flux, therefore, is Φ a = BA cos φ = B 2 2 2 2 ( 12 π rlarger − 12 π rsmaller ) cos 0° = B ( 12 π rlarger − 12 π rsmaller ) (2) which is smaller than the flux in part b of the drawing. According to Lenz’s law the induced magnetic field associated with the induced current opposes the change in flux. To oppose the increase in flux from part a to part b of the drawing, the induced field must point downward. In this way, it will reduce the effect of the increasing coil area available to the upward-pointing external field. An application of RHR-2 shows that the induced current must flow clockwise in the larger semicircle of wire (when Chapter 22 Problems 1211 viewed from above) in order to create a downward-pointing induced field. The average induced current can be determined according to Ohm’s law as the average induced emf divided by the resistance of the coil. The period T of the rotational motion is related to the angular frequency ω according to 2π ω= (Equation 10.6). The shortest time interval ∆t that elapses between parts a and b of T the drawing is the time needed for one quarter of a turn or 1 T . 4 SOLUTION According to Ohm’s law, the average induced current is I= ξ R where ξ is the average induced emf and R is the resistance of the coil. According to ∆Φ Faraday’s law, Equation 22.3, the average induced emf is ξ = − , where we have set ∆t N = 1. We can determine the change ∆Ф in the flux as ∆Φ = Φ b − Φ a , where the fluxes in parts a and b of the drawing are available from Equations (1) and (2). The time interval ∆t is 1 T , as discussed in the REASONING. Using Equation 10.6, we have for the period that 4 T= 2π ω I= . Thus, we find for the current that ξ R =− = B − ∆Φ ∆t = R − ( Φ b − Φa ) 1T 4 R 2 2 2 2 2 ( 12 π rlarger ) − B ( 12 π rlarger − 12 π rsmaller ) = − B ( 12 π rsmaller ) = − B ( 12 π rsmaller ) R (1T) 4 ( 0.35 T ) 1 π ( 0.20 m )2 2 =− = ( 0.025 Ω ) 1 2π / (1.5 rad/s ) 4 R −0.84 A (1T) 4 R 1 ( 2π / ω ) 4 1212 ELECTROMAGNETIC INDUCTION 31. 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 ∆ ( BA cos 0° ) ∆Φ ∆B = −N = NA ∆t ∆t ∆t We have used the fact that the field within a long solenoid is perpendicular to the crosssectional 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 = µ 0 n I , so the change ∆Β in the field is ∆B = µ 0 n∆I , where ∆Ι is the change in the current. The induced current...
View Full Document

Ask a homework question - tutors are online