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Homework Hints Assignment (1-87) Read (22.3 - 22.4), Do PROBLEMS # (2, 15, 23) Ch. 22 + AP 1995 #7 (handout) 2. Near San Francisco, where the vertically downward component of the earth’s magnetic field is 4.8 x 10-5 T, a car is traveling forward at 25 m/s. An emf of 2.4 x 10-3 volts is induced between the sides of the car. REASONING AND SOLUTION a. Which side of the car is positive ? The right-hand rule shows that positive charge would be forced to the driver's side . b. What is the width of the car ? According to the equation emf = BLv , the width of the car is L= 2.4 × 10 –3 V Emf = = 2.0 m vB ( 25 m/s ) 4.8 × 10 –5 T ( ) ________________________________________________________________________________________________________________________________________________________________________________________________________________________ 15. A five-sided object, whose dimensions are shown, is placed in a uniform magnetic field. The magnetic field of magnitude 0.25 T points along the positive y-direction. Determine the magnetic flux through each of the five sides. 1.20 m .50 m B REASONING The general expression for the magnetic flux through an area A is given by the equation Φ = BA cosφ , where B is the magnitude of the magnetic field and φ is the angle of inclination of the magnetic field with respect to the NORMAL to the surface. .30 m .40 m SOLUTION <i> 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 . <ii> The flux through the 1.2 m by 0.30 m face is Φ = (0.25 T)(1.2 m)(0.30 m) cos 0.0° = 0.090 Wb <ii> 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 φ = 90° – tan −1 comp ?? = 53° 0.40 m Therefore, Φ = (0.25 T)(1.2 m)(0.50 m) cos 53° = 0.090 Wb 23. A 1.5 m aluminum rod is rotating about an axis that is perpendicular to one end. A 0.16 T magnetic field is directed parallel to the axis. The rod rotates through ¼ of a circle in 0.66 sec. What is the magnitude of the average emf generated between the ends of the rod during this time ? REASONING AND SOLUTION The rod may be viewed as sweeping out an area whose initial value is zero (0) and final value is π r2/4. The flux change due to this change in area is: 2 ∆Φ = Bπ r /4 Faraday's law gives the magnitude of the emf to be Emf = ∆Φ/∆t = (1/4)(0.16 T)(π)(1.5 m)2/(0.66 s) = 0.43 V