Physics Solution Manual for 1100 and 2101

# According to ohms law the current i induced in either

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: _______________________________________________ 2. REASONING During its fall, both the length of the bar and its velocity v are perpendicular to the horizontal component Bh of the earth’s magnetic field (see the drawing for an overhead view). Therefore, the emf ξ induced across the length L of the rod is given by West ξ = vBh L (Equation 22.1), where v is the speed of the rod. We will use Equation 22.1 to determine the magnitude Bh of the horizontal component of the earth’s magnetic field, and Right-Hand Rule No.1 from Section 21.2 to determine which end of the rod is positive. North Bh East v F (velocity into page) South Overhead view SOLUTION a. Solving ξ = vBh L (Equation 22.1) for Bh, we find that Bh = ξ vL = 6.5 × 10 −4 V = 3.7 × 10−5 T ( 22 m/s )( 0.80 m ) b. Consider a hypothetical positive charge that is free to move inside the falling rod. The bar is falling downward, carrying the positive charge with it, so that the velocity v of the charge is downward. In the drawing, which shows the situation as seen from above, downward is into the page. Applying Right-Hand Rule No. 1 to the vectors v and Bh, the magnetic force F on the charge points to the east. Therefore, positive charges in the rod would accelerate to the east, and negative charges would accelerate to the west. As a result, the east end of the rod acquires a positive charge. 3. REASONING AND SOLUTION The motional emf ξ generated by a conductor moving perpendicular to a magnetic field is given by Equation 22.1 as ξ = vBL, where v and L are the speed and length, respectively, of the conductor, and B is the magnitude of the magnetic field. The emf would have been ( )( )( ) ξ = vBL = 7.6 × 103 m/s 5.1 × 10−5 T 2.0 × 104 m = 7800 V Chapter 22 Problems 1193 ______________________________________________________________________________ 4. REASONING a. The motional emf generated by the moving metal rod depends only on its speed, its length, and the magnitude of the magnetic field (see Equation 22.1). The motional emf does not depend on the resistance in the circuit. Therefore, the emfs for the circuits are the same. b. According to Equation 20.2, the current I is equal to the emf divided by the resistance R of the circuit. Since the emfs in the two circuits are the same, the circuit with the smaller resistance has the larger current. Since circuit 1 has one-half the resistance of circuit 2, the current in circuit 1 is twice as large. c. The power P is P = ξ 2 / R (Equation 20.6c), where ξ is the emf (or voltage) and R is the resistance. The emf produced by the moving bar is directly proportional to its speed (see Equation 22.1). Thus, the bar in circuit 1 produces twice the emf, since it’s moving twice as fast. Moreover, the resistance in circuit 1 is half that in circuit 2. As a result, the power delivered to the bulb in circuit 1 is 22 / ( 1 ) = 8 times greater than in circuit 2. 2 SOLUTION a. The ratio of the emfs is, according to Equation 22.1 ξ1 vBL = =1 ξ 2 vBL b. Equation 20.2 states that the current is equal to the emf divided by the resistance. The ratio of the currents is I1 ξ1 /R1 R2 110 Ω = = = =2 I 2 ξ 2 /R2 R1 55 Ω c. The power, according to Equation 20.6c, is P = ξ 2/R. The motional emf is given by Equation 22.1 as ξ = vBL . The ratio of the powers is ξ12 P 1 = R1 = ξ1 2 ξ P2 ξ 2 2 R2 2 2 R2 v1BL = R1 v2 BL 2 2 R2 R1 v R 2v R = 1 2 = 2 1 2 = 8 v R v R 2 1 2 2 2 ______________________________________________________________________________ 1194 ELECTROMAGNETIC INDUCTION 5. SSM WWW REASONING AND SOLUTION For the three rods in the drawing in the text, we have the following: Rod A: The motional emf is zero , because the velocity of the rod is parallel to the direction of the magnetic field, and the charges do not experience a magnetic force. Rod B: The motional emf ξ is, according to Equation 22.1, ξ = vBL = (2.7 m/s)(0.45 T)(1.3 m) = 1.6 V The positive end of Rod B is end 2 . Rod C: The motional emf is zero , because the magnetic force F on each charge is directed perpendicular to the length of the rod. For the ends of the rod to become charged, the magnetic force must be directed parallel to the length of the rod. ______________________________________________________________________________ 6. REASONING The situation in the drawing given with the problem statement is analogous to that in Figure 22.4b in the text. The blood flowing at a speed v corresponds to the moving rod, and the diameter of the blood vessel corresponds to the length L of the rod in the figure. The magnitude of the magnetic field is B, and the measured voltage is the emf ξ induced by the motion. Thus, we can apply ξ = BvL (Equation 22.1). SOLUTION Using Equation 22.1, we find that ( ) ξ = BvL = ( 0.60 T )( 0.30 m/s ) 5.6 ×10−3 m = 1.0 ×10−3 V 7. REASONING The average power P delivered by the hand is given by P = W / t , where W is the work done by the hand and t is the time interval during which the work is done. The work done by the hand is equal to the product of the magnitude Fha...
View Full Document

## 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.

Ask a homework question - tutors are online