Physics Solution Manual for 1100 and 2101

Thus the angle is 0 b the maximum value of y is twice

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Unformatted text preview: sin 90.0° ( 8. ) REASONING According to Equation 21.1, the magnetic force has a magnitude of F = q vB sin θ. The field B and the directional angle θ are the same for each particle. Particle 1, however, travels faster than particle 2. By itself, a faster speed v would lead to a greater force magnitude F. But the force on each particle is the same. Therefore, particle 1 must have a smaller charge to counteract the effect of its greater speed. SOLUTION Applying Equation 21.1 to each particle, we have F = q1 v B sin θ 144 1 244 3 Particle 1 and F = q2 v2 B sin θ 1442443 Particle 2 Dividing the equation for particle 1 by the equation for particle 2 and remembering that v1 = 3v2 gives Chapter 21 Problems q v B sin θ F = 11 F q2 v2 B sin θ 9. or 1= q1 v1 q2 v2 or q1 q2 = v2 v1 = v2 3v2 = 1141 1 3 REASONING The positive plate has a charge q and is moving downward with a speed v at right angles to a magnetic field of magnitude B. The magnitude F of the magnetic force exerted on the positive plate is F = q vB sin 90.0°. The charge on the positive plate is related to the magnitude E of the electric field that exists between the plates by (see Equation 18.4) q = ε0AE, where A is the area of the positive plate. Substituting this expression for q into F = q vB sin 90.0° gives the answer in terms of known quantities. SOLUTION F = ( ε 0 AE ) vB ( ) = 8.85 × 10−12 C2 /(N ⋅ m 2 ) 7.5 × 10−4 m 2 (170 N/C )( 32 m/s )( 3.6 T ) = 1.3 × 10−10 N An application of Right-Hand Rule No. 1 shows that the magnetic force is perpendicular to the plane of the page and directed out of the page , toward the reader. 10. REASONING a. If the particle is stationary, only the electric field Ex exerts a force on it; this force is FE = q Ex (Equation 18.2), where q is the charge. A magnetic field does not exert a force on a stationary particle. b. If the particle is moving along the +x axis, the electric field exerts a force on it. This force is the same as that in part (a) above. The magnetic field Bx does not exert a force on the particle, because the particle’s velocity is parallel to the field. The magnetic field By does exert a force on the particle, because the particle’s velocity is perpendicular to this field. c. The particle experiences a force from each of the three fields. The electric field exerts a force on it, and this force is the same as that in part (a) above. It does not matter in which direction the particle travels, for the electric force is independent of the particle’s velocity. The particle experiences magnetic forces from both Bx and By. When the particle moves along the +z axis, its velocity is perpendicular to both Bx and By, so each field exerts a force on the particle. . 1142 MAGNETIC FORCES AND MAGNETIC FIELDS SOLUTION a. The electric force exerted on the particle is FE = q Ex (Equation 18.2), so ( ) FE = qE x = +5.60 × 10−6 C ( +245 N/C ) = +1.37 × 10−3 N where the plus sign indicates that the force points along the + x axis . The magnitude F of the magnetic force is given by Equation 21.1 as F = q0 vB sin θ . Since v = 0 m/s, the magnetic forces exerted by Bx and By are zero: FB = 0 N x FB = 0 N y b. The electric force is the same as that computed in part (a), because this force does not depend on the velocity of the particle: FE = +1.37 × 10−3 N , where the plus sign indicates that the force points along the + x axis . Since the velocity of the particle and Bx are along the +x axis (θ = 0° in Equation 21.1) , the magnitude of the magnetic force is FB = q0 vBx sin θ = q0 vBx sin 0° = 0 N x The magnetic force exerted by the magnetic field By on the charge has a magnitude of ( ) FB y = q0 vB y sin θ = 5.60 × 10−6 C ( 375 m/s )(1.40 T ) sin 90.0° = 2.94 × 10−3 N An application of Right-hand Rule No. 1 shows that the direction of the magnetic force is along the + z axis . c. The electric force is the same as that computed in part (a), because this force does not depend on the velocity of the particle: FE = +1.37 × 10−3 N , where the plus sign indicates that the force points along the + x axis . When the particle moves along the +z axis, the magnetic field Bx exerts a force on the charge that has a magnitude of ( ) FB x = q0 vBx sin θ = 5.60 ×10−6 C ( 375 m/s )(1.80 T ) sin 90.0° = 3.78 × 10−3 N Chapter 21 Problems 1143 An application of Right-hand Rule No. 1 shows that the direction of the magnetic force is along the + y axis . When the particle moves along the +z axis, the magnetic field By exerts a force on the charge that has a magnitude of ( ) FB y = q0 vB y sin θ = 5.60 × 10−6 C ( 375 m/s )(1.40 T ) sin 90.0° = 2.94 × 10−3 N An application of Right-hand Rule No. 1 shows that the direction of the magnetic force is along the − x axis . 11. SSM WWW REASONING The direction in which the electrons are deflected can be determined using Right-Hand Rule No. 1 and reversing the direction of the force (RHR-1 applies to positive charges, and electrons are negatively charged). Each electron experiences an acceleration a given by Newton’s second law of motion,...
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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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