Physics Solution Manual for 1100 and 2101

# The area of the loop is 070 m 050 m 035 m2 solution a

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Unformatted text preview: N 3 1 − = 5.25 ×103 m/s v = v0 1 − = 6.50 ×10 m/s q0 E 4.00 ×10−12 C ( 2470 N/C ) ( ) ( 25. SSM REASONING From the discussion in Section 21.3, we know that when a charged particle moves perpendicular to a magnetic field, the trajectory of the particle is a circle. The drawing at the right shows a particle moving in the plane of the paper (the magnetic field is perpendicular to the paper). If the particle is moving initially through the coordinate origin and to the right (along the +x axis), the subsequent circular path of the particle will intersect the y axis at the greatest possible value, which is equal to twice the radius r of the circle. ) y ymax r x v Particle SOLUTION a. From the drawing above, it can be seen that the largest value of y is equal to the diameter (2r) of the circle. When the particle passes through the coordinate origin its velocity must be parallel to the +x axis. Thus, the angle is θ = 0 ° . b. The maximum value of y is twice the radius r of the circle. According to Equation 21.2, the radius of the circular path is r = mv / ( q B ) . The maximum value ymax is, therefore, ymax ( ) ) 3.8 ×10 –8 kg ( 44 m/s ) mv = = 2r = 2 = 2 q B 7.3 ×10 –6 C (1.6 T ) ( 0.29 m 26. REASONING AND SOLUTION The magnitudes of the magnetic and electric forces must be equal. Therefore, FB = FE or q vB = q E This relation can be solved to give the speed of the particle, v = E /B. We also know that when the electric field is turned off, the particle travels in a circular path of radius r = mv/( q B). Substituting v = E /B into this equation and solving for q /m gives 1154 MAGNETIC FORCES AND MAGNETIC FIELDS q m = E 3.80 × 103 N/C = = 6.8 × 105 C/kg 2 2 −2 rB 4.30 × 10 m ( 0.360 T ) ( ) 27. REASONING a. When the particle moves in the magnetic field, its path is circular. To keep the particle moving on a circular path, it must experience a centripetal force, the magnitude of which is given by Equation 5.3 as Fc = mv 2 / r . In the present situation, the magnetic force F furnishes the centripetal force, so Fc = F. The mass m and speed v of the particle are known, but the radius r of the path is not. However, the particle travels at a constant speed, so in a time t the distance s it travels is s = vt. But the distance is one-quarter of the circumference ( 2π r ) of a circle, so s = 1 ( 2π r ) . By combining these three relations, we can determine the 4 magnitude of the magnetic force. b. The magnitude of the magnetic force is given by Equation 21.1 as F = q vB sin θ . Since F, v, B, and θ are known, this relation can be used to determine the magnitude q of the charge. SOLUTION a. The magnetic force, which provides the centripetal force, is F = mv 2 / r . Solving the relation s = 1 ( 2π r ) for the radius and substituting s = vt into the result gives 4 r= 2s π = 2 ( vt ) π Using this expression for r in Equation 5.3, we find that the magnitude of the magnetic force is ( ) −8 mv 2 mv 2 π mv π 7.2 × 10 kg (85 m/s ) F= = = = = 4.4 × 10−3 N −3 2vt r 2t 2 2.2 × 10 s π ( ) b. Solving the relation F = q vB sin θ for the magnitude q of the charge and noting that θ = 90.0° (since the velocity of the particle is perpendicular to the magnetic field), we find that F 4.4 × 10−3 N q= = = 1.7 × 10−4 C vB sin 90.0° ( 85 m/s )( 0.31 T ) sin 90.0° Chapter 21 Problems 1155 28. REASONING When the electron travels perpendicular to a magnetic field, its path is a circle. The radius of the circle is given by Equation 21.2 as r = mv / ( q B ) . All the variables in this relation are known, except the speed v. However, the speed is related to the electron’s kinetic energy KE by KE = 1 mv 2 (Equation 6.2). By combining these two relations, we 2 will be able to find the radius of the path. SOLUTION Solving the relation KE = r = mv / ( q B ) give r= mv = qB m 1 mv 2 2 for the speed and substituting the result into 2 ( KE ) 2 9.11 × 10−31 kg 2.0 ×10−17 J m = 2m ( KE ) = = 0.71 m qB qB 1.60 × 10−19 C 5.3 × 10−5 T ( ( )( )( ) ) Values for the mass and charge of the electron have been taken from the inside of the front cover. 29. REASONING AND SOLUTION The drawings show the two circular paths leading to the target T when the proton is projected from the origin O. In each case, the center of the circle is at C. Since the target is located at x = –0.10 m and y = –0.10 m, the radius of each circle is r = 0.10 m. The speed with which the proton is projected can be obtained from –19 Equation 21.2, if we remember that the charge and mass of a proton are q = +1.60 × 10 C and m = 1.67 × 10–27 kg, respectively: v= rqB m ( 0.10 m ) (1.60 × 10 –19 C ) ( 0.010 T ) = 1.67 × 10 –27 kg +y O T C = 9.6 × 104 m/s +y +x C T O +x 1156 MAGNETIC FORCES AND MAGNETIC FIELDS 30. REASONING A magnetic field exerts no force on a current-carrying wire that is directed along the same direction as the field. This follows directly from F = ILB sin θ (Equation 21.3), which gives the magnitude F of the magne...
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