Physics Solution Manual for 1100 and 2101

# 5 to the magnetic fields on the right side of 2 r

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Unformatted text preview: LB sin θ − ILB sin θ = ( I1 − I ) LB sin θ Solving for the unknown current I gives I = I1 − Fnet LB sin θ = 7.00 A − 3.13 N = 3.00 A ( 2.40 m )( 0.360 T ) sin 65.0° 37. REASONING A maximum magnetic force is exerted on the wire by the field components that are perpendicular to the wire, and no magnetic force is exerted by field components that are parallel to the wire. Thus, the wire experiences a force only from the x- and ycomponents of the field. The z-component of the field may be ignored, since it is parallel to the wire. We can use the Pythagorean theorem to find the net field in the x, y plane. This net field, then, is perpendicular to the wire and makes an angle of θ = 90° with respect to the wire. Equation 21.3 can be used to calculate the magnitude of the magnetic force that this net field applies to the wire. SOLUTION According to Equation 21.3, the magnetic force has a magnitude of F = ILB sin θ, where I is the current, B is the magnitude of the magnetic field, L is the length of the wire, and θ is the angle of the wire with respect to the field. Using the Pythagorean theorem, we find that the net field in the x, y plane is B= 2 2 Bx + By Using this field in Equation 21.3, we calculate the magnitude of the magnetic force to be Chapter 21 Problems 1159 2 2 F = ILB sinθ = IL B x + B y sinθ bg b = 4 .3 A 0.25 m 0.10 0.15 gb T g+ b T g sin 90° = 2 2 0.19 N 38. REASONING There are four forces that act on the wire: the magnetic force (magnitude F), the weight mg of the wire, and the tension in each of the two strings (magnitude T in each string). Since there are two strings, the following drawing shows the total tension as 2T. The magnitude F of the magnetic force is given by F = I LB sin θ (Equation 21.3), where I is the current, L is the length of the wire, B is the magnitude of the magnetic field, and θ is the angle between the wire and the magnetic field. In this problem θ = 90.0°. The direction of the magnetic force is given by Right-Hand Rule No. 1 (see Section 21.5). The drawing shows an end view of the wire, where it can be seen that the magnetic force (magnitude = F ) points to the right, in the +x direction. I (out of page) φ B +y 2T F +x mg In order for the wire to be in equilibrium, the net force ΣFx in the x-direction must be zero, and the net force ΣFy in the y-direction must be zero: ΣFx = 0 (Equation 4.9a) and ΣFy = 0 (Equation 4.9b). These equations will allow us to determine the angle φ and the tension T. SOLUTION Since the wire is in equilibrium, the sum of the forces in the x direction is zero: −2T4 φ +3 = 0 sin F 14 244 ΣFx Substituting in F = I LB sin 90.0° for the magnitude of the magnetic force, this equation becomes −2T sin φ + I LB sin 90.0° = 0 (1) 1444 24444 4 3 ΣFx The sum of the forces in the y direction is also zero: 1160 MAGNETIC FORCES AND MAGNETIC FIELDS +2T cos φ − mg = 0 14 244 4 3 (2) ΣFy Since these two equations contain two unknowns, φ and T, we can solve for each of them. mg a. To obtain the angle φ, we solve Equation (2) for the tension T = and substitute 2 cos φ the result into Equation (1). This gives mg −2 sin φ + I LB sin 90.0° = 0 2 cos φ sin φ I LB = cos φ mg 123 tan φ or Thus, I LB −1 ( 42 A )( 0.20 m )( 0.070 T ) φ = tan = 37° = tan mg ( 0.080 kg ) 9.80 m/s2 −1 ( ) b. The tension in each wire can be found directly from Equation (2): ( ) ( 0.080 kg ) 9.80 m/s 2 mg T= = = 0.49 N 2 cos φ 2 cos 37° 39. SSM REASONING Since the rod does not rotate about the axis at P, the net torque relative to that axis must be zero; Στ = 0 (Equation 9.2). There are two torques that must be considered, one due to the magnetic force and another due to the weight of the rod. We consider both of these to act at the rod's center of gravity, which is at the geometrical center of the rod (length = L), because the rod is uniform. According to Right-Hand Rule No. 1, the magnetic force acts perpendicular to the rod and is directed up and to the left in the drawing. Therefore, the magnetic torque is a counterclockwise (positive) torque. Equation 21.3 gives the magnitude F of the magnetic force as F = ILB sin 90.0°, since the current is perpendicular to the magnetic field. The weight is mg and acts downward, producing a clockwise (negative) torque. The magnitude of each torque is the magnitude of the force times the lever arm (Equation 9.1). Thus, we have for the torques: τ magnetic = + ( ILB ) ( L / 2 ) {1 3 2 force lever arm and τ weight = − ( mg ) ( L / 2 ) cos θ { 14 244 4 3 force lever arm Setting the sum of these torques equal to zero will enable us to find the angle θ that the rod makes with the ground. Chapter 21 Problems 1161 SOLUTION Setting the sum of the torques equal to zero gives Στ = τmagnetic + τweight = 0, and we have ILB + ( ILB )( L / 2 ) − ( mg ) ( L / 2 ) cos θ = 0 or cos θ = mg 4.1 A 0.45 m 0.36 T ( )( )( ) = ( 0.094 kg ) 9.80 m/s 2 θ = cos –1 ( ) 44° 40. REASONING AND SOLUTION a. From Right-Hand Rule No. 1, if we extend the right hand so that the fingers point in the direction of the magnetic field, and the thumb points in the direction o...
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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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