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1 33.3. Model: The magnetic field is that of a moving charged particle. Visualize: The first point is on the x-axis, with θ a = 90. The second point is on the y-axis, with θ b = 180, and the third point is on the y-axis with θ c = 0. Solve: (a) Using Equation 33.1, the Biot-Savart law, the magnetic field strength is ( )( )( ) ( ) μ0 qvsinθ 10 T m/a C m/s sin 90 Ba = = = 4π r m T To use the right-hand rule for finding the direction of B, point your thumb in the direction of v. The magnetic field vector B is perpendicular to the plane of r and v and points in the same direction that your fingers point. In the present case, the fingers point along the ˆk 15 direction. Thus, B = kˆ T. (b) B b = 0 T because sin θ b = sin 180 = 0. (c) B c = 0 T because sin θ c = sin 0 = 0. a

2 33.9. Model: The magnetic field is that of an electric current in a long straight wire. Solve: From Example 33.3, the magnetic field strength of a long, straight wire carrying current I at a distance d from the wire is μ0 I B = π d The distance d at which the magnetic field is equivalent to Earth s magnetic field is calculated as follows: B ( ) 10 A d 5 7 earth surface = 5 10 T = 10 T m/a d = 4.0 cm Likewise, the corresponding distances for a refrigerator magnet, a laboratory magnet, and a superconducting magnet are 0.40 mm, 0 μm to.0 μm, and 0.0 μm.

3 Model: Assume the wires are infinitely long. Visualize: The field vectors are tangent to circles around the currents. The net magnetic field is the vectorial sum of the fields B top and B. bottom Points a and c are at a distance d = cm from both wires and point b is at a distance d = 1 cm. Solve: The magnetic field at points a, b, and c are μ I πd πd 7 μ ( 10 T m/a)( 10 A) 0I ˆ 1 ˆ 4 = cos 45 i = i.0 10 iˆ T π d = 10 m ( ˆ ˆ μ I cos45 sin 45 ) ( cos45 ˆ sin 45 ˆ) 0 0 Ba = Btop + Bbottom = i j + i + j ( 10 7 T m/a)( 10 A) μ0i ˆ μ0i ˆ ˆ 4 B ˆ b = i + i = i = i T πd πd 1 10 m μ I 4 ( ) ( ) μ I cos 45 ˆ sin 45 ˆ cos 45 ˆ sin 45 ˆ.0 10 ˆ T πd πd 0 0 Bc = i + j + i j = i

4 Model: Assume that the 10 cm distance is much larger than the size of the small bar magnet. Solve: (a) From Equation 33.9, the on-axis field of a magnetic dipole is μ ( )( ) 0 μ 4π Bz T 0.10 m B = μ = = = 0.05 A m 3 7 4π z μ 10 T m/a (b) The on-axis field strength 15 cm from the magnet is B 0 ( ) ( ) μ μ 0.05 A m ( 10 T m/a) T 1.48μT 4π z 0.15 m = = = = 3 3 ( )

5 33.1. Model: The magnetic field is that of the three currents enclosed by the loop. Visualize: Please refer to Figure EX33.1. Solve: Ampere s law gives the line integral of the magnetic field around the closed path: 6 B ds = μ0ithrough = T m = μ I I + I = 4π 10 7 T m/a 6.0 A 4.0 A + I ú ( ) ( )( ) T m 3 + = 7 ( I.0 A) I 3 = 1.0 A 4π 10 T m/a Assess: The right-hand rule was used above to assign positive signs to I 1 and I 3 and a negative sign to I.

6 33.7. Model: A magnetic field exerts a magnetic force on a moving charge. Visualize: Please refer to Figure EX33.7. Solve: (a) The force is C ˆ m/s 0.50 ˆ F = qv B= j i T = kˆ N on q ( )( ) ( ) (b) The force is 19 7 ˆ ˆ ˆ F = C m/s cos 45 j+ sin 45 k 0.50 i T = ˆjkˆ N on q ( )( )( ) ( ) ( )

7 Model: Assume the magnetic field is uniform over the Hall probe. Visualize: Please refer to Figure EX33.4(a). The thickness is t = m. Solve: The Hall voltage is given by Equation 33.4: IB IB ( 15 A)( 1.0 T) Δ VH = n = = =.9 10 m tne teδ V m C V H ( )( )( ) 8 3 Assess: The conduction electron density in metals is of the order of m 3 (Table 31.1). The value obtained for the charge carrier density is reasonable.

8 Model: Two parallel wires carrying currents in the same direction exert attractive magnetic forces on each other. Visualize: Please refer to Figure EX The current in the circuit on the left is I 1 and has a clockwise direction. The current in the circuit on the right is I and has a counterclockwise direction. Solve: Since I 1 = 9 V Ω= 4.5 A, the force between the two wires is 5 μ0li1i ( 10 7 T m/a)( 0.10 m)( 4.5 A) F = N = = π d m 9 V I = 3.0 A R = = 3.0 Ω 3.0 A I

9 Model: The torque on the current loop is due to the magnetic field produced by the current-carrying wire. Assume that the wire is very long. Visualize: Please refer to Figure EX Solve: (a) From Equation 33.7, the magnitude of the torque on the current loop is τ = μbsin θ, where μ = I loop A and B is the magnetic field produced by the current I wire in the wire. The magnetic field of the wire is tangent to a circle around the wire. At the position of the loop, B points up and is θ = 90 from the axis of the loop. Thus, 7 ( ) π ( ) ( )( ) μ I 0.0 A m 10 T m/a.0 A sin90 τ = ( I A) = = π d.0 10 m 0 wire 11 loop sinθ N m Note that the magnetic field produced by the wire on the current loop is up so that the angle θ between B and the normal to the loop is 90. (b) The loop is in equilibrium when θ = 0 or 180. That is, when the coil is rotated by ± 90.

10

11 Model: Use the Biot-Savart law for a current carrying segment. Visualize: Please refer to Figure P The distance from P to the inner arc is r 1 and the distance from P to the outer arc is r. Solve: As given in Equation 33.6, the Biot-Savart law for a current carrying small segment Δs is μ ˆ 0 IΔ s r B = 4π r For the linear segments of the loop, B Δ s = 0 T because Δ s rˆ = 0. Consider a segment arc. Because Δs is perpendicular to the ˆr vector, we have I s Ir I π 0 Δ 0 1Δ 0 Δ B Id I I arc 1 r1 r1 r π 1 π r1 r1 r1 μ μ θ μ θ μ θ μ μ B= = = = = = 4π 4π 4π 4π 4π 4 Δs on length on the inner A similar expression applies for B arc. The right-hand rule indicates an out-of-page direction for B arc and an intopage direction for B arc 1. Thus, μ0i μ0i μ0i 1 1 B =, into page +, out of page =, into page 4r1 4r 4 r1 r The field strength is 7 ( 4π 10 T m/a)( 5.0 A) B = = T m 0.00 m Thus B = ( T, into page).

12 Model: A 1000-km-diameter ring makes a loop of diameter 3000 km. Visualize: Solve: (a) The current loop has a diameter of 3000 km, so its nominal area, ignoring curvature effects, is A loop = π r = π ( m) = m Because the magnetic dipole moment of the earth is modeled to be due to a current flowing in such a loop, μ = IA loop. The current in the loop is μ A m 10 I = = = A A m loop (b) The current density J in the above loop is (c) The current density in the wire is J J 10 I A loop = = = A/m A π 1 3 ( m ) I 1.0 A = = = A/m A π 1 3 ( m ) wire 6 You can see that J loop << J wire. The current in the earth s core is large, but the current density is actually quite small.

13 Model: The magnetic field is that of a current in the wire. Visualize: Please refer to Figure P Solve: As given in Equation 33.6 for a current carrying small segment Δs, the Biot-Savart law is μ ˆ 0 IΔ s r B = 4π r For the straight sections, Δ s rˆ = 0 because both Δs and ˆr point along the same line. That is not the case with the curved section over which Δs and r are perpendicular. Thus, μ0 IΔs μ0 IR dθ μ0i dθ B = = = 4π r 4π R 4πR where we used Δ s = RΔθ Rdθ for the small arc length Δs. Integrating to obtain the total magnetic field at the center of the semicircle, π / μ0idθ μ0i μ0i B = = π = 4πR 4πR 4R π /

14 Model: The magnetic field is that of the current which is distributed uniformly in the hollow wire. Visualize: Ampere s integration paths are shown in the figure for the regions 0 m < r < R 1, R 1 < r < R, and R < r. Solve: For the region 0 m < r < R 1, B ds =μ0ithrough. Because the current inside the integration path is zero, ú B = 0 T. To find I through in the region R 1 < r < R, we multiply the current density by the area inside the integration path that carries the current. Thus, I Ithrough = ( r R 1 ) π ( R R1 ) π where the current density is the first term. Because the magnetic field has the same magnitude at every point on the circular path of integration, Ampere s law simplifies to ( ) ( R R1 ) I r R1 μ0i r R 1 ú B ds = B ds= B( πr) = μ0 B= π r R R1 For the region R < r, I through is simply I because the loop encompasses the entire current. Thus, μ0i úb ds= B ds= Bπ r= μ0i B = π r Assess: The results obtained for the regions r > R and R 1 < r < R yield the same result at r = R. Also note that a hollow wire and a regular wire have the same magnetic field outside the wire.

15 Model: Energy is conserved as the electron moves between the two electrodes. Assume the electron starts from rest. Once in the magnetic field, the electron moves along a circular arc. Visualize: The electron is deflected by 10 after moving along a circular arc of angular width 10. Solve: Energy is conserved as the electron moves from the 0 V electrode to the 10,000 V electrode. The potential energy is U = qv with q = e, so 1 K + U = K + U mv ev = 0+ 0 f f i i 19 ev ( C)(10,000 V) v = = = 31 m kg m/s The radius of cyclotron motion in a magnetic field is r = mv/eb. From the figure we see that the radius of the circular arc is r = (.0 cm)/sin10. Thus 31 7 mv ( kg)( m/s) 3 B = = =.9 10 T 19 er ( C)(0.00 m)/sin10

16 Model: Charged particles moving perpendicular to a uniform magnetic field undergo uniform circular motion at constant speed. Solve: (a) The magnetic force on a proton causes a centripetal acceleration: mv ebr evb = v = r m Maximum kinetic energy is achieved when the diameter of the proton s orbit matches the diameter of the cyclotron: 19 ( C) ( 0.75 T) ( 0.35 m) ( ) 1 ebr 13 mv 7 K = = = = J m kg (b) The proton accelerates through a potential difference of 500 V twice during one revolution. The energy gained per cycle is qδv = e (500 V) = J Using the maximum kinetic energy of the proton from part (a), the number of cycles before the proton attains this energy is J = J

17 Model: Assume that the magnetic field is uniform over the Hall probe. Solve: Equation 33.4 gives the Hall voltage and Equation 33.0 gives the cyclotron frequency in terms of the magnetic field. We have IB π mf Δ VH = B = cyl tne e π mfcyci π ( kg)( Hz)( A) tne = = = T A/V 19 3 eδ VH ( C)( V) With this value of tne, we can once again use the Hall voltage equation to find the magnetic field: 3 Δ VH V B= tne= 3 ( TA/V ) =.10 T I A

18 Model: A magnetic field exerts a magnetic force on a length of current-carrying wire. Visualize: Please refer to Figure P Solve: The above figure shows a side view of the wire, with the current moving into the page. From the righthand rule, the magnetic field B points down to give a leftward force on the current. The wire is hanging in static equilibrium, so F net = F mag + FG + T = 0 N. Consider a segment of wire of length L. The wire s linear mass density is μ = kg/m, so the mass of this segment is m = μl and its weight is F G = mg = μlg. The magnetic force on this length of wire is F mag = ILB. In component form, Newton s first law is ( ) F = Tsinθ F = Tsinθ ILB= 0 N Tsinθ = ILB net ( ) Dividing the first equation by the second, x mag Fnet = Tcosθ FG = Tcosθ μlg = 0 N Tcosθ = μlg y ( )( ) T sin tan ILB IB θ μg tanθ kg/m 9.8 m/s tan10 = θ = = Tcosθ B = = = T μlg μg I 10 A B = 8.6 mt, down. The magnetic field is ( )

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