THE MAGNETIC FIELD THE MAGNETIC FIELD 32.1. Model: A magnetic field is caused by an electric current

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  • THE MAGNETIC FIELD

    32.1. Model: A magnetic field is caused by an electric current. Visualize: Please refer to Figure Ex32.1. Solve: Because the north poles of the magnets point counterclockwise, the magnetic force is counterclockwise. When you point fingers of your right hand counterclockwise, the thumb points up. Thus, the current in the wire is out of the page.

    32.2. Model: A magnetic field is caused by an electric current. Visualize: Please refer to Figure Ex32.2. The magnetic field is into the page on the left of the wire and it is out of the page on the right of the wire. Solve: Grab the wire with your right hand in such a way that your fingers point out of the page to the right of the wire. Since the thumb now points down, the current in the wire is down.

    32.3. Model: Visualize: Solve: The magnitude of the magnetic field at point 1 is 2.0 mT and its direction can be determined by using the right hand rule. Grab the current carrying wire so that your thumb points in the direction of the current. Because your fingers at point 1 point into the page, gl = (2.0 mT, into the page). At point 2, the magnetic field due to the bottom wire is into the page. The right-hand rule tells us that the magnetic field from the top wire is also into the page. At point 2, & = (4.0 mT, into the page). 32.4. Model: Visualize: Solve: The current in the wire is directed to the right. B2 = 20 mT + 20 mT = 40 mT because the two overlapping wires are carrying current in the same direction and each wire produces a magnetic field having the same direction at point 2. B, = 20 mT - 20 mT = 0 mT, because the two overlapping wires cany currents in opposite directions and each wire produces a field having opposite directions at point 3. The currents at 4 are also in opposite directions, but the point is to the right of one wire and to the left of the other. From the right-hand rule, the field of both currents is out of the page. Thus B,= 20 mT + 20 mT = 40 mT. 32.5. Model: The magnetic field is that of a moving charged particle. Visualize: V

    A magnetic field is caused by an electric current. Please refer to Figure Ex32.3.

    A magnetic field is caused by an electric current. Please refer to Figure 32.4.

    32-1

  • 32-2 Chapter 32

    The first point is on the x-axis, with 6, = 90". "he second point is on the y-axis, with 0, = 180", and the third point is on the -y-axis with 0, = 0". Solve: (a) Using Equation 32.1, the Biot-Savart law, the magnetic field strength is

    pa qvsine (lo-' T m / A ) ( 1 . 6 0 ~ 1 0 - ~ ~ C) (1 .0~10 '~ m/s)sin90" = 1.60 x 10-1~ T B, = -- - -

    4n r2 (1.0 x lo-? m)'

    To use the right-hand rule for finding the direction of 13, point your thumb in the direction of $. The magnetic field vector 13 is perpendicular to the plane of r' and ? and points in the same direction that your fingers point. In the present case, the fmgers point along the i direction. Thus, El = 1.60 x l O - I 5 i 2'. (b) B2 = 0 T because sin 0, = sin 180" = 0. (c) B, = 0 T because sin 6, = sin 0" = 0.

    32.6. Model: The magnetic field is that of a moving charged particle. Visualize: Y

    1

    = O T

    I -1 - 1 The first point is on the y-axis, the second point is on the x-axis, and the third point is in the xy plane. Solve:

    (lo" T m / A)(1.60 x (a) Using Equation 32.1, the Biot-Savart law, the magnetic field strength is

    po gvsine 4z r2 ( 1 . 0 ~ IO-' m)'

    C)(l.O x lo7 m / s)sinO" B, =--=

    (b) At point 2, 0, = 90" and the magnetic field is i2 = 1.60 x l O - " i T. We applied the right-hand rule to the negative charge, to find that the direction of Bz is -( -i) = i . (c) The field has the same direction as in part (b). The magnitude is

    T m / A)(1.60x C)(l.O x lo7 m / s)sin45" B3 = = 5.66 x T

    (1.0 x IO-' m)' + (1.0 x IO-' m)' and SO = 5.66 x 1O-I6i T.

    32.7. Model: Visualize: Solve: Using the Biot-Savart law,

    The magnetic field is that of a moving charged particle. Please refer to Figure Ex32.7.

    pa qvsine 4z r2

    T m / A)(1.60 ~ 1 O - l ~ C)(2.0 x lo7 m / s)sin135" = 1.13 x lo-'' T B = - -- -

    (1.0 x IO-' m)' + (1.0 x 10-2 m)* The right-hand rule applied to the proton points

    32.8. Model: Visualize: Solve: The Biot-Savart law is

    out of the page. Thus, 5 = 1.13 x l O - " i T . The magnetic field is that of a moving charged particle.

    Please refer to Figure Ex32.8.

    p vsine (10" T m / A ) ( 1 . 6 0 ~ 1 0 - ' ~ C)(2.0x1O7 m/s)sin135" 4n r2 B = A L = = 2.83 x T (2.0 x IO-? m)' + (2.0 x lo-' m)'

    The right-hand rule for the positive charge indicates the field points out of the page. Thus, = 2.83 x 10-l6i T.

  • The Magnetic Field 32-3

    32.9. Model: The magnetic field is that of a moving proton. Visualize: Z

    The magnetic field lies in the xy-plane. Solve: Using the Biot-Savart law,

    Using the right-hand rule, the charge is moving along the +z-direction. That is, V = v i .

    po qvsin8 (lo-’ T m / A)(1.60 x C)vsin90°

    4~ r2 (1.0 x 10-~ m)’ B=-- 1.0 x 10-l3 T = -v=6.25x1O6 m / s

    32.10. Model: The magnetic field is that of an electric current in a long straight wire. Solve: From Example 32.3, the magnetic field strength of a long straight wire carrying current I at a distance d from the wire is

    Po I B,, = -- 2a d

    The current needed to produce the earth’s magnetic field is calculated as follows:

    ( 2 x T m / A)Z Beanhsurfsce = 5 x 10” T = I = 2.5 A 0.01 m

    Likewise, the currents needed for a refrigerator magnet, a laboratory magnet, and a superconducting magnet are 250 A, 5000-50,000 A, and 500,000 A.

    32.11. Model: Solve: From Example 32.3, the magnetic field strength of a long straight wire carrying current I at a distance d from the wire is

    The magnetic field is that of an electric current in a long straight wire.

    B=-- Po * 2 z d

    The distance d at which the magnetic field is equivalent to Earth’s magnetic field is calculated as follows:

    BeanhrMacr = 5 x lo-’ T = (2 x T m / A)= * d = 4.0 cm d

    Likewise, the corresponding distances for a refrigerator magnet, a laboratory magnet, and a superconducting magnet are 0.4 mm, 20 p to 2 p, and 0.20 pm.

    32.12. Model: The magnetic field is that of a current loop. Solve: From Example 32.5, the on-axis magnetic field of a current loop is

    Also, B , , , (: = 0 m) = p01/2R. The ratio of these two fields is z3/’ = 0.354.

    32.13. Model: The magnetic field is that of a current loop. Solve: (a) From Equation 32.7, the magnetic field strength at the center of a loop is

    2 RBI,, crnler - 2(0.5 x lo-’ m)(2.5 x T) - = 2 0 A Pol 2R PO 4z(10-’ T m / A)

    B,oo,,”~r = - 3 I =

    - .

  • 32-4 Chapter 32

    (b) For a long, straight wire that carries a current I , the magnetic field strength is

    =3d=1 .60~10-~ m P I 2nd

    4n(10-' T m / A)(20 A) 2nd

    B,, = 0 3 2 . 5 0 ~ 1 0 - ~ T =

    32.14. Model: Assume the wires are infinitely long. Visualize: Please refer to Figure Ex3 1.14. Solve: The magnetic field strength at point 1 is

    - 1 2 x lo-' m 6 x lo-' m

    = ( 2 x lo-' T m / A)(10 A) 27z 2cm (4+2)cm

    3 Bat, =- -- Pol 3 E , , = (6.67 x lo9 T, out of page)

    At points 2 and 3,

    ia2 = (z, into page i,, = (z, into page

    = (2.0 x 10-4 T, into page)

    = (6.67 x T, out of page)

    32.15. Model: Assume the wires are infinitely long. Visualize: @ 10A

    @ 10A The field vectors are tangent to circles around the currents. The net magnetic field is the vectorial sum of the fields imp and Bbamm. Points 1 and 3 are at a distance d = f i cm from both wires and point 3 is at a distance d = 1 cm. Solve: The magnetic field at points 1, 2 , and 3 are

    B, = Bmp + &,,, = -(cos45"i^ POI - sin45"j) + -(cos45"1+ POI sin45"j) - - 2nd 27ld

    & x IO-' m

    ( 2 x 10" T m / A)(10 A) - B - - 0 1 P I : + - I POI c = 2 i = 4.0 x 10"; T ' - 2 n d 2nd 1 x IO-' m

    B3 - = A ( c 0 s 4 5 ~ 1 P I + s i n 4 5 " j ) + ~ ( c o s 4 5 " ~ * P I -sin45'j) = 2.OX lo4; T 2 d 2nd

  • The Magnetic Field 32-5

    32.16. Solve: (a) The magnetic dipole moment of the superconducting ring is

    ,~=((ltR~)I=n(l.OxlO-~ m)2(100A)=3.14x104 A m 2

    (b) From Example 32.5, the on-axis magnetic field of the superconducting ring is

    B , =yo 1R2 2Z(lO-’ T m / A)(100 A)(1.0x10-3 m)’ = 5.02 x T -

    M~ 2 ( z ~ + p ) 3 ’ 1 - [(O.OS mI2 + (0.001 m)‘]yL

    32.17. Model: Solve:

    Assume that the 10 cm distance is much larger than the size of the small bar magnet. (a) From Equation 32.9, the on-axis field of a magnetic dipole is

    4~ Bz3 (5.0 X IO4 T)(0.10 m)3 = 0.025Am’ B = --a P o 2P p =--=

    4n z3 Po 2 2(10-’ T m A)

    (b) As the bar magnet is rotated end-over-end by 180°, the magnitude of the magnetic field remains the same 5.0 pT, but the vector points in the opposite direction.

    32.18. Model: Solve:

    The radius of the earth is much larger than the size of the current loop. (a) From Equation 32.9, the magnetic field strength at the surface of the earth at the earth’s north pole is

    (2 x 10” T m / A) (~ .o x IO?’ A m2)

    (6.38 x lo6 m)’

    T given in Table 32.1.

    B = b k ? = = 6.16 x 10” T 2K z3

    This value is close to the value of 5 x (b) The current required to produce a dipole moment like that on the earth is

    p = A I = ( ~ ~ ) I = $ 8 x 1 O 2