ANSWER TO CHAPTER 28 HOMEWORK

1.  We find the direction of the field for each wire from the tangent to the circle around the wire, as shown 

          For their magnitudes, we have

                       B1 = (mo/4p)2I/r1

                                   = (10-7T.m/A)2(25A)/(0.120m) = 4.17 x 105T.

                       B2 = (mo/4p)2IB/r2   

                          = (10-7T.m/A)2(25A)/(0.050m) = 1.00 x 10-4T.

          We use the property of the triangle to find the angles shown:

                      r22 = r12 + d2 - 2r1dcosq1;

                     (5.0cm)2 = (1.20cm)2 + (15.0cm)2 - 2(12.0cm)(15.0cm)cosq1,

           which gives cosq1 = 0.956, q1 = 17.1°;

                      r12 = r22 + d2 - 2r2dcosq2;

                     (12.0cm)2 = (5.0cm)2 + (15.0cm)2 - 2(5.0cm)(15.0cm)cosq2,

           which gives cosq2 = 0.707, q2 = 45.0°.

           From the vector diagram, we have

                      B = B1(-cosq1i + sinq1j)+ B1(cosq2i + sinq2j)

                          = (4.17 x 10-5T)(-cos17.1°i +sin17.1°j) + (1.00 x 10-4T)(cos45.0°i + sin45.0°j)

                          =  (3.1 x 10-5T)i + (8.3 x 10-5T)j.

            We find the direction from

                     tan a = By/Bx = (8.39 x 10-5T)/(3.09 x 10-5T) = 2.68, a = 70.1°.

            We find the magnitude from

                     B = Bx/cosa = (3.09 x 10-5T)/cos70.1° =  8.9 x 10-5T, 70° above horizontal.

  
2.   Because the currents are in the same direction, between the wires the field will be in opposite directions.

          For the net field we have

                      B = B1 - B2 = [(mo/4p)2I1/x]j -[(mo/4p)2I2/(d- x)]j.

                          = (mo/4p)2I{[(d -x) - x]/x(d - x)}j

                          =  -[(mo/4p)2I(2d -x)/x(d - x)]j

 

  

3.  (a) If D is the diameter of the solenoid, the length of a coil is pD. Thus the number of turns is

                       N = Lwire/Lcoil = Lwire/pD.

              Because the coils are tightly wrapped, the length of the solenoid is

                       Lsolenoid = Nd = Lwired/pd = (20.0m)(2.00 x 10-3m)/p(2.50 x 10-2m) = 0.509m = 50.9cm.

         (b) The magnetic field at the center of the solenoid is 

                       B = moNI/L = moId = (4p x 10-7T.m/A)/(2.00 x 10-3m) =  1.26 x 10-2T.

                      

4.  Because the point C is along the line of the two straight segments of the wire, there is no magnetic field from these segments.

         The magnetic field at the point C is the sum of two fields:

                      B =  Binner arc + Bouter arc.

         Each field is a portion of a circular loop, with the field of the inner arc out the page and that of the outer arc into the page,

         so we subtract the two magnitudes:

                      B = (q/2p)(moI/2R1) - (q/2p)(moI/2R2)

                         = (moIq/4p)[(1/R1) - (1/R2)]

                         =  moIq(R2 - R1)/4pR1R2 out of the page.

  
 5.  (a)  We choose the y-axis along the wire. We choose the differential element dy and use the angle f indicated on the figure to specify the location of the point where we want to find the magnetic field.

                From the figure, we see that

                        tanf = y/R, cosf = R/r, and sinf = y/r, and q = f + p/2.

               The angle f will vary from -fo to fo = tan-1(l/2R)

               We relate the change in angle to the change in y from

                        y = Rtanf;

                       dy = Rsec2fdf = (R/cos2f)df.

                The field from each of the differential elements will be circular about the wire, that is, into the page, so the integration of dB becomes a scalar integration of the magnitude:

                       B = ò dB = ò y = -l/2y= l/2 (moI/4p)|dy x r|/r3 = (moI/4p)òy=-l/2y=l/2dy|sin(f + p/2)|/r3

                          = moI/4p ò-fofo[R/cos2fdf(cosf)]/(R/cosf)2

                                  = moI/4p(sinf) |-fofo, which gives

                      B = (moI/4p)(2sinfo). with sinfo= (l/2)/[l/2)2 + R2]1/2 = l/(l2 + 4R2)1/2, we get

                      B = moI/2pR(l2 + 4R2)1/2  circular.

            (b) We approximate an infinite wire by having l >> R:

                     B = (moI/2pR)/[1 + (2R/l)2]1/2 » moI/2pR,

                  which is the field for an infinite wire.

 

 		 
6*.(a) The figure shows a view looking directly at the current. the sheet may be thought of as an infinite number of parallel wires. we choose a differential element of width dx a distance x from the center of the strip. this element has a current dI = (I/L)dx which produces a magnetic field dB =(mo/4p)2dI/r = (moI/2pL)dx/r, in the direction shown. Because a differential element at -x will produce a field of the same magnitude but below the horizontal, the symmetry means the resultant field will be parallel to the strip in the x-direction. we find the total field by adding (integrating) the x-components of the differential fields:

               B = òsinq dB  = ò-L/2L/2(moI/2pL)[y/(x2 + y2)]dx

                  = (moIy/2pL)[(1/y)tan-1(x/y)|-L/2L/2 = (moI/2pL)tan-1(L/2y).

(b) If y >> L, or L/2y<< 1, the angle is small and equal to the tangent, so we have

               tan-1(L/2y) = L/2y, and 

               B = (moI/2pL)(L/2y) = (mo/4p)(2I/y),

which is the magnetic field produced by a long wire. This is what the strip will appear to be when we are far from the strip.

 

7*. We find the constants in the current densities in the wires by requiring that the total current in each wire be I0. Fopr the inner wire we have

                I0 = ò0R1j12prdr = 2pC1ò0R1r2dr = 2pC1R13/3,

which gives C1 =3I0/2pR13.

For the outer wire we have

                I0 = òR2R3j22prdr = 2pC2òR2R3r2dr = 2pC2(R33 - R23)/<3.

which gives C1 = (3I0/2p)/(R33 - R23).

>

Because of the cylindrical symmetry, we know that the magnetic fields will be circular. In each case we apply Ampere's law to a circulaar path of radiusr.

(a) Inside the inner wire, r < R1;

                òB · ds = moIenclosed;

                B2pr = moò0rC1r2prdr = C12pmor3/3, which gives

                B = moC1r2/3 = (mor2/3)(3I0/2pR13) = (moI0/2pR13)r2 circular CCW, R < R1.

(b) Between the wires, R1 < r < R2:

                òB · ds = moIenclosed;

                B2pr = moI0 = (moI0/2pr circular CCW, R1 < r < R2.

(c) Inside the outer wire, R2 < r < R3:

                òB · ds = moIenclosed;

                B2pr = mo(I0 - ò>R2rC2r2prdr) = mo[I0 - (C22p/3)(r3 - R23)], whhich gives

                B = (moI0/2pr)[1 - (r3 - R223)/(R33 - R23)]

>

                   = (moI0/2pr)(R33 - r3)/(R3<3 - R23) circular CCW, R2 < r < R3.

(d) Outside the outer wire, r > R3:

                 òB · ds = moIenclosed;

                   B2pr = mo(I0 - I0), which gives&nbssp; B = 0, r > R3.

8*. The magnetic field from each side of the loop will be out of the page. We could find the field from each side by selecting a differential element and using the Biot-Savart law. We will use the result from Problem 37 by considering each side to be made of two pieces and adding the fields at P. We have labeled the currents to distinguish the side. Thus we have

               B1 = (moI0/4p)[x/y(x2 + y2)]1/2 + (moI0/4p)[(b - x)/y[(b -x)2 + y2)]1/2;

               B2 = (moI0/4p)y/{(b - x)[y2 + (b - x)2]1/2} + (moI0/4p)(a - y)/{(b - x)[(a - y)2 + (b - x)2)]1/2;

               B3 = (moI0/4p)x/{(a - y)[x2 + (a - y)2]1/2} + (moI0/4p)(b - x)/{(a - y)[(b - x)2 + (a - y)2)]1/2;

               B4 = (moI0/4p)[y/x(y2 + x2)]1/2 + (moI0/4p)[(a - y)/y[(a -y)2 + x2)]1/2.

We can simplify the algebra by putting the constants with the field. thus the net field is

               4pB/moI0 = x/y(x2 + y2)1/2 + (b - x)/y[(b -x)2 + y2)]1/2 +y/(b - x)[y2 + (b - x)2]1/2 + (a - y)/(b - x)[(a - y)2 + (b - x)2]1/2 

                                            + x/(a - y)[x2 + (a - y)2]1/2 + (b - x)/(a - y)[(b - x)2 + (a - y)2)]1/2 + y/x(x2 + y2)1/2 + (a -y)/x[(a - y)2 + x2]1/2

                              = [(x/y) +(y/x)]/(x2 + y2)1/2 + {[(b - x)/y] + y/(b - x)}/[(b -x)2 + y2)]1/2 + {[(a - y)/(b -x)] + [(b -x)/(a - y)]}/[(a - y)2 + (b - x)2)]1/2 + {[x/(a - y)] + [(a - y)/x}/[x2 + (a - y)2]1/2, so we get

                B = (moI/4p){(x2 + y2)1/2/xy + [(b -x)2 + y2)]1/2/y(b -x) + 

                       [(a - y)2 + (b - x)2)]1/2 /(a - y)(b - x) + [x2 + (a - y)2]1/2/x(a - y)} out of the page.

 

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