Halliday ♦ Resnick ♦Walker

FUNDAMENTALS OF PHYSICSSIXTH EDITION

Selected Solutions

Chapter 30

30.1130.2530.3730.47

11. Our x axis is along the wire with the origin at the midpoint. The current flows in the positive xdirection. All segments of the wire produce magnetic fields at P1 that are out of the page. According tothe Biot-Savart law, the magnitude of the field any (infinitesimal) segment produces at P1 is given by

dB =µ0i

4πsin θ

r2 dx

where θ (the angle between the segment and a line drawn from the segment to P1 ) and r (the lengthof that line) are functions of x. Replacing r with

√x2 + R2 and sin θ with R/r = R/

√x2 + R2, we

integrate from x = −L/2 to x = L/2. The total field is

B =µ0iR

4π

∫ L/2

−L/2

dx

(x2 + R2)3/2 =µ0iR

4π1

R2

x

(x2 + R2)1/2

∣∣∣∣∣L/2

−L/2

=µ0i

2πR

L√L2 + 4R2

.

If L � R, then R2 in the denominator can be ignored and

B =µ0i

2πR

is obtained. This is the field of a long straight wire. For points very close to a finite wire, the field isquite similar to that of an infinitely long wire.

25. Each wire produces a field with magnitude given by B = µ0i/2πr, where r is the distance from thecorner of the square to the center. According to the Pythagorean theorem, the diagonal of the squarehas length

√2a, so r = a/

√2 and B = µ0i/

√2πa. The fields due to the wires at the upper left and lower

right corners both point toward the upper right corner of the square. The fields due to the wires at theupper right and lower left corners both point toward the upper left corner. The horizontal componentscancel and the vertical components sum to

B total = 4µ0i√2πa

cos 45◦ =2µ0i

πa

=2(4π × 10−7T·m/A)(20A)

π(0.20m)= 8.0× 10−5 T .

In the calculation cos 45◦ was replaced with 1/√2. The total field points upward.

37. (a) The the magnetic field at a point within the hole is the sum of the fields due to two currentdistributions. The first is that of the solid cylinder obtained by filling the hole and has a currentdensity that is the same as that in the original cylinder (with the hole). The second is the solidcylinder that fills the hole. It has a current density with the same magnitude as that of the originalcylinder but is in the opposite direction. If these two situations are superposed the total current inthe region of the hole is zero. Now, a solid cylinder carrying current i, uniformly distributed overa cross section, produces a magnetic field with magnitude

B =µ0ir

2πR2

a distance r from its axis, inside the cylinder. Here R is the radius of the cylinder. For the cylinderof this problem the current density is

J =i

A=

i

π(a2 − b2),

where A = π(a2 − b2) is the cross-sectional area of the cylinder with the hole. The current in thecylinder without the hole is

I1 = JA = πJa2 =ia2

a2 − b2

and the magnetic field it produces at a point inside, a distance r1 from its axis, has magnitude

B1 =µ0I1r1

2πa2 =µ0ir1a

2

2πa2(a2 − b2)=

µ0ir1

2π(a2 − b2).

The current in the cylinder that fills the hole is

I2 = πJb2 =ib2

a2 − b2

and the field it produces at a point inside, a distance r2 from the its axis, has magnitude

B2 =µ0I2r2

2πb2 =µ0ir2b

2

2πb2(a2 − b2)=

µ0ir2

2π(a2 − b2).

At the center of the hole, this field is zero and the field there is exactly the same as it would be ifthe hole were filled. Place r1 = d in the expression for B1 and obtain

B =µ0id

2π(a2 − b2)

for the field at the center of the hole. The field points upward in the diagram if the current is outof the page.

(b) If b = 0 the formula for the field becomes

B =µ0id

2πa2 .

This correctly gives the field of a solid cylinder carrying a uniform current i, at a point inside thecylinder a distance d from the axis. If d = 0 the formula gives B = 0. This is correct for the fieldon the axis of a cylindrical shell carrying a uniform current.

(c) Consider a rectangular path with two long sides (side 1 and 2, each with length L) and two shortsides (each of length less than b). If side 1 is directly along the axis of the hole, then side 2 would bealso parallel to it and also in the hole. To ensure that the short sides do not contribute significantlyto the integral in Ampere’s law, we might wish to make L very long (perhaps longer than the length

of the cylinder), or we might appeal to an argument regarding the angle between �B and the shortsides (which is 90◦ at the axis of the hole). In any case, the integral in Ampere’s law reduces to

∮rectangle

�B · d�s = µ0ienclosed

∫side 1

�B · d�s +∫

side 2

�B · d�s = µ0iin hole

(Bside 1 − Bside 2)L = 0

where Bside 1 is the field along the axis found in part (a). This shows that the field at off-axis points(where Bside 2 is evaluated) is the same as the field at the center of the hole; therefore, the field inthe hole is uniform.

47. (a) We denote the �B-fields at point P on the axis due to the solenoid and the wire as �Bs and �Bw,respectively. Since �Bs is along the axis of the solenoid and �Bw is perpendicular to it, �Bs ⊥ �Bw,respectively. For the net field �B to be at 45◦ with the axis we then must have Bs = Bw. Thus,

Bs = µ0isn = Bw =µ0iw2πd

,

which gives the separation d to point P on the axis:

d =iw

2πisn=

6.00A2π(20.0× 10−3 A)(10 turns/cm)

= 4.77 cm .

(b) The magnetic field strength is

B =√2Bs

=√2(4π × 10−7 T·m/A)(20.0× 10−3 A)(10 turns/0.0100m)

= 3.55× 10−5 T .