Chapter29 - 1. (a) The field due to the wire, at a point...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
1. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 µ T and must be directed due south. Since Bi r = 0 2 π , i rB == × ×⋅ = 2 23 9 1 0 4 16 0 6 π π 0.080 π1 0 −7 mT TmA A. bg ch (b) The current must be from west to east to produce a field which is directed southward at points below it.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2. The straight segment of the wire produces no magnetic field at C (see the straight sections discussion in Sample Problem 29-1). Also, the fields from the two semi-circular loops cancel at C (by symmetry). Therefore, B C = 0.
Background image of page 2
3. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by B i r = µ 0 2 π . With r = 20 ft = 6.10 m, we have B = ×⋅ = 4 100 2 33 10 33 6 π1 0 π6 .10 −7 TmA A m TT . ch b g b g .. (b) This is about one-sixth the magnitude of the Earth’s field. It will affect the compass reading.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4. Eq. 29-1 is maximized (with respect to angle) by setting θ = 90º ( = π /2 rad). Its value in this case is 0 max 2 4 i ds dB R µ π = . From Fig. 29-36(b), we have 12 max 60 10 T. B We can relate this B max to our dB max by setting “ ds ” equal to 1 × 10 6 m and R = 0.025 m. This allows us to solve for the current: i = 0.375 A. Plugging this into Eq. 29-4 (for the infinite wire) gives B = 3.0 µ T.
Background image of page 4
the current in the large radius arc contributes µθ 0 4 ia π (into the page) to the field there. Thus, the net field at P is 0 1 1 (4 T m A)(0.411A)(74 /180 ) 1 1 4 4 0.107m 0.135m 1.02 T. i B ba π ×⋅ ° ⋅° ⎛⎞ =− = ⎜⎟ ⎝⎠ −7 −7 π1 0 10 (b) The direction is out of the page. 5. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with P do not contribute to the field at that point. Using Eq. 29-9 (with φ = θ ) and the right-hand rule, we find that the current in the semicircular arc of radius b contributes 0 4 ib π (out of the page) to the field at P . Also,
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
6. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in segments AH and JD do not contribute to the field at point C . Using Eq. 29-9 (with φ = π ) and the right-hand rule, we find that the current in the semicircular arc H J contributes µ 01 4 iR (into the page) to the field at C . Also, arc D A contributes 02 4 (out of the page) to the field there. Thus, the net field at C is 0 12 1 1 (4 T m A)(0.281A) 1 1 1.67 T. 4 4 0.0315m 0.0780m i B RR ⎛⎞ ×⋅ ⎛⎞ =− = = × ⎜⎟ ⎝⎠ −7 −6 π1 0 10 (b) The direction of the field is into the page.
Background image of page 6
7. (a) The currents must be opposite or antiparallel, so that the resulting fields are in the same direction in the region between the wires. If the currents are parallel, then the two fields are in opposite directions in the region between the wires. Since the currents are the same, the total field is zero along the line that runs halfway between the wires.
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 94

Chapter29 - 1. (a) The field due to the wire, at a point...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online