ch29 - 1(a The magnitude of the magnetic field due to the...

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1. (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 π 10 π 6.10 −7 T m A A m T T. c hb g b g . . µ (b) This is about one-sixth the magnitude of the Earth’s field. It will affect the compass reading.
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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.
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3. (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 B i r = µ 0 2 π , i rB = = × × = 2 2 39 10 4 16 0 6 π π 0.080 π 10 −7 µ m T T m A A. b gc h (b) The current must be from west to east to produce a field which is directed southward at points below it.
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4. (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 0 1 4 i R (into the page) to the field at C . Also, arc D A contributes µ 0 2 4 i R (out of the page) to the field there. Thus, the net field at C is 0 1 2 1 1 (4 T m A)(0.281A) 1 1 1.67 T. 4 4 0.0315m 0.0780m i B R R µ § · × § · = = = × ¨ ¸ ¨ ¸ © ¹ © ¹ −7 −6 π 10 10 (b) The direction of the field is into the page.
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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 µ θ 0 4 i b π (out of the page) to the field at P . Also, the current in the large radius arc contributes µ θ 0 4 i a 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 b a µ θ π π × °⋅ ° § · § · = = ¨ ¸ ¨ ¸ © ¹ © ¹ = × −7 −7 π 10 10 (b) The direction is out of the page.
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6. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with C do not contribute to the field at that point. Eq. 29-9 (with φ = π ) indicates that the current in the semicircular arc contributes µ 0 4 i R to the field at C . Thus, the magnitude of the magnetic field is 0 (4 T m A)(0.0348A) 1.18 T. 4 4(0.0926m) i B R µ × = = = × −7 −7 π 10 10 (b) The right-hand rule shows that this field is into the page.
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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.
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