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Lecture

# problemset8

4 Pages
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Department
Physics
Course
PHY 303L
Professor
Storr
Semester
Summer

Description
rashid (sar3553) – HW8 Sources of Magnetic Field – storr – (37537) 1 This print-out should have 6 questions. Biot-Savart Law. However, in this part of the Multiple-choice questions may continue on wire, r is pointing towards O as well, so s▯ the next column or page – ﬁnd all choices and r are parallel meaning s × r = 0 for this before answering. part of the wire. It is now easy to see that the right part, having a s antiparallel tr, also 001 10.0 points gives no contribution to B at O. The wire is carrying a current I. Let us go through the semicircle C. The y element d▯s, which is along the wire, will now be perpendicular tor, which is pointing along 180◦ the radius towards O. Therefore |d▯s ×r| = ds using the fact thatr is a unit vector. So the I I Biot-Savart Law gives for the magnitude B of r the magnetic ﬁeld at O:Z ▯0I ds O I x B = 2 . ▯ 4π C r Find the magnitude of the magnetic ﬁeld B Since the distance r to the elements is con- at O due to a current-carrying wire shown in the ﬁgure, where the semicircle has radius r, stant everywhere on the semicircle C, we will be able to pull it out of the integral.The and the straight parts to the left and to the integral is right extend to inﬁnity. Z Z ds 1 1 ▯0I r2 = r2 ds = r2L C 1. B = C C 3r where L C = π r is the length of the semicircle. ▯0I Thus the magnitude of the magnetic ﬁeld is 2. B = r ▯ 0 1 ▯0I ▯0I B = 2 π r = . 3. B = correct 4π r 4r 4r ▯0I 002 10.0 points 4. B = At what distance from a long, straight wire 4π r ▯0I carrying a current of 5 A is the magnetic ﬁeld 5. B = due to the wire equal to the strength of the 2π r Earth’s ﬁeld, approximately 2 × 105 T? ▯0I 6. B = 3π r Correct answer: 5 cm. ▯0I 7. B = π r Explanation: ▯0I 8. B = 2r Let : I = 5 A and −5 Explanation: B = 2 × 10 T. By the Biot-Savart Law, ▯0I Z We use B = 2π r to ﬁnd ▯0I ds × r B = 2 . ▯0I 4π r r = 2π B −6 2 Consider the left straight part of the wire. = (1.25664 × 10 N/A )(5 A) The line element ds at this part, if we come 2π (2 × 10−5 T) in from ∞, points towards O, i.e., in the x- = 0.05 m = 5 cm . direction. We need to ﬁnd d▯s ×r to use the rashid (sar3553) – HW8 Sources of Magnetic Field – storr – (37537) 2 I gral B ▯ dℓ around any closed path equals 003 10.0 points ▯ I, where I is the total steady current pass- The ﬁgure below shows a straight cylindrical 0 coaxial cable of radii a, b, and c in which ing through any surface bounded by the closed equal, uniformly distributed, but antiparallel path. currents i exist in the two conductors. Considering the symmetry of this problem, a we choose a circular path, so Ampere’s Law simpliﬁes to iout⊙ b c B (2π r1) = ▯ 0 en, where r1is the radius of the circle and Ienis iin⊗ the current enclosed. For c < r2< b, O F ▯0 en E B = 2π r 2 D r1 ▯0(i) = r2 2π r2 C r3 = ▯ 0 . r4 2π r2 Which expression gives the magnitude B(r ) E of the magnetic ﬁeld in the region 004 10.0 points 2 c < r2< b? Niobium metal becomes a superconductor (with zero electrical
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