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Homework 9 Solutions

Homework 9 Solutions - Chapter 29 1(a The magnitude of the...

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1127 Chapter 29 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 p . With r = 20 ft = 6.10 m, we have B = × = × = 4 100 2 33 10 33 6 p 10 p 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. 2. Equation 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. 29-34(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. 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 p , i rB = = × × = 2 2 39 10 4 16 0 6 p p 0.080 p 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 that is directed southward at points below it. 4. The straight segment of the wire produces no magnetic field at C (see the straight sections discussion in Sample Problem — “Magnetic field at the center of a circular arc of current”). Also, the fields from the two semicircular loops cancel at C (by symmetry). Therefore, B C = 0. 5. (a) We find the field by superposing the results of two semi-infinite wires (Eq. 29-7) and a semicircular arc (Eq. 29-9 with φ = π rad). The direction of G B is out of the page, as can be checked by referring to Fig. 29-6(c). The magnitude of G B at point a is therefore
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