861_Physics ProblemsTechnical Physics

861_Physics ProblemsTechnical Physics - Chapter 30 F N I I...

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Chapter 30 201 P30.40 Bn I N r I == F H G I K J µµ π 2 so I rB N ×⋅ = 2 2 0 100 1 30 5 000 4 10 470 277 7 µ bg a fa f ej af .. m T Wb A m mA P30.41 Assuming a uniform B inside the toroid is equivalent to assuming rR << ; then B NI R 00 2 as for a tightly wound solenoid. B 630 3 00 20 2 0 0 000189 a fa f . . . T With the steel, BB B m + = κχ 1 101 0 001 89 T B = 0191 T FIG. P30.41 P30.42 C TM B ×× 4 00 10 0% 8 00 10 5 00 9 27 10 500 297 10 27 24 4 . . . . . K a t o m s J T T KJ Tm 32 23 a f P30.43 BH M =+ 0 so H B M =− = × 0 6 262 10 A m P30.44 In M 0 we have 2 00 0 T = M . But also Mx n B = . Then Bx n B = 0 where n is the number of atoms per volume and x is the number of electrons per atom contributing. Then x B n B × = −− 0 4 7 200 927 10 4 10 202 . . . T 8.50 10 m N m T T m A 28 e j e j . P30.45 (a) Comparing Equations 30.29 and 30.30, we see that the applied field is described by = . Then Eq. 30.35 becomes MC B T C T H 0 0 , and the definition of susceptibility (Eq. 30.32) is χµ M H C T 0 . (b) C T × χ 0 4 7 4 270 10 300 41 0 645 10 . . K TmA KA Section 30.9 The Magnetic Field of the Earth
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