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Unformatted text preview: 10−7 T ⋅ m A 8.85 ×10−12 C2 /N ⋅ m 2 3.0 × 10−3 m ( )( ) 2 = 2.4 × 1013 V
.
m ⋅s 6. The integral of the field along the indicated path is, by Eq. 3218 and Eq. 3219, equal
to
(4.0 cm)(2.0 cm)
⎛ enclosed area ⎞
μ0id ⎜
= 52 nT ⋅ m .
⎟ = μ0 (0.75 A)
12 cm 2
⎝ total area ⎠ 7. (a) Inside we have (by Eq. 3216) B = μ0id r1 / 2π R 2 , where r1 = 0.0200 m,
R = 0.0300 m, and the displacement current is given by Eq. 3238 (in SI units):
id = ε 0 dΦE
= (8.85 × 10−12 C2 /N ⋅ m 2 )(3.00 × 10−3 V/m ⋅ s) = 2.66 × 10−14 A .
dt Thus we find μ0id r1 (4π ×10−7 T ⋅ m/A)(2.66 × 10−14 A)(0.0200 m)
B=
=
= 1.18 × 10−19 T .
2
2
2π R
2π (0.0300 m)
(b) Outside we have (by Eq. 3217) B = μ0id / 2π r2 where r2 = 0.0500 cm. Here we
obtain
μ0id (4π ×10−7 T ⋅ m/A)(2.66 ×10−14 A)
=
= 1.06 ×10−19 T
B=
2π r2
2π (0.0500 m)
8. (a) Application of Eq. 323 along the circle referred to in the second sentence of the
problem statement (and taking the derivative of the flux expression given in that sentence)
leads to
r
B (2π r ) = ε 0 μ0 ( 0.60 V ⋅ m/s ) .
R Using r = 0.0200 m (which, in any case, cancels out) and R = 0.0300 m, we obtain 1245 B= ε 0 μ0 (0.60 V ⋅ m/s) (8.85 × 10−12 C2 /N ⋅ m 2 )(4π×10−7 T ⋅ m/A)(0.60 V ⋅ m/s)
=
2π R
2π (0.0300 m) = 3.54 × 10−17 T .
(b) For a value of r larger than R, we must note that the flux enclosed has already reached
its full amount (when r = R in the given flux expression). Referring to the equation we
wrote in our solution of part (a), this means that the final fraction ( r / R ) should be
replaced with unity. On the left hand side of that equation, we set r = 0.0500 m and solve.
We now find B= ε 0 μ0 (0.60 V ⋅ m/s) (8.85 ×10−12 C2 /N ⋅ m 2 )(4π×10−7 T ⋅ m/A)(0.60 V ⋅ m/s)
=
2π r
2π (0.0500 m) = 2.13 × 10−17 T .
9. (a) Application of Eq. 327 with A = πr2 (and taking the derivative of the field
expression given in the problem) leads to
B (2π r )...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.
 Fall '08
 schuller
 Magnetism, Work

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