Física I, 7ªED Halliday Fundamentos da física mecânica, cap.32

Física I, 7ªED Halliday Fundamentos da física mecânica, cap.32

(Parte 1 de 6)

1. (a) The flux through the top is +(0.30 T)πr2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the total of the previously mentioned fluxes. Thus (in magnitude) the flux though the sides is 1.1 mWb.

(b) The fact that it is negative means it is inward.

2. We use

3. (a) We use Gauss’ law for magnetism:z⋅= G G BdA0. Now, z⋅=++

Bd A C ΦΦΦ12, where Φ1 is the magnetic flux through the first end mentioned, Φ2 is the magnetic flux through the second end mentioned, and ΦC is the magnetic flux through the curved surface. Over the first end the magnetic field is inward, so the flux is Φ1 = –25.0 µWb. Over the second end the magnetic field is uniform, normal to the surface, and outward, so the flux is Φ2 = AB = πr2 B, where A is the area of the end and r is the radius of the cylinder. It value is

2 350120 160 10 724 10 724=× = + × = +−−πm T Wb Wbb g c h µ

Since the three fluxes must sum to zero,

=− − = − = −12 250 724 474µ µ µWb Wb Wb

(b) The minus sign in c Φindicates that the flux is inward through the curved surface.

4. From Gauss’ law for magnetism, the flux through S1 is equal to that through S2, the portion of the xz plane that lies within the cylinder. Here the normal direction of S2 is +y. Therefore, r r

B r r i S S B x Ld x B x L dx Ld x iL µ µ

5. We use the result of part (b) in Sample Problem 32-1:

R dE B rR rd t µε=≥ to solve for dE/dt:

dE Br

6. From Sample Problem 32-1 we know that B ∝ r for r ≤ R and B ∝ r–1 for r ≥ R. So the maximum value of B occurs at r = R, and there are two possible values of r at which the magnetic field is 75% of Bmax. We denote these two values as r1 and r2, where r1 < R and r2 > R.

(a) Inside the capacitor, 0.75 Bmax/Bmax = r1/R, or r1 = 0.75 R = 0.75 (40 m)=30 m.

(b) Outside the capacitor, 0.75 Bmax/Bmax = (r2/R)–1 , or r2 = R/0.75 = 4R/3 = (4/3)(40 m)

= 53 m..

(c) From Eqs. 32-15 and 32-17,

iR i R d max

Tm A A

T. c hb g bg

7. (a) Noting that the magnitude of the electric field (assumed uniform) is given by E = V/d (where d = 5.0 m), we use the result of part (a) in Sample Problem 32-1 rd Edt rd dV dt rR== ≤µ ε µ ε0 022 forbg.

We also use the fact that the time derivative of sin (ωt) (where ω = 2πf = 2π(60) ≈ 377/s in this problem) is ω cos(ωt). Thus, we find the magnetic field as a function of r (for r ≤ R; note that this neglects “fringing” and related effects at the edges):

B r

Vt B rV

max max maxcosbg where Vmax = 150 V. This grows with r until reaching its highest value at r = R = 30 m:

0 max

(b) For r ≤ 0.03 m, we use the 00max max rV B µ εω = expression found in part (a) (note the

B ∝ r dependence), and for r ≥ 0.03 m we perform a similar calculation starting with the result of part (b) in Sample Problem 32-1:

max max max

for 2

R RdE dV BV t rd t rd dt rd

RV rR

== =¸

(note the B ∝ r–1 dependence — See also Eqs. 32-16 and 32-17). The plot (with SI units understood) is shown below.

8. (a) Inside we have (by Eq. 32-16) B = µoid r1 /2πR2 , where r1 = 0.0200, R = 0.0300, and the displacement current is given by Eq. 32-38: id = εo dΦE /dt = εo(0.00300), in SI units. Thus we find B = 1.18 × 10 −19 T.

(b) Outside we have (by Eq. 32-17) B = µoid /2πr2 where r2 = 0.0500 in SI units. Here we obtain B = 1.06 × 10 −19 T.

9. (a) Application of Eq. 32-3 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

B (2πr ) = εo µo (0.60 V. m/s)

Using r = 0.0200 (which, in any case, cancels out) and R = 0.0500 (SI units understood) leads to B = 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 = 2.13 × 10 −17 T.

10. (a) Application of Eq. 32-7 with A = πr2 (and taking the derivative of the field expression given in the problem) leads to

s)

B (2πr) = εo µo πr2 (0.00450 V/m.

With r = 0.0200 m, this gives B = 5.01 × 10 −2 T.

(b) With r > R, the expression above must replaced by

s)

B (2πr) = εo µo πR2 (0.00450 V/m.

Substituting r = 0.050 m and R = 0.030 m, we obtain B = 4.51 × 10 −2 T.

1. (a) Here, the enclosed electric flux is found by integrating

ΦE = ³ r o E 2πr dr = t(0.500 V/m.s)(2π) ³ r o ()1 –

R r dr = t 𠩧¹ r2 – with SI units understood. Then (after taking the derivative with respect to time) Eq. 32-3 leads to

B (2πr) = εo µo 𠩧¹ · 1 r2 –

(Parte 1 de 6)

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