Assignement 9 - solutions

FGT 32.43

a) We use the dipole formula,

$$B(d) = {\mu_0 m \over 2\pi d^3}$$ (1)

where m is the dipole moment, d is the distance from the dipole center to the point at which the field is measured. We assume that the dipole is aligned along the z-axis and the field is measured on the same axis. Isolating m yields,

$$m = {2\pi B d^3 \over \mu_0} = 7.86\times 10^{22}Am^2$$ (2)

where we have taken the radius of the earth to be d=R_e = 6,400km.

b) If we assume that the dipole is due to atomic dipoles, then the magnetisation is given by,

M = m/V = 573A/m (3)

where we have taken the volume to be $V=4\pi R_e^3/3\times 8$.

c) If we assume that the dipole is due to a circulating current, then the current is found from m = IA, so that,

$$I = m/A = 2.44\times 10^9 A$$ (4)

where we have taken $A = \pi R_e^2/4$.

FGT 33:23

The magnetic field due the two current carrying wires aligned with the x-axis are,

$$B_1(r) = {\mu_0 I \over 2\pi r}\ \ \ ;\ \ \ B_2(r) = {-\mu_0 I \over 2\pi( r-a)}$$ (5)

The flux through the second loop is then

$$\phi_B = L \int_{b+a}^{b+2a} {\mu_0 I\over 2\pi}[{1\over r-a} - {1\over r}]$$

$$= {L\mu_0 I \over 2\pi} ln({r-a\over r})\vert _{b+a}^{b+2a}$$

$$= {L\mu_0 I \over 2\pi}[ ln({b+a\over b+2a} - ln({b\over b+a})] = {L\mu_0 I \over 2\pi} ln[{(b+a)^2\over b(b+2a)}]$$ (6)

The mutual inductance is defined by,

$$M= \phi_B/I$$ (7)

FGT 33.67

a) As $t \rightarrow \infty$, the inductor has no resistance, so all of the current goes through the inductor. So we have,

$$I = {12A\over 5000} = 2.4mA$$ (8)

b) If the switch open at t=0, the the time dependence of the current is

$$I(t) = I_0 e^{-t\over \tau}$$ (9)

where $\tau = L/R$. The time at which the current has reduced to half its initial value is found from,

$${I_{1/2}\over I_0} = {1\over 2} = e^{-t_{1/2}/\tau}$$ (10)

We are given $t_{1/2} = 8\mu s$. Solving for $\tau$,

$$\tau = {-t_{1/2} \over [ln(1/2)]} = 11.5\mu s$$ (11)

Now we use $\tau = L/R$ and solve for L gives $L = \tau R = 0.28mH$.

c) The current at $t=12\mu s$ is given by

I = 2.4mA e^-12/11.5 = 0.845mA (12)

FGT 33.72

a) Immediately after the switch is closed, the inductor does not carry any current. The current flow is only in the part $6k\Omega$ resistor. This current is given by,

$$i(0) = {1.5V\over 8k\Omega} = 0.1875mA$$ (13)

b) At long times, the inductor is has no resistance, so we must combine the two resistors $3k\Omega$ and $6k\Omega$ is parallel. This parallel combintation yields $2k\Omega$. Adding this in series with the $2k\Omega$ resistor gives a $4k\Omega$ resistor. The current supplied by the 1.5V source is then,

$$I = {1.5V\over 4 k\Omega} = 0.375 mA$$ (14)

The parallel combination of the $3k\Omega$ and $6k\Omega$ resistors acts as a current divider, with 2/3 of the current passing through the $3k\Omega$ resistor. Thus the current in the $6k\Omega$ resistor is 0.125mA and the current in the $3k\Omega$ resistor is 0.25mA.