Homework Solutions

135
Draw all currents down or right. Then

V_{0} = R_{1} I_{1} + R_{3} I_{3}, I_{1} = I_{3} + I_{5}, I_{2} + I_{5} = I_{4}

0 = I_{2} R_{2} - I_{5} R_{5} - I_{1} R_{1}, 0 = I_{5} R_{5} + I_{4} R_{4} - I_{3} R_{3}

Use the second equation to eliminate $I_{5} = I_{1} - I_{3}$ ;

V_{0} = R_{1} I_{1} + R_{3} I_{3}, I_{2} + (I_{1} - I_{3}) = I_{4}

0 = I_{2} R_{2} - (I_{1} - I_{3}) R_{5} - I_{1} R_{1}, 0 = (I_{1} - I_{3}) R_{5} + I_{4} R_{4} - I_{3} R_{3}

Use the second equation to eliminate $I_{4}$ ;

V_{0} = R_{1} I_{1} + R_{3} I_{3}, 0 = I_{2} R_{2} - (I_{1} - I_{3}) R_{5} - I_{1} R_{1}, 0 = (I_{1} - I_{3}) R_{5} + (I_{1} + I_{2} - I_{3}) R_{4} - I_{3} R_{3}

re-arrange;

V_{0} = R_{1} I_{1} + R_{3} I_{3}, 0 = I_{2} R_{2} - I_{1} (R_{1} + R_{5}) + I_{3} R_{5}, 0 = I_{1} (R_{5} + R_{4}) + I_{2} R_{4} - I_{3} (R_{3} + R_{4} + R_{5})

use the second of these to eliminate $I_{2} = I_{1} \frac{R_{1} + R_{5}}{R_{2}} - I_{3} \frac{R_{5}}{R_{2}}$

V_{0} = R_{1} I_{1} + R_{3} I_{3}, 0 = I_{1} (R_{5} + R_{4}) + R_{4} (I_{1} \frac{R_{1} + R_{5}}{R_{2}} - I_{3} \frac{R_{5}}{R_{2}}) - I_{3} (R_{3} + R_{4} + R_{5})

re-arrange the second;

0 = I_{1} (\frac{R_{5} R_{2} + R_{4} R_{2} + R_{4} R_{1} + R_{4} R_{5}}{R_{2}}) - I_{3} (\frac{R_{4} R_{5} + R_{2} R_{3} + R_{2} R_{5} + R_{2} R_{4}}{R_{2}})

I_{3} = I_{1} \frac{R_{5} R_{2} + R_{4} R_{2} + R_{4} R_{1} + R_{4} R_{5}}{R_{4} R_{5} + R_{2} R_{3} + R_{2} R_{5} + R_{2} R_{4}}

ﬁnally

I_{1} = \frac{V_{0}}{R_{1} + R_{3} (\frac{R_{5} R_{2} + R_{4} R_{2} + R_{4} R_{1} + R_{4} R_{5}}{R_{4} R_{5} + R_{2} R_{3} + R_{2} R_{5} + R_{2} R_{4}})}

If $I_{5} = 0$ then $I_{1} = I_{3}$ and $I_{2} = I_{4}$ , the fourth and ﬁfth original loop equations read

I_{2} R_{2} = I_{1} R_{1}, I_{4} R_{4} = I_{3} R_{3}, or I_{2} R_{2} = I_{1} R_{1}, I_{2} R_{4} = I_{1} R_{3}

Divide these

\frac{R_{2}}{R_{4}} = \frac{R_{1}}{R_{3}}

for a balanced bridge.

136
Put all currents pointing down, positive charge on the left capacitor plate.

V_{0} = I_{1} R_{1} + I_{3} R_{3}, 0 = I_{2} R_{2} - \frac{Q}{C} - R_{1} I_{1}, 0 = \frac{Q}{C} + R_{4} I_{4} - R_{3} I_{3}

The current from the battery is at all times

I_{t o t a l} = I_{1} + I_{2} = I_{3} + I_{4}

Instantly after closing switch we have $Q = 0$ so

V_{0} = I_{1} R_{1} + I_{3} R_{3}, 0 = I_{2} R_{2} - R_{1} I_{1}, 0 = R_{4} I_{4} - R_{3} I_{3}, I_{1} + I_{2} = I_{3} + I_{4}

eliminate

I_{2} = I_{1} \frac{R_{1}}{R_{2}}, I_{4} = I_{3} \frac{R_{3}}{R_{4}}

to get

V_{0} = I_{1} R_{1} + I_{3} R_{3}, I_{1} (1 + \frac{R_{1}}{R_{2}}) = I_{3} (1 + \frac{R_{3}}{R_{4}})

I_{1} = \frac{V_{0}}{R_{1} + R_{3} \frac{(1 + \frac{R_{1}}{R_{2}})}{(1 + \frac{R_{3}}{R_{4}})}}, I_{3} = \frac{(1 + \frac{R_{1}}{R_{2}})}{(1 + \frac{R_{3}}{R_{4}})} \frac{V_{0}}{R_{1} + R_{3} \frac{(1 + \frac{R_{1}}{R_{2}})}{(1 + \frac{R_{3}}{R_{4}})}}

Long afterwards $I_{1} = I_{3}$ , $I_{2} = I_{4}$ so

I_{1} = I_{3} = \frac{V_{0}}{R_{1} + R_{3}}, I_{2} = I_{4} = \frac{V_{0}}{R_{2} + R_{4}}

Q = C (I_{2} R_{2} - R_{1} I_{1}) = C (R_{2} \frac{V_{0}}{R_{2} + R_{4}} - R_{1} \frac{V_{0}}{R_{1} + R_{3}})

137
Put all currents pointing down, positive charge on the top capacitor plates.

V_{0} = R_{1} I_{1} + \frac{Q_{1}}{C_{1}}, 0 = \frac{Q_{2}}{C_{2}} - R_{3} I_{3} - R_{1} I_{1}, 0 = R_{3} I_{3} + R_{2} I_{2} - \frac{Q_{1}}{C_{1}}

Instantly after closure we have $Q_{1} = Q_{2} = 0$ , so that

V_{0} = R_{1} I_{1} 0 = R_{3} I_{3} - R_{1} I_{1}, 0 = R_{3} I_{3} + R_{2} I_{2}

then

I_{1} = \frac{V_{0}}{R_{1}}, I_{3} = - \frac{R_{1}}{R_{3}} I_{1} = - \frac{R_{1}}{R_{3}} \frac{V_{0}}{R_{1}} = - \frac{V_{0}}{R_{3}}, I_{2} = - \frac{R_{3}}{R_{2}} I_{3} = \frac{V_{0}}{R_{2}}

so the current $I_{3}$ actually ﬂows left, not right.
Long after closure we have $I_{1} = I_{2} = I_{3} = \frac{V_{0}}{R_{1} + R_{2} + R_{3}}$ from the one large loop through all resistors, since no currents will be diverted at any junction to feed a full capacitor. Then

Q_{1} = C_{1} (R_{3} I_{3} + R_{2} I_{2}) = C_{1} (R_{2} + R_{3}) \frac{V_{0}}{R_{1} + R_{2} + R_{3}}, Q_{2} = C_{2} (R_{3} I_{3} + R_{1} I_{1}) = C_{2} (R_{1} + R_{3}) \frac{V_{0}}{R_{1} + R_{2} + R_{3}}

You can see that we just don’t use the junction equations for the “instantly after” case.

138
Put all currents pointing down or right, positive charge on the top capacitor plates.

V_{0} = R_{0} I_{0} + R_{1} I_{1} + R_{2} I_{2}, 0 = \frac{Q_{1}}{C_{1}} - I_{3} R_{3} - I_{1} R_{1}, 0 = R_{3} I_{3} + \frac{Q_{2}}{C_{2}} - R_{2} I_{2}, I_{1} = I_{2} + I_{3}

Instantly after closure we have $Q_{1} = Q_{2} = 0$ , so that

V_{0} = R_{0} I_{0} + R_{1} I_{1} + R_{2} I_{2}, 0 = - I_{3} R_{3} - I_{1} R_{1}, 0 = R_{3} I_{3} - R_{2} I_{2}, I_{1} = I_{2} + I_{3}

The lastr three equations are three homogeneous linear equations in three unknowns, which means that $I_{1} = I_{2} = I_{3} = 0$ , and therefore $I_{0} = \frac{V_{0}}{R_{0}}$ .
Long after closure we have $I_{0} = I_{1} = I_{2} = \frac{V_{0}}{R_{0} + R_{1} + R_{2}}$ , $I_{3} = 0$ from the one large loop through all resistors, since no currents will be diverted at any junction to feed a full capacitor. Then

Q_{1} = C_{1} (R_{1} I_{1}) = C_{1} R_{1} \frac{V_{0}}{R_{0} + R_{1} + R_{2}}

Q_{2} = C_{2} (R_{2} I_{2}) = C_{2} R_{2} \frac{V_{0}}{R_{0} + R_{1} + R_{2}}

You can see that we just don’t use the junction equations for the “instantly after” case.

155
The ﬁeld at a point on the axis ( $z$ -axis) of a coil is

B_{z} = \frac{μ_{0} N I R^{2}}{2 {(R^{2} + z^{2})}^{\frac{3}{}}}

For two coils seperated by $R$ , the ﬁeld at $\frac{R}{2}$ from each will be

B_{z} = \frac{2 μ_{0} N I R^{2}}{2 {(R^{2} + {(\frac{R}{2})}^{2})}^{\frac{3}{}}} = \frac{2 N I R^{2} μ_{0}}{2 R^{3} {(\frac{125}{64})}^{\frac{1}{2}}} = \frac{8 N I μ_{0}}{\sqrt{125} R}

Using the ﬁeld above as a function of $z$ , place one coil parallel to $x y$ -plane with its center at $(0, 0, - \frac{R}{2})$ , another coaxial to it but with its center at $(0, 0, - \frac{R}{2})$ . A point at $z$ is $∣ \frac{R}{2} - z ∣$ from one coil center, $∣ \frac{R}{2} + z ∣$ from the other, so

B_{z} = \frac{μ_{0} N I R^{2}}{{(R^{2} + {(\frac{R}{2} - z)}^{2})}^{\frac{3}{2}}} + \frac{μ_{0} N I R^{2}}{{(R^{2} + {(\frac{R}{2} + z)}^{2})}^{\frac{3}{2}}}

Calling $f (z) = \frac{μ_{0} N I R^{2}}{{(R^{2} + {(\frac{R}{2} - z)}^{2})}^{\frac{3}{2}}}$ , we have

B_{z} = f (z) + f (- z)

Expand these in Taylor series about $z = 0$ ;

B_{z} = (f (0) + z f^{'} (0) + \frac{1}{2} z^{2} f^{''} (0) + \dots) + (f (0) + (- z) f^{'} (0) + \frac{1}{2} {(- z)}^{2} f^{''} (0) + \dots) \approx 2 f (0) + z^{2} f^{''} (0) + \dots

The derivative of this is

\frac{d B_{z}}{d z} = 2 z f^{''} (0) + \dots

which is zero for $z = 0$ , and so $B_{z}$ is very nearly uniform and constant at $z = 0$ , the center of the pair of coils. This is a reasonably good method in the lab for creating a nearly uniform ﬁeld in a small volume of space.

173

\frac{d z}{R d θ} = \frac{\frac{d z}{d t}}{R \frac{d θ}{d t}} = \frac{v_{z}}{R ω} = \frac{v_{z}}{\sqrt{v_{z}^{2} + v_{y}^{2}}} = \frac{m v_{z}}{q R B}

178

0 = q (E + v \times B), E = - v \times B

(v_{x} i + v_{y} j + v_{z} k) \times (B_{x} i + B_{y} j) = v_{x} B_{y} k + v_{y} B_{x} (- k) + v_{z} B_{x} j + v_{z} B_{y} (- i) = - E

B_{y} = \frac{E_{x}}{v_{z}} = \frac{1 \times 1 0^{3}}{5 \times 1 0^{4}} = 0.02 T, B_{x} = - \frac{E_{y}}{v_{z}} = - \frac{3 \times 1 0^{3}}{5 \times 1 0^{4}} = - 0.06 T

E_{z} = v_{x} B_{y} - v_{y} B_{x} = (2 \times 1 0^{4}) (0.02) + (3 \times 1 0^{4}) (0.06) = 2.2 \times 1 0^{4} \frac{m}{s}