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Introduction to Electrodynamics

David J. Griffiths

Physics

4th edition Edition 路 613 pages

ISBN: 9780321856562

Chapter 7: Q7.53P (page 349)

The current in a long solenoid is increasing linearly with time, so the flux is proportional $t$ :. $蠒 = 伪 t$ Two voltmeters are connected to diametrically opposite points (A and B), together with resistors ( $R_{1}$ and $R_{2}$ ), as shown in Fig. 7.55. What is the reading on each voltmeter? Assume that these are ideal voltmeters that draw negligible current (they have huge internal resistance), and that a voltmeter register - $- {鈭�}_{a}^{b} E 脳 d l$ between the terminals and through the meter. [Answer: $V_{1} = 伪 R_{1} / (R_{1} + R_{2})$ . Notice that $V_{1} 鈮� V_{2}$ , even though they are connected to the same points]

Short Answer

Expert verified

The expression for voltage across the resistance is $\frac{伪 R_{1}}{R_{1} + R_{2}}$ and is $鈭� \frac{伪 R_{2}}{R_{1} + R_{2}}$ .

Step by step solution

01

Write the given data from the question.

The relationship between current and flux, $蠒 = 伪 t$

The two registers are $R_{1}$ and $R_{2}$ .

02

Determine the formulas to calculate the voltmeter reading.

The expression for the induced emf is given as follows.

$蔚 = - \frac{d 蠒}{d t}$ 鈥︹€�. (1)

03

Draw the expression for the voltmeter reading.

Calculate the induced emf.

Substitute $伪 t$ for $蠒$ into equation (1).

$\begin{array}{l} 蔚 = \frac{d}{d t} (伪 t) \\ 蔚 = 伪 \end{array}$

The current in the registers is given by

$I = \frac{蔚}{R_{1} + R_{2}}$

Substitute $伪$ for $蔚$ into above equation.

$I = \frac{伪}{R_{1} + R_{2}}$

The voltage across the register $R_{1}$ is given by,

$V_{1} = I R_{1}$

Substitute $\frac{伪}{R_{1} + R_{2}}$ for $I$ into above equation.

$\begin{array}{l} V_{1} = \frac{伪}{R_{1} + R_{2}} R_{1} \\ V_{1} = \frac{伪 R_{1}}{R_{1} + R_{2}} \end{array}$

The voltage across the register $R_{1}$ is given by,

$V_{2} = 鈭� I R_{2}$

Substitute $\frac{伪}{R_{1} + R_{2}}$ for $I$ into above equation.

$\begin{array}{l} V_{2} = 鈭� \frac{伪}{R_{1} + R_{2}} R_{2} \\ V_{2} = 鈭� \frac{伪 R_{2}}{R_{1} + R_{2}} \end{array}$

Hence, the expression for voltage across the resistance $R_{1}$ is $\frac{伪 R_{1}}{R_{1} + R_{2}}$ and $R_{2}$ is $鈭� \frac{伪 R_{2}}{R_{1} + R_{2}}$ .

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Most popular questions from this chapter

Question: Assuming that "Coulomb's law" for magnetic charges ( q_m) reads

$F = \frac{渭_{0}}{4 蟺} \frac{{qm}_{1} {qm}_{2}}{r^{2}} \hat{r}$

Work out the force law for a monopole moving with velocity through electric and magnetic fields E and B.

(a) Two metal objects are embedded in weakly conducting material of conductivity $蟽$ (Fig. 7 .6). Show that the resistance between them is related to the capacitance of the arrangement by

$R = \frac{{鈭�}_{0}}{蟽 C}$

(b) Suppose you connected a battery between 1 and 2, and charged them up to a potential difference $V_{0}$ . If you then disconnect the battery, the charge will gradually leak off. Show that $V (t) = V_{0} e^{- t / r}$ , and find the time constant, $蟿$ , in terms of ${鈭�}_{0}$ and . $蟽$

A long solenoid with radius a and n turns per unit length carries a time-dependent current $l (t)$ in the $\hat{蠒}$ direction. Find the electric field (magnitude and direction) at a distance s from the axis (both inside and outside the solenoid), in the quasistatic approximation.

A square loop of wire (side a) lies on a table, a distance s from a very long straight wire, which carries a current I, as shown in Fig. 7.18.

(a) Find the flux of B through the loop.

(b) If someone now pulls the loop directly away from the wire, at speed, V what emf is generated? In what direction (clockwise or counter clockwise) does the current flow?

(c) What if the loop is pulled to the right at speed V ?

Prove Alfven's theorem: In a perfectly conducting fluid (say, a gas of free electrons), the magnetic flux through any closed loop moving with the fluid is constant in time. (The magnetic field lines are, as it were, "frozen" into the fluid.)

(a) Use Ohm's law, in the form of Eq. 7.2, together with Faraday's law, to prove that if $蟽 = 鈭�$ and is J finite, then

$\frac{鈭� B}{鈭� t} = 鈭� 脳 (v 脳 B)$

(b) Let S be the surface bounded by the loop $(P)$ at time t , and $S'$ a surface bounded by the loop in its new position $(P')$ at time t+dt (see Fig. 7.58). The change in flux is

$诲桅 = {鈭�}_{S'} B (t + dt) 鈥� da - {鈭�}_{S} B (t) 鈥� da$

Use $鈭� 路 B = 0$ to show that

${鈭�}_{S'} B (t + dt) 鈥� da + {鈭�}_{R} B (t + dt) 鈥� da = {鈭�}_{S} B (t + dt) 鈥� da$

(Where R is the "ribbon" joining P and P' ), and hence that

$诲桅 = dt {鈭�}_{S} \frac{鈭� B}{鈭� t} 路 da - {鈭�}_{R} B (t + dt) 鈥� da$

(For infinitesimal dt ). Use the method of Sect. 7.1.3 to rewrite the second integral as

$dt {鈭�}_{P} (B 脳 v) 路 dI$

And invoke Stokes' theorem to conclude that

$\frac{诲桅}{dt} = {鈭�}_{S} (\frac{鈭� B}{鈭� t} - 鈭� 脳 (v 脳 B)) 路 da$

Together with the result in (a), this proves the theorem.

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