Exercise 1: The force on a charge, by components

Three point charges are fixed at the vertices of a right triangle: $q_{1}=+4.0$ nC at the origin, $q_{2}=-6.0$ nC at $(0.30\ \mathrm{m}, 0)$ and $q_{3}=+2.0$ nC at $(0, 0.40\ \mathrm{m})$ . $k=8.99\times 10^{9}$ N·m²/C².

a) Calculate the magnitude of the force exerted by $q_{1}$ on $q_{3}$ .
b) Calculate the magnitude of the force exerted by $q_{2}$ on $q_{3}$ .
c) Calculate the components $F_{x}$ and $F_{y}$ of the net force on $q_{3}$ .
d) Calculate the magnitude of the net force.
e) Calculate the angle of the net force above the positive $x$ axis.

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Answers

a) $F_{13}=4.50\times 10^{-7}$ N
b) $F_{23}=4.32\times 10^{-7}$ N
c) $F_{x}=2.59$ , $F_{y}=1.04$ ( $\times 10^{-7}$ N)
d) $F=2.79\times 10^{-7}$ N
e) $21.9°$

a)

$F_{13}=\dfrac{k|q_{1}q_{3}|}{r^{2}}=\dfrac{8.99\times 10^{9}\times 4.0\times 10^{-9}\times 2.0\times 10^{-9}}{0.40^{2}}=4.50\times 10^{-7}$ N, repulsive, so directed upward ( $+y$ ).

b)

Distance $\sqrt{0.30^{2}+0.40^{2}}=0.50$ m. $F_{23}=\dfrac{8.99\times 10^{9}\times 6.0\times 10^{-9}\times 2.0\times 10^{-9}}{0.50^{2}}=4.32\times 10^{-7}$ N, attractive, so directed from $q_{3}$ toward $q_{2}$ .

c)

The unit vector from $q_{3}$ to $q_{2}$ is $\left(\dfrac{0.30}{0.50}, -\dfrac{0.40}{0.50}\right)=(0.6, -0.8)$ . $F_{x}=0.6\times 4.32\times 10^{-7}=2.59\times 10^{-7}$ N; $F_{y}=4.50\times 10^{-7}-0.8\times 4.32\times 10^{-7}=1.04\times 10^{-7}$ N. We add VECTORS: adding the magnitudes would give $8.8\times 10^{-7}$ N, which is wrong.

d)

$F=\sqrt{(2.59)^{2}+(1.04)^{2}}\times 10^{-7}=2.79\times 10^{-7}$ N.

e)

$\theta=\arctan\dfrac{1.04}{2.59}=21.9°$ above the positive $x$ axis.

On

q_{3}

:

\vec{F}_{13}

upward (repulsion),

\vec{F}_{23}

toward

q_{2}

(attraction), and their sum in black, built with the parallelogram rule. All three arrows are to the same scale: the sum,

2.79\times 10^{-7}

N at

21.9°

from the

x

axis, is much shorter than

F_{13}+F_{23}

.

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Exercise 2: Potential of a dipole, work and the energy of a system

Two charges, $+5.0$ nC and $-5.0$ nC, are fixed at $x=-0.10$ m and $x=+0.10$ m on the $x$ axis. Point $P$ is at $(0, 0.10\ \mathrm{m})$ and point $Q$ at $(0.20\ \mathrm{m}, 0)$ . $k=8.99\times 10^{9}$ N·m²/C².

a) Calculate the electric potential at $P$ .
b) Calculate the electric potential at $Q$ .
c) Calculate the work an external agent must do to move a $+2.0$ nC charge slowly from $P$ to $Q$ .
d) Calculate the electric potential energy of the system of the two fixed charges.
e) What does the sign of this energy mean?

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Answers

a) $V_{P}=0$
b) $V_{Q}\approx -300$ V
c) $W_{ext}=-6.0\times 10^{-7}$ J
d) $U=-1.12\times 10^{-6}$ J
e) Bound system

a)

$P$ is the same distance from both charges, $\sqrt{0.10^{2}+0.10^{2}}=0.141$ m, and the charges are opposite: $V_{P}=\dfrac{k(5.0\times 10^{-9})}{0.141}+\dfrac{k(-5.0\times 10^{-9})}{0.141}=0$ V. Potential is a SCALAR: we add signed numbers, and the field at $P$ , for its part, is not zero.

b)

Distances: $0.30$ m from the positive charge, $0.10$ m from the negative one. $V_{Q}=8.99\times 10^{9}\times 5.0\times 10^{-9}\left(\dfrac{1}{0.30}-\dfrac{1}{0.10}\right)=-300$ V.

c)

$W_{ext}=q(V_{Q}-V_{P})=2.0\times 10^{-9}\times(-300-0)=-6.0\times 10^{-7}$ J. The work is negative: the positive charge “goes downhill” toward a lower potential, and the external agent has to hold it back.

d)

$U=\dfrac{kq_{1}q_{2}}{r}=\dfrac{8.99\times 10^{9}\times(5.0\times 10^{-9})(-5.0\times 10^{-9})}{0.20}=-1.12\times 10^{-6}$ J.

e)

A negative energy means that $1.12\times 10^{-6}$ J would have to be SUPPLIED to pull the two charges apart to infinity: the system is bound, as is always the case for charges of opposite signs.

Equipotentials in dashed gray: the perpendicular bisector of the two charges is the

V=0

equipotential, and

P

lies on it. The field at

P

is not zero: it is perpendicular to the equipotential and points toward the negative charge.

Q

, closer to the negative charge, is below zero:

V_{Q}=-300

V.

Exercise 3: The deflected ink drop of a printer

In a continuous inkjet printer, a drop of mass $1.3\times 10^{-10}$ kg carries a charge of $-1.5\times 10^{-13}$ C. It enters horizontally at $20$ m/s between two plates $1.6$ cm long, where there is a uniform vertical electric field of $1.2\times 10^{6}$ N/C.

a) Calculate the electric force on the drop.
b) Calculate its acceleration.
c) Calculate the time it takes to cross the plates.
d) Calculate the vertical deflection as it leaves the plates, in mm.
e) Why can the weight of the drop be neglected?

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Answers

a) $F=1.8\times 10^{-7}$ N
b) $a\approx 1.38\times 10^{3}$ m/s²
c) $t=8.0\times 10^{-4}$ s
d) $y\approx 0.44$ mm
e) Weight about 140 times smaller

a)

$F=|q|E=1.5\times 10^{-13}\times 1.2\times 10^{6}=1.8\times 10^{-7}$ N.

b)

$a=\dfrac{F}{m}=\dfrac{1.8\times 10^{-7}}{1.3\times 10^{-10}}=1.38\times 10^{3}$ m/s².

c)

The horizontal velocity is constant (no horizontal force): $t=\dfrac{L}{v}=\dfrac{0.016}{20}=8.0\times 10^{-4}$ s.

d)

Uniformly accelerated vertical motion, with no initial vertical velocity: $y=\dfrac{1}{2}at^{2}=\dfrac{1}{2}\times 1385\times(8.0\times 10^{-4})^{2}=4.4\times 10^{-4}$ m $=0.44$ mm. This is exactly projectile motion, with $a$ in place of $g$ .

e)

Weight: $mg=1.3\times 10^{-10}\times 9.81=1.3\times 10^{-9}$ N, more than a hundred times smaller than the electric force (about a hundred and forty times): it changes the deflection by less than $1$ %.

Positive plate on top:

\vec{E}

points down, and the force on the negative drop points up. The path is an arc of a parabola between the plates and then a straight line; the deflection is exaggerated in the drawing, it is only

0.44

mm.

Exercise 4: A two-source network: Kirchhoff's rules

In the circuit below, both ideal sources have their positive terminal at the top. Let $I_{1}$ be the current in $R_{1}$ and $I_{2}$ the current in $R_{2}$ , both directed toward the top junction, and let $I_{3}$ be the current flowing down through $R_{3}$ .

a) Write the junction rule at the top junction.
b) The loop rule gives $9.0=3I_{1}+2I_{3}$ and $6.0=6I_{2}+2I_{3}$ . Solve: calculate $I_{2}$ .
c) Calculate $I_{1}$ and $I_{3}$ .
d) Calculate the voltage across $R_{3}$ .
e) Calculate the total power delivered by the two sources, and check that it equals the power dissipated.

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Answers

a) $I_{1}+I_{2}=I_{3}$
b) $I_{2}=0.333$ A
c) $I_{1}=1.67$ A, $I_{3}=2.00$ A
d) $V_{3}=4.0$ V
e) $17.0$ W delivered and dissipated

a)

What arrives at the junction leaves it: $I_{1}+I_{2}=I_{3}$ .

b)

Substituting for $I_{3}$ : $9.0=5I_{1}+2I_{2}$ and $6.0=2I_{1}+8I_{2}$ . From the first, $I_{1}=\dfrac{9.0-2I_{2}}{5}$ ; in the second: $30=18-4I_{2}+40I_{2}$ , so $I_{2}=\dfrac{12}{36}=0.333$ A.

c)

$I_{1}=\dfrac{9.0-0.667}{5}=1.67$ A and $I_{3}=1.67+0.33=2.00$ A. Both currents are positive: both sources push current in the assumed direction.

d)

$V_{3}=R_{3}I_{3}=2.0\times 2.00=4.0$ V. Check with the left loop: $9.0-3\times 1.67=4.0$ V.

e)

Delivered: $9.0\times 1.67+6.0\times 0.333=15.0+2.0=17.0$ W. Dissipated: $3\times 1.67^{2}+6\times 0.333^{2}+2\times 2.00^{2}=8.33+0.67+8.00=17.0$ W. The power balance is the best check on a Kirchhoff calculation.

The gray arcs give the direction in which each loop is traversed: clockwise on the left (

9.0=3I_{1}+2I_{3}

), counterclockwise on the right (

6.0=6I_{2}+2I_{3}

). All three currents are positive,

I_{1}=1.67

A,

I_{2}=0.333

A and

I_{3}=2.00

A: they flow in the assumed direction, and

V_{3}=4.0

V.

Exercise 5: Resistivity: sizing the element of a heater

We want to make the heating element of a $1500$ W heater operating at $120$ V from nichrome wire of diameter $0.50$ mm and resistivity $1.10\times 10^{-6}$ Ω·m. Ignore the change of resistivity with temperature.

a) Calculate the resistance needed.
b) Calculate the current drawn.
c) Calculate the cross-sectional area of the wire, in mm².
d) Calculate the length of wire needed.
e) By mistake, the wire is cut to half that length. What power does the heater dissipate at $120$ V?

Show the solution

Answers

a) $R=9.6$ Ω
b) $I=12.5$ A
c) $A=0.196$ mm²
d) $L\approx 1.71$ m
e) $3000$ W

a)

$P=\dfrac{V^{2}}{R}$ , so $R=\dfrac{120^{2}}{1500}=9.6$ Ω.

b)

$I=\dfrac{P}{V}=\dfrac{1500}{120}=12.5$ A.

c)

$A=\pi r^{2}=\pi\times(0.25\times 10^{-3})^{2}=1.96\times 10^{-7}$ m² $=0.196$ mm². The RADIUS is half the diameter.

d)

$R=\dfrac{\rho L}{A}$ , so $L=\dfrac{RA}{\rho}=\dfrac{9.6\times 1.96\times 10^{-7}}{1.10\times 10^{-6}}=1.71$ m.

e)

The resistance is halved, $4.8$ Ω, and at CONSTANT voltage the power $\dfrac{V^{2}}{R}$ doubles: $3000$ W. Counterintuitive but correct: shortening a heating element makes it run hotter, until it burns out.

Exercise 6: Charging, then discharging through another resistor

A $500$ µF capacitor, initially uncharged, charges from a $12.0$ V source through a $2.0$ kΩ resistor. At $t=2.0$ s, a two-way switch disconnects it from the source and connects it to a $4.0$ kΩ resistor, through which it discharges.

a) Calculate the time constant of the charging.
b) Calculate the capacitor voltage at $t=2.0$ s.
c) Calculate the current in the $4.0$ kΩ resistor just after the switch is thrown, in mA.
d) Calculate the capacitor voltage at $t=5.0$ s.
e) Calculate the energy stored at the instant the switch is thrown, in mJ.

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Answers

a) $\tau_{1}=1.0$ s
b) $v\approx 10.4$ V
c) $i\approx 2.6$ mA
d) $v\approx 2.3$ V
e) $U\approx 27$ mJ

a)

$\tau_{1}=R_{1}C=2.0\times 10^{3}\times 500\times 10^{-6}=1.0$ s.

b)

$v=12.0\left(1-e^{-2.0/1.0}\right)=12.0\times 0.865=10.4$ V.

c)

The capacitor voltage cannot jump: just after the switch is thrown, it is still $10.4$ V, applied across the new resistor. $i=\dfrac{10.38}{4000}=2.6\times 10^{-3}$ A $=2.6$ mA.

d)

New time constant: $\tau_{2}=4.0\times 10^{3}\times 500\times 10^{-6}=2.0$ s. The discharge has been going on for $3.0$ s: $v=10.38\,e^{-3.0/2.0}=2.3$ V. The pitfall is to measure the time from $t=0$ instead of from the switch.

e)

$U=\dfrac{1}{2}Cv^{2}=\dfrac{1}{2}\times 500\times 10^{-6}\times 10.38^{2}=2.7\times 10^{-2}$ J $=27$ mJ.

After the switch is thrown, the two-way switch connects the capacitor to

R_{2}

alone: the top plate, positively charged (

u_{C}=10.4

V), empties through

R_{2}

: the current flows down through it,

i=2.6

mA just after the switch is thrown. The time constant becomes

\tau_{2}=R_{2}C=2.0

s.

Exercise 7: The current balance: levitating a wire in a magnetic field

A horizontal conducting rod $20.0$ cm long with a mass of $8.0$ g hangs from two flexible conducting wires, in a uniform horizontal magnetic field of $0.050$ T perpendicular to the rod. A current $I$ is passed through it so that the wires are no longer under tension. $g=9.81$ m/s².

a) Calculate the weight of the rod.
b) Calculate the current that cancels the tension in the wires.
c) The field points north and the magnetic force must point upward. Which way must the current flow?
d) The field is now rotated in the horizontal plane so that it makes an angle of $30°$ with the rod. What current is needed then?
e) Why could a field parallel to the rod never make it levitate?

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Answers

a) $mg=7.85\times 10^{-2}$ N
b) $I=7.85$ A
c) From west to east
d) $I=15.7$ A
e) $\sin 0°=0$ : zero force

a)

$mg=0.0080\times 9.81=7.85\times 10^{-2}$ N.

b)

$F=ILB$ must equal the weight: $I=\dfrac{mg}{LB}=\dfrac{0.0785}{0.200\times 0.050}=7.85$ A.

c)

$\vec F=I\vec L\times\vec B$ must point up with $\vec B$ pointing north. With the $x$ axis toward the east, $y$ toward the north and $z$ upward: $\hat x\times\hat y=\hat z$ . So $\vec L$ must point EAST. The current flows from west to east; in the other direction, the force would push the rod down.

d)

$F=ILB\sin\theta$ : at $30°$ the force is halved, and we need $I=\dfrac{7.85}{\sin 30°}=15.7$ A.

e)

With $\vec L$ parallel to $\vec B$ , $\sin\theta=0$ : the cross product is zero, so there is no force, whatever the current.

Seen from the south: north is behind the page, so

\vec{B}

points into the page. With the current flowing west to east,

I\vec{L}\times\vec{B}

points up and exactly balances the weight for

I=7.85

A: the wires are no longer under tension.

Exercise 8: The solenoid, the wire and the compass

A solenoid $25.0$ cm long has $500$ turns carrying $2.00$ A. Farther away, a long straight vertical wire carries $10.0$ A. The horizontal component of Earth's magnetic field is $2.0\times 10^{-5}$ T. $\mu_{0}=4\pi\times 10^{-7}$ T·m/A.

The diagram below shows the wire and the compass, seen from above.

a) Calculate the magnetic field at the center of the solenoid, in mT.
b) Calculate the field produced by the wire at $5.0$ cm, in µT.
c) How many times larger is the solenoid's field than the wire's?
d) A compass is placed $5.0$ cm from the wire, at a point where the wire's field is perpendicular to Earth's field. By what angle does its needle deflect from north?
e) Why is the field inside a long solenoid independent of the radius of the turns?

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Answers

a) $B=5.03$ mT
b) $B=40$ µT
c) About $126$ times
d) $\alpha\approx 63°$
e) $B=\mu_{0}nI$ , independent of the radius

a)

$B=\mu_{0}nI=4\pi\times 10^{-7}\times\dfrac{500}{0.250}\times 2.00=5.03\times 10^{-3}$ T $=5.03$ mT. What matters is the number of turns PER METRE, $n=2000$ m⁻¹.

b)

$B=\dfrac{\mu_{0}I}{2\pi r}=\dfrac{4\pi\times 10^{-7}\times 10.0}{2\pi\times 0.050}=4.0\times 10^{-5}$ T $=40$ µT.

c)

$\dfrac{5.03\times 10^{-3}}{4.0\times 10^{-5}}=126$ : the winding concentrates the field.

d)

The needle lines up with the net field, the vector sum of two perpendicular fields: $\tan\alpha=\dfrac{40}{20}=2$ , so $\alpha=63°$ . A current of ten amperes completely upsets a compass five centimetres away.

e)

In an ideal solenoid the interior field is uniform, and Ampère's law applied to a rectangle gives $BL=\mu_{0}NI$ , with no transverse dimension at all: only the density of turns along the axis matters. The radius only comes into play near the ends, where the approximation stops being valid.

Top view, current flowing up in the wire:

5.0

cm south of the wire, its field (

40

µT) points east, perpendicular to Earth's field (

20

µT, toward the north). The needle lines up with the sum, at

63°

from north (vectors to scale).

Exercise 9: Deflecting an electron beam with a region of magnetic field

Electrons, accelerated from rest through $2.00$ kV, enter, perpendicular to its edge, a region $3.0$ cm wide where there is a uniform magnetic field of $1.0$ mT, perpendicular to their velocity. Outside this region the field is zero. $e=1.60\times 10^{-19}$ C, $m_{e}=9.11\times 10^{-31}$ kg.

a) Calculate the speed of the electrons as they enter the region.
b) Calculate the radius of their path in the field, in cm.
c) Calculate the angle of deflection as they leave the region.
d) What is the speed of the electrons as they leave?
e) Why does the magnetic force not change this speed?

Show the solution

Answers

a) $v=2.65\times 10^{7}$ m/s
b) $r=15.1$ cm
c) $\theta\approx 11.5°$
d) Unchanged
e) Perpendicular force: zero work

a)

$eV=\dfrac{1}{2}m_{e}v^{2}$ , so $v=\sqrt{\dfrac{2\times 1.60\times 10^{-19}\times 2000}{9.11\times 10^{-31}}}=2.65\times 10^{7}$ m/s, about $9$ % of the speed of light: the non-relativistic calculation remains acceptable.

b)

The magnetic force plays the role of the centripetal force: $evB=\dfrac{m_{e}v^{2}}{r}$ , so $r=\dfrac{m_{e}v}{eB}=\dfrac{9.11\times 10^{-31}\times 2.65\times 10^{7}}{1.60\times 10^{-19}\times 1.0\times 10^{-3}}=0.151$ m $=15.1$ cm.

c)

The electron follows an arc of a circle of radius $r$ whose projection on the initial direction equals the width $d$ of the region: $\sin\theta=\dfrac{d}{r}=\dfrac{3.0}{15.1}=0.199$ , so $\theta=11.5°$ . It then leaves in a straight line.

d)

$2.65\times 10^{7}$ m/s, the same as on entry: only the DIRECTION has changed.

e)

$\vec F=q\vec v\times\vec B$ is always perpendicular to the velocity: its power $\vec F\cdot\vec v$ is zero, so it does no work and does not change the kinetic energy. A magnetic field bends a path; it never speeds a particle up.

With

\vec{B}

pointing out of the page, the force

-e\,\vec{v}\times\vec{B}

deflects the electron upward. In the region it follows an arc of radius

15.1

cm, drawn to scale, then leaves in a straight line at

11.5°

from its initial direction.

Exercise 10: Faraday's law on a coil pulled out of the field

A flat coil of $200$ turns and area $0.012$ m² is placed perpendicular to a uniform magnetic field of $0.30$ T. It is pulled out of the field in $0.050$ s, the flux dropping steadily to zero. The coil has a resistance of $4.0$ Ω and forms a closed circuit.

a) Calculate the initial magnetic flux through ONE turn, in mWb.
b) Calculate the average induced electromotive force.
c) Calculate the average induced current.
d) Calculate the energy dissipated in the coil during the withdrawal, in J.
e) According to Lenz's law, the induced current creates a field that...

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Answers

a) $\Phi_{1}=3.6$ mWb
b) $|\mathcal{E}|=14.4$ V
c) $I=3.6$ A
d) $E\approx 2.6$ J
e) Same direction as the decreasing field

a)

$\Phi_{1}=BA=0.30\times 0.012=3.6\times 10^{-3}$ Wb $=3.6$ mWb.

b)

$|\mathcal{E}|=N\dfrac{|\Delta\Phi|}{\Delta t}=200\times\dfrac{3.6\times 10^{-3}}{0.050}=14.4$ V. The factor $N$ is essential: each turn adds its own emf.

c)

$I=\dfrac{\mathcal{E}}{R}=\dfrac{14.4}{4.0}=3.6$ A.

d)

$E=RI^{2}\Delta t=4.0\times 3.6^{2}\times 0.050=2.6$ J. This energy comes from the work done by the hand pulling the coil: you feel a resistance to the withdrawal.

e)

The flux DECREASES: the induced current creates a field in the SAME direction as the external field, to oppose this decrease. Lenz's law does not say “opposite to the field”, it says “opposing the change in flux”.

With

\vec{B}

pointing into the page: during the withdrawal, the inward flux decreases. The induced current flows clockwise, which creates inside the coil a field pointing inward, in the SAME direction as the external field.

Exercise 11: A field that grows nonlinearly

A coil of $50$ turns and area $0.040$ m² is crossed perpendicularly by a uniform magnetic field that varies as $B(t)=0.020+0.0040\,t^{2}$ , in teslas, with $t$ in seconds. The coil is connected to a resistor, for a total circuit resistance of $2.5$ Ω.

a) Calculate the flux through one turn at $t=3.0$ s, in mWb.
b) Calculate $\dfrac{d\Phi}{dt}$ for one turn at $t=3.0$ s, in mWb/s.
c) Calculate the induced emf at $t=3.0$ s, in mV.
d) Calculate the induced current at that instant, in mA.
e) Why is the induced emf zero at $t=0$ , even though the field is already $0.020$ T there?

Show the solution

Answers

a) $\Phi=2.24$ mWb
b) $0.96$ mWb/s
c) $48$ mV
d) $I\approx 19$ mA
e) $\dfrac{dB}{dt}=0$ at $t=0$

a)

$B(3.0)=0.020+0.0040\times 9.0=0.056$ T and $\Phi=BA=0.056\times 0.040=2.24\times 10^{-3}$ Wb $=2.24$ mWb.

b)

$\dfrac{d\Phi}{dt}=A\dfrac{dB}{dt}=0.040\times(2\times 0.0040\times 3.0)=9.6\times 10^{-4}$ Wb/s $=0.96$ mWb/s.

c)

$|\mathcal{E}|=N\dfrac{d\Phi}{dt}=50\times 9.6\times 10^{-4}=4.8\times 10^{-2}$ V $=48$ mV.

d)

$I=\dfrac{0.048}{2.5}=1.9\times 10^{-2}$ A $=19$ mA.

e)

The emf depends on the CHANGE in flux, not on its value. At $t=0$ , $\dfrac{dB}{dt}=2\times 0.0040\times 0=0$ : the field is momentarily stationary, so no emf is induced. It is the derivative, not the field, that appears in Faraday's law.

With

\vec{B}

pointing into the page and growing, the inward flux increases: the induced current flows counterclockwise, which creates an outward field in the coil and opposes the increase.

Exercise 12: Lab work: checking a resistor with a V(I) line

The voltage across a resistor marked $47$ Ω $\pm 5$ % is measured for five currents.

$V$ (V): $1.0$ ; $2.0$ ; $3.0$ ; $4.0$ ; $5.0$ . $I$ (mA): $21.5$ ; $42.2$ ; $64.0$ ; $85.1$ ; $106.0$ .

a) Which law predicts a straight line through the origin?
b) The linear regression of $V$ against $I$ (in amperes) gives a slope. Calculate it, in Ω.
c) Calculate the percent difference from the nominal value, in %.
d) Calculate the manufacturer's tolerance interval: lower and upper limits, in Ω.
e) Calculate the power dissipated at $5.0$ V, and say whether a $\dfrac{1}{4}$ W resistor is suitable.

Show the solution

Answers

a) Ohm's law $V=RI$
b) $R\approx 47.2$ Ω
c) $+0.4$ %
d) $[44.65, 49.35]$ Ω
e) $P\approx 0.53$ W: no, a 1 W resistor is needed

a)

Ohm's law, $V=RI$ : for an ohmic conductor, the voltage is proportional to the current, and the slope of $V$ against $I$ is the resistance.

b)

In amperes: $0.0215$ ; $0.0422$ ; $0.0640$ ; $0.0851$ ; $0.1060$ . Regression: slope $47.2$ Ω, intercept $-0.009$ V, negligible. Convert the milliamperes BEFORE the regression, otherwise the slope comes out a thousand times too small.

c)

$\dfrac{47.19-47}{47}=+0.4$ %.

d)

$5$ % of $47$ Ω is $2.35$ Ω: the interval is $[44.65, 49.35]$ Ω. The measurement lies well within it: the resistor is within tolerance.

e)

$P=\dfrac{V^{2}}{R}=\dfrac{25}{47.2}=0.53$ W, more than double $0.25$ W: the resistor would overheat and could burn out. A $1$ W resistor is needed, or the voltage in the experiment must be limited.

Corrected practice final exam: electricity and magnetism (203-SN2-RE, formerly 203-NYB)

Course recap

Part A: Electrostatics (/25)

Exercise 1: The force on a charge, by components

Exercise 2: Potential of a dipole, work and the energy of a system

Exercise 3: The deflected ink drop of a printer

Part B: Direct-current circuits (/25)

Exercise 4: A two-source network: Kirchhoff's rules

Exercise 5: Resistivity: sizing the element of a heater

Exercise 6: Charging, then discharging through another resistor

Part C: Magnetism (/25)

Exercise 7: The current balance: levitating a wire in a magnetic field

Exercise 8: The solenoid, the wire and the compass

Exercise 9: Deflecting an electron beam with a region of magnetic field

Part D: Induction and lab work (/25)

Exercise 10: Faraday's law on a coil pulled out of the field

Exercise 11: A field that grows nonlinearly

Exercise 12: Lab work: checking a resistor with a V(I) line

See also

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