Introductory Physics: Problems solving

D. A. Garanin

30 April 2020

Introduction

Solving problems is an inherent part of the physics course that requires a more active

approach than just reading the theory or listening to lectures. Making only the latter, the

student can have an illusion of having understood the material but it is not the case until

s/he becomes able to apply one’s knowledge to solving problems that is, working actively

with the material.

The main purpose of our Introductory Physics course, for the majority of our students, is to

acquire a conceptual understanding of physics, to develop a scientific way of thinking. The

latter means relying on the scientific definitions, simple logic, and the common sense,

opposed to making wild assumptions at every step that leads to wrong results and loss of

points.

PHY166 and PHY167 courses are algebra based, while PHY168 and PHY169 are calculus

based. Both types of courses require that problems are solved algebraically and an algebraic

result, that is, a formula is obtained. Only after that the numbers are plugged in the

resulting formula and the numerical result is obtained. One should understand that physics

is mainly about formulas, not about numbers, thus the main result of problem solving is the

algebraic result, while the numerical result is secondary.

Unfortunately, most of the students taking part in our physics courses reject algebra and try

to work out the solution numerically from the very beginning. Probably, bad teachers at the

high school taught the students that problem solving consists in finding the “right” formula

and plugging the numbers into it. This is fundamentally wrong.

There are several arguments for why the algebraic approach to problem solving is better

than the numeric approach.

1. Algebraic manipulations leading to the solution are no more difficult than the

corresponding operations with numbers. In fact, they are easier as a single symbol,

such as a, stands for a number that usually requires much more efforts to write

without mistakes.

2. Numerical calculations are for computers, while algebraic calculations are for

humans. Computers do not understand what they are computing, and they are

proceeding blindly along prescribed routes. The same does a human trying to

operate with numbers. However, the human forgets what do these numbers stand

for and loses the clue very soon. If a human operates with algebraic symbols, s/he is

not losing the clue as the symbols speak for themselves. For instance, a usually is an

acceleration or a distance, m usually is a mass, etc.

3. The value of a formula is much higher than that of the numerical answer because the

formula can be used with another set of input values while the numerical result

cannot. In all more or less intelligent devices, formulas are implemented that work

as “black boxes”: one supplies the input values and collects the output values.

4. Formulas allow analysis of their dependence on the input values or parameters. This

is important for understanding the formula and for checking its validity on simple

particular cases in which one can obtain the result in a simpler way. This is

impossible to do with numerical answers. Actually, one can hardly understand them.

Probably, the reasons given above are sufficient to abandon attempts to ignore the

algebraic approach, especially as the absence of the algebraic result does not give a full

score, even if the numerical answer is correct.

In this collection, the reader will find some exemplary solutions of Introductory Physics

problems that show the efficient methods and approaches. It is recommended to read my

collection of math used in our course, “REFRESHING High-School Mathematics”.

This collection of physics problems solutions does not intend to cover the whole

Introductory Physics course. Its purpose is to show the right way to solve physics problems.

In selecting the problems, preference was given to those that are not artificial and make

sense.

Here some useful tips.

1. Always try to find out what a problem is about, which part of the physics course is in

question

2. Drawings are very helpful in most cases. They help to understand the problem and

its solution

3. Write down basic formulas that will be used in the solution

4. Write comments in a good scientific language. It will make the solution more

readable and will help you to understand it. Solution that consists only of formulas

and numbers is not good.

5. Frame your resulting formulas. This shows to the grader that you really understand

where your results are.

Physics part I

Physics part II

Electrostatics

1. Electric field from a collection of charges

Electric charges Q1 = Q, Q2 = 2Q, and Q3 = 3Q are placed at r1 = (1,0,0)a, r2 = (0,1,0)a, and r3 =

(0,0,1)a. Find the electric field E at r = (1,1,1)a.

Solution. This problem deals the electric field in the general vector form in 3D with vectors

given by their components, such as A = (Ax,Ay,Az). In the notations above a is a length in

meters. The basic formula for the electric field of a point charge Q following from the

Coulomb’s law reads

E = 𝑘𝑄

𝑟2

r

𝑟.

Here the first part gives the magnitude of the electric-field vector E while the unit vector r/r

gives its direction. (Check that it is indeed a unit vector by calculating its magnitude that

should be 1). This formula implies that the charge Q is in the origin of a coordinate system

and the position of the observation point is given by the vector r that goes from the origin to

the observation point. However, if there are several charges, one cannot put them all in the

origin. A more general formula for the electric field E1 created by the charge Q1 that is not

necessarily in the origin reads

𝑬1 = 𝑘𝑄1

|r − r1|2r − r1

|r − r1|.

Here the vector r − r1 goes from the charge Q1 located at r1 to the observation point r. This

formula can be generalized for several charges put at different positions:

𝐄 = ∑𝑘𝑄𝑖

|r − r𝑖|2r − r𝑖

|r − r𝑖|𝑖

= 𝑘 ∑𝑄𝑖

r − r𝑖

|r − r𝑖|3𝑖

.

This formula can be used as a starting point for solving our problem. It is convenient to pre-

calculate

r − r1 = (1,1,1)𝑎 − (1,0,0)𝑎 = (0,1,1)𝑎

r − r2 = (1,1,1)𝑎 − (0,1,0)𝑎 = (1,0,1)𝑎

r − r3 = (1,1,1)𝑎 − (0,0,1)𝑎 = (1,1,0)𝑎

and

|r − r1| = √02 + 12 + 12𝑎 = √2𝑎

|r − r2| = √2𝑎

|r − r3| = √2𝑎.

Now

𝐄 = 𝑘 [𝑄

𝑎2

(0,1,1)

232

+2𝑄

𝑎2

(1,0,1)

232

+3𝑄

𝑎2

(1,1,0)

232

]

= 𝑘𝑄

𝑎2

(0,1,1) + 2(1,0,1) + 3(1,1,0)

232

= 𝑘𝑄

𝑎2

(0 + 2 + 3,1 + 0 + 3,1 + 2 + 0)

232

= 𝑘𝑄

𝑎2

(5,4,3)

232

.

Understanding this problem gives a student a tool to find the electric field from a collection

of point charges in the most general form.

2. Forces on charges Q put in corners of a rectangle with sides a and b

Charges Q are put in the corners of a rectangle with sides a and b. Find the magnitude of the

force acting on each charge.

Solution. Here we use the Coulomb’s law for the interaction of two point charges Q1 and Q2

at positions r1 and r2

F12 = 𝑘𝑄1𝑄2

|𝐫1 − r2|2𝐫1 − r2

|𝐫1 − r2|.

This is the force acting on charge 1 from charge 2. The force is directed along the line

connecting the two charges. It is repulsive if the charges have the same sign and attractive if

the charges have different signs. Here position vectors are not explicitly given but we know

that the lines connecting the charges are sides and diagonals of the rectangle.

y

x F13

F14 1

2 3

4

F12

a

b

We have to introduce the axes x and y and project all forces on these axes to and add the

forces component by component. The magnitudes of the forces acting on each of the

charges are the same, only their directions are different. We will consider the force acting

on the charge 1,

F1 = 𝐅12 + 𝐅13 + 𝐅14.

The force from charge 2 has only y component, while the force from charge 4 has only x-

component. The force from charge 3 has both x- and y-components and is defined by the

distance √𝑎2 + 𝑏2, as well as by

cos 𝜃 =𝑎

√𝑎2 + 𝑏2, sin 𝜃 =

𝑏

√𝑎2 + 𝑏2.

One obtains

𝐹1,𝑥 = −𝑘𝑄2

𝑎2− 𝑘

𝑄2

𝑎2 + 𝑏2cos 𝜃 = −𝑘

𝑄2

𝑎2− 𝑘

𝑄2𝑎

(𝑎2 + 𝑏2)32

𝐹1,𝑦 = −𝑘𝑄2

𝑏2− 𝑘

𝑄2

𝑎2 + 𝑏2sin 𝜃 = −𝑘

𝑄2

𝑏2− 𝑘

𝑄2𝑏

(𝑎2 + 𝑏2)32

.

Now the magnitude of the force is given by

𝐹1 = √𝐹1,𝑥2 + 𝐹1,𝑦

2 = 𝑘𝑄2√(1

𝑎2+

𝑎

(𝑎2 + 𝑏2)32

)

2

+ (1

𝑏2+

𝑏

(𝑎2 + 𝑏2)32

)

2

.

Forces acting on other charges have the same magnitude, so that one can discard the

subscript 1. In the case b=a the result simplifies to

𝐹 = 𝑘𝑄2

𝑎2√2(1 +

1

23/2)2

= 𝑘𝑄2

𝑎2(√2 +

1

2).

This result can be obtained directly for the problem with the square instead of the rectangle

that is much simpler (the next problem).

3. Forces on charges Q put in corners of a square

This problem is a particular case of the more general preceding problem. In the case of the

square with all charges the same, there is a symmetry that allows one to obtain the solution

without invoking the full vector formalism.

Solution. By the symmetry is is clear that the force on charge 1 is directed along the diagonal

shown in the drawing. Thus, one has to project all three forces on this direction that yields

𝐹 = 2𝑘𝑄2

𝑎2cos 45° +𝑘

𝑄2

(√2𝑎)2.

Here the factor 2 accounts for the two contributions from charges 2 and 4. The second term

accounts for charge 3. With cos 45° = √2/2 one obtains

𝐹 = 𝑘𝑄2

𝑎2(√2 +

1

2).

This result has been obtained as a particular case of the preceding problem and it can be

used for testing the validity of the latter.

4. Forces on different charges at an equilateral triangle

Charges Q, Q, and 2Q are put at the corners of an equilateral triangle with side a. Find the

magnitudes of the forces acting on each charge.

5. Electric field at the center line between two equal charges

Find the electric field E(z) along the straight line going between two equal charges Q put at

the distance a from each other. Analyze particular cases.

3

Q

E

r a/2

a/2

z

0

0

O

Q

a

F12

2

F13

F14 1 4

a

Solution. Put the origin of the coordinate system O in the point between the two charges

and direct z-axis in the right horizontal direction. From the symmetry it follows that the

electric field on the center line is directed horizontally. It is given by

𝐸 = 2𝑘𝑄

𝑟2cos 𝜃,

where the factor 2 takes care for the two contributions into the result from the upper and

lower charges. Using

𝑟 = √𝑧2 + (𝑎

2)2

, cos 𝜃 =𝑧

𝑟 ,

one finally obtains

𝐸 = 𝑘2𝑄𝑧

(𝑧2 + (𝑎2)

2

)3/2

.

One particular case is the point between the charges, z=0. Here the electric field from the

two charges cancel each other and E=0. The formula above reproduces this result that

serves as its check.

Another particular case is the region far away from the charges, z>>a. Here one can neglect

the term (a/2)2 in the denominator of the formula after which the result becomes

𝐸 = 𝑘2𝑄

𝑧2.

This is is nothing else than the electric field of the charge 2Q at the distance z. This is an

expected result as from large distances the two charges close to each other are looking as

one double charge. This is another check on our general formula.

The plot of the dependence E(z) is shown below. There is a maximum of E at the distance z

of order a.

0 z

E

6. Electric field at the center line between two opposite charges

Find the electric field E(z) along the straight line going between two opposite charges Q and

–Q put at the distance a from each other. Analyze particular cases.

Solution. As the upper charge is negative, its electric field at the observation point is

directed toward it, as shown in the drawing. The resulting electric field E is thus directed up.

It is given by

𝐸 = 2𝑘𝑄

𝑟2sin 𝜃.

Using

𝑟 = √𝑧2 + (𝑎

2)2

, sin 𝜃 =𝑎/2

𝑟 ,

one finally obtains

𝐸 = 𝑘𝑄𝑎

(𝑧2 + (𝑎2)

2

)3/2

.

One particular case of this formula is z=0, the observation point between the charges. In this

case one obtains

𝐸 = 𝑘8𝑄

𝑎2

that is the doubled electric field at the distance a/2 from the charge Q.

At large distances , z>>a, one can neglect the term (a/2)2 in the denominator of the formula

after which the result becomes

𝐸 = 𝑘𝑄𝑎

𝑧3.

Q

-Q

Q

E

r a/2

a/2

z

0

0

O

As the power in the denominator is three rather than two, one concludes that the electric

field produced by two opposite charges decreases faster at large distances than the field

produces by one charge. Indeed, looking from large distances, one cannot see the system of

two opposite charges as one effective charge (as in the preceding problem) because the

sum of the two charges is zero. Such a system is called “electric dipole”.

7. Electric potentials in the middle of the equilateral triangle of charges and

in the middle of a side.

Consider an equilateral triangle with side a having point charges Q at its corners. Compare

electric potentials in the center of the triangle and in the middle of its side. What is your

expectation? Which potential is higher?

Solution. Let us denote the electric potential in the center of the triangle Vc and the electric

potential in the middle of the side Vm. At the center one has

𝑉𝑐 = 3𝑘𝑄

𝑙,

while in the middle of the side

𝑉𝑚 = 2𝑘𝑄

𝑎/2+ 𝑘

𝑄

ℎ.

With

ℎ = 𝑎 cos 30° = 𝑎√3

2, 𝑙 =

𝑎/2

cos 30°=

𝑎

√3

one finally obtains

Vm

Vc l

a h

a/2

Q Q

Q

z

E

𝑉𝑐 = 𝑘𝑄

𝑎3√3 ≈ 𝑘

𝑄

𝑎× 5.196

and

𝑉𝑚 = 𝑘𝑄

𝑎(4 +

2

√3) ≈ 𝑘

𝑄

𝑎× 5.155.

These two values are so close to each other that no intuition in the world can figure out

what potential is higher without the actual calculating the potentials.

8. Electric potentials in the middle of a square of charges and in the middle of

a square’s side.

Consider a square with side a having point charges Q at its corners. Compare electric

potentials in the center of the square and in the middle of its side. What is your

expectation? Which potential is higher?

9. The electroscope

Electroscope consists of two thin filaments of length l hanging from the common fixation

point. At the end of each filament there is a light ball of mass m. If the balls are charged with

the charge Q each, they repel via the Coulomb force, so that in the equilibrium state the

filaments make an angle with the vertical. Find the charge Q.

Solution. The mechanical equilibrium of each ball is defined by the condition of the zero

total force:

𝐅 + 𝑚𝐠 + 𝐓 = 0.

mg

T

F

l

x

y

Here mg is the gravity force, T is the filament tension force, and F is the Coulomb repulsion

force given by

𝐹 = 𝑘𝑄2

(2𝑙 sin 𝜃)2.

In projections onto the x- and y-axes the equilibrium condition becomes

𝑇 cos 𝜃 − 𝑚𝑔 = 0

𝑇 sin 𝜃 − 𝐹 = 0.

Multiplying the first equation by sin 𝜃 and the second one by sin 𝜃 and then subtracting one

equation from the other, one can eliminate T and obtains the relation

𝐹 cos 𝜃 = 𝑚𝑔 sin 𝜃.

Using the expression for F above, one can find Q from this equation as

𝑄 = 2𝑙 sin 𝜃√𝑚𝑔

𝑘tan𝜃 .

Electric current, electric circuits

10. Scaling of the resistance

The length of the electric cable was increased by a factor of three and its diameter was

increased by a factor of two. By which factor has the resistance changed?

Solution. We use the basic formula for the resistance

𝑅 = 𝜌𝑙

𝑆,

Where l is the cable’s length and S is the cable’s cross-sectional area. For the circular cross-

section, S=d2/4, where d is the cable’s diameter. We use two instances of the formula

above,

𝑅 = 𝜌4𝑙

𝜋𝑑2, 𝑅′ = 𝜌

4𝑙′

𝜋𝑑′2,

where

𝑙′ = 3𝑙, 𝑑′ = 2𝑑.

Dividing one resistance by the other, one gets rid of the non-essential and unknown entries

and obtains

𝑅′

𝑅=

𝑙′

𝑙

𝑑2

𝑑′2=

3

22=

3

4.

This means

𝑅′ =3

4𝑅.

11. A circuit with two batteries and three resistors

Find the currents and voltages for each of three resistors in the following circuit

Solution. Choose the positive directions of the currents according to the directions of the

EMF’s of the batteries. According to the first Kirchhoff’s law,

𝐼3 = 𝐼1 + 𝐼2

(charges are not accumulating in the nodes). The second Kirchhoff’s law states that for each

closed loop in the circuit the sum of voltages is zero that reflects the fact that electric

potential is defined unambiguously (and the work of the electric field over each closed

trajectory is zero):

∑𝑉𝑖

𝑖

= 0.

To the Kirchhoff’s laws, one has to add the Ohm’s law

𝑉𝑖 = 𝑅𝑖𝐼𝑖

for each resistor. On the top of it, there can be EMF’s acting within resistors (batteries have

their own internal resistance and thus can be considered as resistors) and pushing the

current through them. With the EMF’s, the Ohms law becomes

𝑉𝑖 + E𝑖 = 𝑅𝑖𝐼𝑖.

Substituting 𝑉𝑖 = 𝑅𝑖𝐼𝑖 − E𝑖 into the second Kirchhoff’s law, one obtains

I3

I2 I

1

R3 R

2 R

1

E2 E

1

∑𝑅𝑖𝐼𝑖𝑖

= ∑ E𝑖

𝑖

.

In the circuit above, we neglect the internal resistances of the batteries. The Kirchhoff-

Ohm’s law above for the left and right loops becomes

𝑅1𝐼1 + 𝑅3(𝐼1 + 𝐼2) = E1

𝑅2𝐼2 + 𝑅3(𝐼1 + 𝐼2) = E2.

This is a system of two linear equations with two unknowns that has a solution. To solve this

system of equations in a most elegant way, one can first rewrite it in the form collecting

terms with I1 and I2

(𝑅1+𝑅3)𝐼1 + 𝑅3𝐼2 = E1

𝑅3𝐼1 + (𝑅2+𝑅3)𝐼2 = E2.

Then one can eliminate, say, I2 by multiplying the first equation by R2+R3, the second

equation by R3 and then subtracting the second equation from the first one. This yields

(𝑅1 + 𝑅3)(𝑅2 + 𝑅3)𝐼1 − 𝑅32𝐼1 = (𝑅2 + 𝑅3)E1 − 𝑅3E2.

From here one finds

𝐼1 =(𝑅2 + 𝑅3)E1 − 𝑅3E2

𝑅1𝑅2 + (𝑅1+𝑅2)𝑅3.

Since this circuit is symmetric, one can obtain the formula for I2 without calculations just by

replacing 1 → 2, 2 → 1. This yields

𝐼2 =(𝑅1 + 𝑅3)E2 − 𝑅3E1

𝑅1𝑅2 + (𝑅1+𝑅2)𝑅3.

Now

𝐼3 = 𝐼1 + 𝐼2 =𝑅2E1 + 𝑅1E2

𝑅1𝑅2 + (𝑅1+𝑅2)𝑅3.

Voltages on the three resistors can be now found from Ohm’s law. One can see that the

currents can flow both in positive and negative directions. For instance, if E1=E2, then both

currents are positive. If E1 is sufficiently stronger than E2 (work out the exact condition!),

then I1>0 but I2<0. If E2 is sufficiently stronger than E1, then I2>0 but I1<0.

In the particular case R3=0 (short circuiting) there are two independent circuits for which

one obtains

𝐼1 =E1

𝑅1, 𝐼2 =

E2

𝑅2

that follows from the formulas above if one sets R3=0. This provides a check for the formulas

obtained.

If one removes R3, there is only one loop for which one obtains

𝐼1 ≡ 𝐼 =E1 − E2

𝑅1 + 𝑅2.

This result follows from the limit 𝑅3 → ∞ as follows

𝐼1 =(𝑅2/𝑅3 + 1)E1 − E2

𝑅1𝑅2/𝑅3 + 𝑅1+𝑅2→E1 − E2

𝑅1 + 𝑅2.

(one discards the terms containing R3 in the denominator). This is another check of our main

result.

Magnetic field created by electric currents

12. Magnetic field at the center line between two long wires with the currents

in the same direction

Two long straight wires are going parallel to each other at the distance a from each other.

The wires carry the currents I that go in the same direction. Find the magnetic field created

by this system at the center line between the wires (the vertical line on the drawing).

Solution. We set the origin of the coordinate system in the point O in the middle between

the wires and the z-axis going up along the center line. Let the currents be directed inside

the paper sheet, then, according to the screw rule, the magnetic field created by each wire

is directed clockwise. The currents 1 and 2 create magnetic fields B1 and B2 directed as

shown in the drawing. By symmetry, the resulting magnetic field B = B1 + B2 is directed

r r

z

a/2 1 2

B1

B2

B

a/2 O

horizontally to the right. Using the basic formula for the magnetic field created by the long

wire

𝐵 =𝜇0

2𝜋

𝐼

𝑟,

and projecting the vectors B1 and B2 on the horizontal direction x, one obtains

𝐵 = 𝐵1,𝑥 + 𝐵2,𝑥 =𝜇0

2𝜋

𝐼

𝑟2 sin 𝜃.

With

𝑟 = √𝑧2 + (𝑎/2)2 , sin 𝜃 =𝑧

𝑟

one finally obtains

𝐵 =𝜇0𝐼

2𝜋

2𝑧

𝑧2 + (𝑎/2)2.

Here, it does not make sense to cancel 2 in the numerator and denominator.

Let us analyze particular and limiting cases. First, our formula yields B=0 for z=0 (at the point

O). This is an expected result, as in this case fields B1 and B2 are opposite and cancel each

other.

At large distances, z>>a, one can neglect (𝑎/2)2 in the denominator that yields

𝐵 =𝜇0

2𝜋

2𝐼

𝑧.

This is the field produced by one wire with the current 2I. Indeed, from large distances these

two wires are seen as one wire with the double current. This limiting case serves as one of

the checks of the general formula.

13. Magnetic field at the center line between two long wires with the currents

in the opposite directions

Two long straight wires are going parallel to each other at the distance a from each other.

The wires carry the currents I that go in the opposite directions. Find the magnetic field

created by this system at the center line between the wires (the vertical line on the

drawing).

Solution. We set the origin of the coordinate system in the point O in the middle between

the wires and the z-axis going up along the center line. Let current 1 be directed inside the

paper sheet and current 2 be directed outside the paper sheet, then, according to the screw

rule, the magnetic fields created by each wire are directed clockwise and counterclockwise,

respectively. The currents 1 and 2 create magnetic fields B1 and B2 directed as shown in the

drawing. By symmetry, the resulting magnetic field B = B1 + B2 is directed vertically down.

Using the basic formula for the magnetic field created by the long wire

𝐵 =𝜇0

2𝜋

𝐼

𝑟,

and projecting the vectors B1 and B2 on the vertical direction x, one obtains

𝐵 = 𝐵1,𝑧 + 𝐵2,𝑧 =𝜇0

2𝜋

𝐼

𝑟2 cos 𝜃.

With

𝑟 = √𝑧2 + (𝑎/2)2 , cos 𝜃 =𝑎/2

𝑟

one finally obtains

𝐵 =𝜇0𝐼

2𝜋

𝑎

𝑧2 + (𝑎/2)2.

Let us consider particular and limiting cases of this formula. At z=0 the result reads

𝐵 =𝜇0𝐼

2𝜋

4

𝑎

that is twice the magnetic field created by a wire at the distance a/2, an expected result.

At large distances, z>>a, one can neglect (𝑎/2)2 in the denominator that yields

. O B 1

2 B

2 B

1

z

r r

a/2 a/2

𝐵 =𝜇0𝐼

2𝜋

𝑎

𝑧2.

This decreases at large distances faster than the magnetic field from one wire since B1 and

B2 become almost opposite and nearly cancel each other.

14. Forces on parallel long wires forming an equilateral triangle

Three long wires go parallel to each other, forming an equilateral triangle with the side a in

the cross-section (sheet of paper). The currents I are the same in each wire and flow in the

same direction. Find the forces acting on the wires.

Solution. A we know, long parallel wires with currents flowing in the same direction are

attracting and the formula for the force has the form

𝐹 =𝜇0

2𝜋

𝐼1𝐼2𝑟

𝐿,

where r is the distance between the wires and L is the length of the wires. In our case,

𝐼1 = 𝐼2 = 𝐼 and r=a. The forces between the wires are directed along the lines connecting

them in the drawing and forming an equilateral triangle. By symmetry, the force on each

wire is directed along the bisector. Thus, it is sufficient to project the forces onto the

bisector direction that yields

𝐹 =𝜇0

2𝜋2

𝐼2

𝑎𝐿 cos 30° =

𝜇0

2𝜋

𝐼2

𝑎𝐿√3.

1

2

3

a

F31

F32

F

3

I

I

I

15. Forces on parallel long wires forming a right triangle with equal catheti

(legs)

Three long wires go parallel to each other, forming a right triangle with equal legs of length

a in the cross-section (sheet of paper). The currents I are the same in each wire and flow in

the same direction. Find the forces acting on the wires.

Solution. Here, by symmetry, the force F2 is directed vertically down and can be found with

the help of the symmetry, similarly to the preceding problem. The magnitudes of the forces

F1 and F3 are equal to each other, only their directions differ. These forces, however, cannot

be found using the symmetry. Thus, for them one has to introduce the coordinate axes x

and y and first compute projections of the forces on these axes.

Using the basic formula for the force between two long parallel wires, one obtains

𝐹2 =𝜇0

2𝜋2

𝐼2

𝑎𝐿 cos 45° =

𝜇0

2𝜋

𝐼2

𝑎𝐿√2.

For F3, one writes

F3 = F31 + F32,

and, in components,

𝐹3,𝑥 = 𝐹31,𝑥 + 𝐹32,𝑥, 𝐹3,𝑦 = 𝐹31,𝑦 + 𝐹32,𝑦.

Using

𝐹31,𝑥 = −𝜇0

2𝜋

𝐼2

𝑎√2𝐿, 𝐹31,𝑦 = 0

and

𝐹32,𝑥 = −𝜇0

2𝜋

𝐼2

𝑎𝐿 cos 45° = −

𝜇0

2𝜋

𝐼2

𝑎𝐿

1

√2, 𝐹32,𝑦 =

𝜇0

2𝜋

𝐼2

𝑎𝐿 sin 45° =

𝜇0

2𝜋

𝐼2

𝑎𝐿

1

√2.

1

2

3

a

F31

F32

F3

I

I

I

a

F23

F21

F2

F1

x

y

Now,

𝐹3 = √𝐹3,𝑥2 + 𝐹3,𝑦

2 =𝜇0

2𝜋

𝐼2

𝑎𝐿√(

1

√2+

1

√2)2

+ (1

√2)2

=𝜇0

2𝜋

𝐼2

𝑎𝐿√

5

2= 𝐹1.

16. Magnetic field in the center between four long wires with current forming a

square

Four long wires go parallel to each other, forming a square of side a in the cross-section

(sheet of paper). The currents I are directed inside the paper in the left and right upper

corners and out of the paper in the left and right bottom corners. Find the magnetic field in

the center of the square.

Solution. Long wires with the current produce the magnetic field given by

𝐵 =𝜇0

2𝜋

𝐼

𝑟.

The direction of the field is defined by the screw rule, as shown in the drawing. There are

four contributions to the magnetic field from each wire:

B = 𝐁𝟏 + 𝐁𝟐 + 𝐁𝟑 + 𝐁𝟒

that are directed along the diagonals of the square as shown. By symmetry, the resulting

field is horizontal and directed to the left. Projecting the fields of each wire onto this

direction, one obtains

𝐵 =𝜇0

2𝜋

𝐼

𝑎/√24 cos 45° =

𝜇0

2𝜋

𝐼

𝑎4.

. .

1 2

3 4

B2

B

B1

B3

a

B4

In all currents are directed in the same directions, magnetic fields of different wires point in

four different directions and cancel each other.

17. Magnetic field in the center of a side of the square formed by four long

wires with current ― I

Four long wires go parallel to each other, forming a square of side a in the cross-section

(sheet of paper). The currents I are directed inside the paper in the left and right upper

corners and out of the paper in the left and right bottom corners. Find the magnetic field in

the centers of the top and bottom sides of the square.

Solution. The situations at the top and bottom sides of the square are similar, and the

resulting magnetic field is the same. It is sufficient to consider, say, only the bottom side.

The magnetic fields of four wires are shown in the drawing. The fields from wires 3 and 4

are opposite and cancel each other. Thus

B = 𝐁𝟏 + 𝐁𝟐.

The fields from wires 1 and 2 are perpendicular to the lines connecting the corresponding

wires with the center of the side. By symmetry, the resulting magnetic field is horizontal and

given by

𝐵 =𝜇0

2𝜋

𝐼

𝑙2 cos 𝜃.

The distance l can be found from the Pythagoras’s theorem:

𝑙 = √𝑎2 + (𝑎/2)2 = 𝑎√5

2.

. .

1 2

3 4

B2

B B

1

B3

B4

a

l

Angle is not a simple angle but its cosine can be found from the triangle:

cos 𝜃 =𝑎

𝑙=

2

√5.

Now, finally one obtains

𝐵 =𝜇0

2𝜋

𝐼

𝑎√5/22

2

√5=

𝜇0

2𝜋

𝐼

𝑎

8

5.

If all four currents are directed in the same direction, the result is the same.

18. Magnetic field in the center of a side of the square formed by four long

wires with current ― II

Four long wires go parallel to each other, forming a square of side a in the cross-section

(sheet of paper). The currents I are directed inside the paper in the left and right upper

corners and out of the paper in the left and right bottom corners. Find the magnetic field in

the centers of the left and right sides of the square.

Solution. The situations at the left and right sides of the square are similar, and the resulting

magnetic field is the same. It is sufficient to consider, say, only the left side. The magnetic

fields of four wires are shown in the drawing. By symmetry, the resulting magnetic field is

directed horizontally to the left. Thus, one has to project all vectors onto this direction. One

obtains

B = 𝐁𝟏 + 𝐁𝟐 + 𝐁𝟑 + 𝐁𝟒

and

. .

1 2

3 4

B2

B

B1

B3

B4

a

l

𝐵 =𝜇0

2𝜋

𝐼

𝑎/22 +

𝜇0

2𝜋

𝐼

𝑙2 cos 𝜃.

Using

𝑙 = 𝑎√5

2, cos 𝜃 =

𝑎/2

𝑙=

1

√5,

one obtains

𝐵 =𝜇0

2𝜋

𝐼

𝑎2 (2 +

2

√5

1

√5) =

𝜇0

2𝜋

𝐼

𝑎4 (1 +

1

5) =

𝜇0

2𝜋

𝐼

𝑎

24

5.

This magnetic field is stronger than that at the centers of the top and bottom sides

calculated in the preceding problem.

19. Interaction of long suspended wires

Two long wires of length L and mass m are suspended in a horizontal position parallel to

each other at the distance a by cords of length r. After the current I started flowing in each

wire in the same direction, the wires displaced so that the cords make the angle q with the

vertical. Find the current I in the wires. Also solve the problem for the currents flowing in

different directions.

Solution. Two long wires with current I in each wire are attracting with the force

𝐹 =𝜇0

2𝜋

𝐼2

𝑑𝐿,

a

r r

r

r

mg

F

T

I I d

where d is the distance between the wires. In our case,

𝑑 = 𝑎 − 2𝑟 sin 𝜃.

Mechanical equilibrium of the system is defined by the condition of the total force equal to

zero:

𝐅 + 𝑚𝐠 + 𝐓 = 0.

Here mg is the gravity force, T is the filament tension force, and F is the magnetic attraction

force. In projections onto the x- and y-axes the equilibrium condition becomes

𝑇 cos 𝜃 − 𝑚𝑔 = 0

𝑇 sin 𝜃 − 𝐹 = 0.

Multiplying the first equation by sin 𝜃 and the second one by sin 𝜃 and then subtracting one

equation from the other, one can eliminate T and obtains the relation

𝐹 cos 𝜃 = 𝑚𝑔 sin 𝜃.

Using the expression for F above, one can obtains

𝜇0

2𝜋

𝐼2

𝑎 − 2𝑟 sin 𝜃𝐿 = 𝑚𝑔 tan𝜃

and, finally,

𝐼 = √2𝜋

𝜇0

𝑚𝑔

𝐿tan𝜃 (𝑎 − 2𝑟 sin 𝜃).

Mechanical equilibrium found here should be an unstable equilibrium if r>a/2 (so that the

wires can touch each other) and the current is strong enough. Clarifying this needs a special

investigation.

If the currents are flowing in different directions, the wires repel each other. In this case,

one obtains a similar formula with the sign + instead of –, and the equilibrium is stable.

Motion of charged particles in magnetic field

20. Cyclotron resonance

An electron is accelerated by the voltage V and after that it enters the region where it is

moving perpendicular to the uniform magnetic field B making a circular trajectory of radius

r. Find the charge-to-mass ratio of the electron measured in this experiment.

Solution. The electron’s speed after the acceleration is defined by the energy conservation

law

𝑒𝑉 =𝑚𝑣2

2.

The radius of the circular trajectory of the electron in the magnetic field is defined from the

condition that the Lorentz force plays the role of the centripetal force:

𝑒𝑣𝐵 =𝑚𝑣2

𝑟.

To find e/m, one has to eliminate v from the two equations above. For this, one can cancel

one v in the second equation, than square this equation, and then divide it by the first

equation. This yields

𝑒2𝐵2

𝑒𝑉=

2𝑚2𝑣2

𝑟2𝑚𝑣2

and, after simplification,

𝑒

𝑚=

2𝑉

𝑟2𝐵2.

21. Compensation of the electric and magnetic forces

An electron is accelerated by the voltage V and after that it enters the capacitor where it is

moving perpendicular to the uniform electric field E. There is a uniform magnetic field B

perpendicular to E and electron’s velocity v. As a result, the electric and magnetic forces

acting on the electron cancel each other, so that the electron is moving straight. Find the

charge-to-mass ratio of the electron measured in this experiment.

Solution. The electron’s speed after the acceleration is defined by the energy conservation

law

𝑒𝑉 =𝑚𝑣2

2.

Cancellation of the electric and magnetic forces implies

𝑒𝐸 = 𝑒𝑣𝐵

that yields

𝑣 =𝐸

𝐵.

Substituting this into the first formula, one obtains

𝑒

𝑚=

𝑣2

2𝑉=

𝐸2

2𝐵2𝑉.

This is another method to measure the charge-to-mass ratio of the electron (the main

method uses the cyclotron resonance that is simpler as only the magnetic field is needed).

Electromagnetic induction

22. Magnetic flux, magnetic moment, magnetic torque for a loop

A loop of wire is a rectangle with sides a and b lying in the x-y plane. A current I flows

clockwise, if seen from below. There is a magnetic field B=(Bx,By,Bz). Calculate the magnetic

flux through the loop, the magnetic moment M of the loop, and magnitude of the torque

acting on the loop. What can you say about the direction of the torque?

Solution. The magnetic flux through a flat loop is given by

Φ = 𝑆n∙B,

where S is the cross-section of the loop and n is the normal to the loop (the unit vector

perpendicular to the plane of the loop). Here, S=ab and it is convenient to direct n along z-

axis, n=(0,0,1). Using the definition of the dot-product of two vectors,

A∙B = 𝐴𝑥𝐵𝑥 + 𝐴𝑦𝐵𝑦 + 𝐴𝑧𝐵𝑧,

one obtains

Φ = 𝑎𝑏(0 × 𝐵𝑥 + 0 × 𝐵𝑦 + 1 × 𝐵𝑧) = 𝑎𝑏𝐵𝑧.

The magnetic moment of a flat loop is given by

M = 𝑆𝐼𝐧.

In this definition, the normal n and thus M are directed so that the direction of the current

in the loop and M form a screw, whereas I>0 in this formula. According to the setting of the

problem (current I flows clockwise, if seen from below), M is directed up along z-axis. The

magnitude of the magnetic moment is M=abI.

The torque acting on the magnetic moment is given by

= M × 𝐁.

The magnitude of the torque is

𝜏 = 𝑀𝐵 sin 𝜃,

where is the angle between these two vectors. This can be rewritten as

𝜏 = 𝑀𝐵⊥,

where 𝐵⊥ is the projection of B on the plane perpendicular to M. As M is directed along z-

axis, the final result reads

𝜏 = 𝑀√𝐵𝑥2 + 𝐵𝑦

2 = 𝑎𝑏𝐼√𝐵𝑥2 + 𝐵𝑦

2.

The direction of the torque is perpendicular to both M and B. Under the influence of the

torque, the loop tends to rotate into the position in which M and B are collinear.

Note that the angle between n and B is defined by

cos 𝜃 =𝐵𝑧

𝐵, sin 𝜃 =

√𝐵𝑥2 + 𝐵𝑦

2

𝐵,

where

𝐵 = |B| = √𝐵𝑥2 + 𝐵𝑦

2 + 𝐵𝑧2.

The angle itself is given by

𝜃 = cos−1𝐵𝑧

𝐵.

At this stage, if the numerical values of the parameters are given, they can be substituted to

the formulas obtained above.

23. Solenoid and EMF in a loop – I

A solenoid has the length l and contains N turns of wire carrying the current I. Inside the

solenoid is a loop of wire, a circle of radius R. In the initial state, the normal to the loop is

rotated away from the solenoid’s axis by the angle . Then this angle is increased by 180°

during the time t. Find the average EMF in the loop.

Solution. We solve this problem starting from the end. The average EMF is given by the

Faraday’s law

E =∆Φ

Δ𝑡.

In the initial state, the magnetic flux is given by

Φ𝑖 = 𝑆𝐵 cos 𝜃.

In the final state, the magnetic flux is

Φ𝑓 = 𝑆𝐵 cos(𝜃 + 180°) = −𝑆𝐵 cos 𝜃.

Using 𝑆 = 𝜋𝑅2 for the area of the loop, for the change of the magnetic flux one obtains

∆Φ = Φ𝑓 − Φ𝑖 = −2𝑆𝐵 cos 𝜃 = −2𝜋𝑅2𝐵 cos 𝜃.

Now, the magnetic field inside the solenoid is given by

𝐵 = 𝜇0

𝑁

𝑙𝐼.

Substituting this into the Faraday’s law above, one finally obtains

E =2𝜋𝑅2𝐵 cos 𝜃

Δ𝑡=

2𝜋𝑅2𝜇0𝑁𝐼 cos 𝜃

𝑙Δ𝑡.

Here we have discarded the minus sign that is not important here.

24. Solenoid and EMF in a loop – II

A solenoid has the length L and contains N turns of wire. Inside the solenoid is a loop of

wire, a circle of radius R. The normal to the loop is rotated away from the solenoid’s axis by

the angle . The current in the solenoid is increased from zero to I during the time t. Find

the average EMF in the loop.

Solution. The average EMF is given by the Faraday’s law

E =∆Φ

Δ𝑡.

The magnetic flux through the loop created by the solenoid is given by

Φ = 𝑆𝐵 cos 𝜃 = 𝜋𝑅2𝜇0

𝑁

𝑙𝐼 cos 𝜃.

The temporal change of is due to the change of I, thus

E =∆Φ

Δ𝑡= 𝜋𝑅2𝜇0

𝑁

𝑙

∆𝐼

∆𝑡cos 𝜃

and, since

∆𝐼

∆𝑡=

𝐼

∆𝑡,

finally one obtains

E =𝜋𝑅2𝜇0𝑁𝐼 cos 𝜃

𝑙Δ𝑡.

25. Changing of the magnetic flux and magnetic field in a coil connected to a

resistor

A coil with the self-inductance L is connected to the resistor R. The external magnetic field

B0 changes in time so that its flux through the coil changes from zero to 0 during the time

t. Estimate the change of the total flux through the coil taking into account the magnetic

flux produced by the coil itself. Further, for the solenoid, relate the change of the external

magnetic field B0 and the change of the total magnetic field B in the solenoid.

Solution. We use the Faraday’s law

E =∆Φ

Δ𝑡,

where the number of turns of the wire N in the coil is encapsulated in the magnetic flux. The

latter consists in the magnetic flux due to the external field and that produced by the coil

itself. The EMF due to the latter can be expressed via the self-inductance L, so that

E =∆Φ

Δ𝑡=

∆Φ0

Δ𝑡− 𝐿

Δ𝐼

Δ𝑡.

The minus in the second term, introduced by Lenz, ensures that the current due to the self-

induction creates the magnetic field and thus the magnetic flux that partially compensates

the change of the magnetic flux due to the external magnetic field. For an estimation, one

can use ∆Φ0 = Φ0 and Δ𝐼 = 𝐼 and then express the electric current via the Ohm’s law

𝐼 =E

𝑅.

This yields the equation for E,

E =Φ0

𝑡− 𝐿

E

𝑅𝑡

that can be solved to give

E =Φ0

𝑡

1

1 +𝐿𝑅𝑡

.

Now for the total magnetic flux from the Faraday’s law one obtains

Φ = E𝑡 =Φ0

1 +𝐿𝑅𝑡

. (1)

The limits of this expression are

Φ ≅ Φ0

𝑅𝑡

𝐿 (short times), Φ ≅ Φ0 (long times). (2)

One can see that because of the self-induction, the change of the total flux is smaller than

the change of the external flux, as the coil tries to compensate the change by creating its

own magnetic flux in the opposite direction. If the self-inductance L is large, the resistance R

is small, or the process is very fast (t is very short) the total flux is not changing, =0. In

particular, this happens in the case of the superconducting coil (R=0).

For the solenoid, the magnetic flux is given by

Φ = 𝑆𝑁𝐵.

Thus, for the magnetic field from the formula for one obtains

𝐵 =𝐵0

1 +𝐿𝑅𝑡

.

The solution above is qualitative and it is only good as an estimation that captures the

essential physics. Let us consider the accurate solution that is possible within the calculus-

based physics course. In terms of derivatives, the basic equations read

E =dΦ0

d𝑡− 𝐿

d𝐼

d𝑡, 𝐼 =

E

𝑅

If the external flux changes linearly with time,

dΦ0

d𝑡= 𝛼 = const,

then the differential equation for the EMF has the form

E = 𝛼 −𝐿

𝑅

dE

d𝑡

or

dE

d𝑡+

𝑅

𝐿E =

𝑅

𝐿𝛼.

The initial condition for the problem is E(0) = 0. This is the only possibility. If at t = 0 the EMF

jumps from zero to a finite value, this causes an infinity in the self-induction term because of

dE/dt.

To solve this differential equation, it is convenient to introduce a new variable

𝑋 = E− 𝛼.

In terms of the latter, the differential equation becomes homogeneous:

d𝑋

d𝑡+

𝑅

𝐿𝑋 = 0, 𝑋(0) = −𝛼.

Its solution has the form

𝑋(𝑡) = −𝛼exp (−𝑅

𝐿𝑡)

that finally yields

E = 𝑋 + 𝛼 = 𝛼 [1 − exp (−𝑅

𝐿𝑡)].

Now, the total magnetic flux can be obtained by integrating the Faraday’s equation

E =𝑑Φ

𝑑𝑡.

This yields

Φ(𝑡) = ∫ E(𝑡′)𝑑𝑡′

𝑡

0

= 𝛼 ∫[1 − exp (−𝑅

𝐿𝑡′)] 𝑑𝑡′

𝑡

0

= 𝛼𝑡 + 𝛼𝐿

𝑅 𝑒𝑥𝑝 (−

𝑅

𝐿𝑡′)|

0

𝑡

= 𝛼𝑡 − 𝛼𝐿

𝑅 [1 − exp (−

𝑅

𝐿𝑡)].

This can be rewritten in terms of the external flux as follows

Φ(𝑡) = Φ0(𝑡) {1 −𝐿

𝑅𝑡[1 − exp (−

𝑅

𝐿𝑡)]}

That should be compared with the approximate Eq.(1). The limits of this expression are

Φ(𝑡) ≅ Φ0(𝑡)𝑅𝑡

2𝐿 (short times), Φ(𝑡) ≅ Φ0(𝑡) (long times).

This is almost the same as Eq.(2), except for the factor 2 in the denominator. Thus, the

approximate approach in the first part of this treatise was not that bad. Note that at short

times Φ(𝑡) increases quadratically in time, whereas Φ0(𝑡) increases linearly in time. In the

superconducting limit R = 0, one has = 0.

26. Energy density in the solenoid

Express the energy density (energy per volume) in a solenoid via the magnetic field

induction B in the solenoid.

Solution. We start with the formulas for the magnetic field in the solenoid

𝐵 = 𝜇0

𝑁

𝑙𝐼

and the energy of the solenoid

𝐸 =1

2𝐿𝐼2,

where the self-inductance is given by

𝐿 =𝜇0𝑁

2𝑆

𝑙

Expressing I via B as

𝐼 =𝐵𝑙

𝜇0𝑁

and substituting this into the formula for the energy, one obtains

𝐸 =1

2

𝜇0𝑁2𝑆

𝑙(

𝐵𝑙

𝜇0𝑁)2

=𝐵2

2𝜇0𝑆𝑙.

Thus, the energy density is

𝑒 ≡𝐸

𝑆𝑙=

𝐵2

2𝜇0.

This is a quite general result for the energy due to the magnetic field in space, not limited to

the solenoid.

27. The LC-circuit

A circuit contains a capacitor C and a coil with the inductance L. In the initial state, the

capacitor has the charge Q0 and the circuit is disconnected by a switch. Then the switch

connects the circuit. What happens after that? Find the maximal current in the circuit. Find

the time dependence of the current at short times.

Solution. In the initial state, the energy of the charged capacitor is

𝐸𝐶 =𝑄2

2𝐶.

After the circuit is connected, the capacitor gradually discharges through the coil and the

energy stored in the capacitor migrates into the coil. The energy of the coil is given by

𝐸𝐿 =𝐿𝐼2

2.

As there is no resistor in the circuit, the total energy is conserved:

𝐸 = 𝐸𝐶 + 𝐸𝐿 = const.

At the moment of time when the capacitor is completely discharged, Q = 0, the whole

energy is in the coil. The maximal current I0 in the coil can be found from the relation

𝐿𝐼02

2=

𝑄02

2𝐶.

(the energy in the final state is equal to the energy in the initial state) that yields

𝐼0 =𝑄0

√𝐿𝐶.

The physical process in this circuit is the oscillation of the charge and current with the

period

𝑇𝐿𝐶 = 2𝜋√𝐿𝐶.

The time dependence of the charge and the current has the form

𝑄(𝑡) = 𝑄0 cos (2𝜋𝑡

𝑇𝐿𝐶) , 𝐼(𝑡) = 𝐼0 sin (

2𝜋𝑡

𝑇𝐿𝐶).

The time dependence of the current at short times can be found from the condition

𝑉 = E

that means that the EMF produced by the coil balances the voltage on the capacitor,

otherwise in the absence of the resistance the current were infinite. This one obtains

𝑉 =𝑄

𝐶= 𝐿

∆𝐼

∆𝑡.

As in the initial state, t = 0, one has I = 0, one has

∆𝐼

∆𝑡=

𝐼

𝑡

at short times. This yields

𝐼(𝑡) ≅𝑄0

𝐿𝐶𝑡

at short times. The general formula for the current above yields the same result at short

times:

𝐼(𝑡) = 𝐼0 sin (2𝜋𝑡

𝑇𝐿𝐶) =

𝑄0

√𝐿𝐶sin (

𝑡

√𝐿𝐶) ≅

𝑄0

√𝐿𝐶

𝑡

√𝐿𝐶=

𝑄0

𝐿𝐶𝑡.

Here we have used the limiting form

sin 𝑥 ≅ 𝑥 for 𝑥 ≪ 1 (in radians).

Geometrical optics

28. Concave spherical mirror: basic case

A 1.5-cm-high diamond ring is placed 20 cm from a concave mirror with radius of curvature

30 cm. Determine (a) the position of the image, and (b) its size. (Example 23-2 in the

Giancoli book).

Solution: To connect to formulas needed to solve the problem, we first formulate the latter

as follows. Given: ho = 1.5 cm, do = 20 cm, R = 30 cm. Find: (a) image distance di; (b) image

height hi. We use several basic formulas: (1) the focal distance

𝑓 = 𝑅/2

for the spherical mirror, that gives f = 15 cm, (2) the mirror-lens equation

1

𝑑𝑜+

1

𝑑𝑖=

1

𝑓, (2)

and the (3) magnification formula

𝑚 =ℎ𝑖

ℎ𝑜= −

𝑑𝑖

𝑑𝑜. (3)

Solving for di, one obtains the answer to (a):

𝑑𝑖 =1

1𝑓

−1𝑑𝑜

=𝑑𝑜𝑓

𝑑𝑜 − 𝑓. (4)

One can see that if do > f, the image distance is positive, that is, the image is in front of the

mirror, as the object, that is, it is real. This is the basic case for the concave mirror.

Substituting numbers, one obtains

𝑑𝑖 =20 × 15

20 − 15= 60 𝑐𝑚.

Now, from (3) one obtains the answer to (b)

ℎ𝑖 = −ℎ𝑜

𝑑𝑖

𝑑𝑜= −

ℎ𝑜𝑓

𝑑𝑜 − 𝑓=

ℎ𝑜𝑓

𝑓 − 𝑑𝑜. (5)

One can see that if do > f, the image height is negative, that is, the image is upside-down.

Substituting numbers, one obtains

ℎ𝑖 = −1.5 × 15

20 − 15= −4.5 𝑐𝑚.

The magnification is given by the formula

𝑚 = −𝑓

𝑑𝑜 − 𝑓=

𝑓

𝑓 − 𝑑𝑜. (6)

In general, for the concave mirror m can be both positive and negative, and, similarly to di, it

jumps between positive and negative infinity at do = f. The magnitude of m is always greater

than 1 for the concave mirror:

|𝑚| =𝑓

|𝑓 − 𝑑𝑜|> 1.

Here, its numerical value is m =3.

This algebra-based solution is much better than the cumbersome solution in the book that

uses intermediate numerical calculations and does not allow analyzing the behavior of the

solution as parameters are changed.

29. Concave spherical mirror: virtual image

A 1-cm-high object is placed 10 cm from the concave mirror whose radius of curvature is 30

cm. Determine the position of the image and the magnification analytically (Example 23-3 in

the Giancoli book).

Solution: To connect to formulas needed to solve the problem, we first formulate the latter

as follows. Given: ho = 1 cm, do = 10 cm, R = 30 cm, thus f = R/2 = 15 cm. Find: (a) image

distance di; (b) magnification. We use the same basic formulas as in the preceding problem

that lead to the same analytical results. The answer to (a) is given by (4). Since, here, do < f,

one has di < 0, that is, the image is virtual and positioned behind the mirror. Substituting

numbers to (4), one obtains

𝑑𝑖 =𝑑𝑜𝑓

𝑑𝑜 − 𝑓=

10 × 15

10 − 15= −30 𝑐𝑚.

Magnification m is given by (6) of the preceding problem. Substituting numbers to (6), one

obtains

𝑚 =𝑓

𝑓 − 𝑑𝑜=

15

15 − 10= 3. (6)

As the magnification is positive, the image is upright.

Again, this solution is better than the solution in the book since it is indeed analytical.

30. Convex mirror

An external rearview car mirror is convex with a radius of curvature of 16 m. Determine the

location of the image and its magnification for an object 10 m from the mirror.

Solution: To connect to formulas needed to solve the problem, we first formulate the latter

as follows. Given: R = 16 m, thus f = R/2 = 8 m, do = 10 m. Find: (a) image distance di; (b)

magnification. We use the same basic formulas as in the preceding problem that lead to the

same analytical results. However, since for the convex mirror the focal distance is negative,

it is convenient to write

𝑓 = −|𝑓|

to make the negativity explicit. After that, the formulas obtained above become

𝑑𝑖 =𝑑𝑜𝑓

𝑑𝑜 − 𝑓= −

𝑑𝑜|𝑓|

𝑑𝑜 + |𝑓|.

Now one can see better that for the convex mirror the image distance is always negative,

whatever is the object distance. For the magnification, similarly one obtains

𝑚 = −𝑓

𝑑𝑜 − 𝑓=

|𝑓|

𝑑𝑜 + |𝑓|. (6)

Thus, for the convex mirror the magnification is always positive (the image is upright) and

smaller than one. Mathematically, one can express it as 0 < m < 1.

Here, substituting numbers, one obtains

𝑑𝑖 = −𝑑𝑜|𝑓|

𝑑𝑜 + |𝑓|= −

10 × 8

10 + 8= −4.44 𝑚

and

𝑚 =|𝑓|

𝑑𝑜 + |𝑓|=

8

10 + 8= 0.444. (6)

31. Refraction in a glass slab

A light ray is strikes a slab of glass, n = 1.5, of a thickness h = 1 cm, at the incidence angle =

45 degrees. find (a) the direction of the ray exiting the glass and (b) lateral displacement of

the ray.

Solution: (a) Combining the Snell’s law for the top and bottom interfaces,

sin 𝜃 = 𝑛 sin 𝜃′ , 𝑛 sin 𝜃′ = sin 𝜃′′,

'

''

h l

x

A

B C

D '

respectively, one obtains 𝜃 = 𝜃′′, that is, the direction of the light ray after exiting the glass

is the same as the direction of the incident ray.

(b) The lateral displacement of the ray is x indicated in the figure. From the triangle ACD one

obtains

𝑥 = 𝑙 sin 𝜙,

where 𝜙 = 𝜃 − 𝜃′. In turn, l cam be found from the triangle ABC as

𝑙 =ℎ

cos 𝜃′.

Combining these formulas, one obtains

𝑥 = ℎsin(𝜃 − 𝜃′)

cos 𝜃′= ℎ

sin (𝜃 − arcsinsin 𝜃

𝑛 )

cos arcsinsin 𝜃

𝑛

,

where 𝜃′ has been found from the Snell’s law. Substituting numbers, one obtains x = 0.329

cm. The analytical result above is sufficient, however, it contains a lot of trigonometric

functions and can be simplified. Using

cos 𝜃′ = √1 − (sin 𝜃′)2 = √1 − (sin 𝜃

𝑛)

2

and

sin(𝜃 − 𝜃′) = sin 𝜃 cos 𝜃′ − cos 𝜃 sin 𝜃′ = sin 𝜃 √1 − (sin 𝜃

𝑛)2

− cos 𝜃sin 𝜃

𝑛,

one obtains

𝑥 = ℎsin 𝜃 √1 − (

sin 𝜃𝑛 )

2

− cos 𝜃sin 𝜃

𝑛

√1 − (sin 𝜃

𝑛 )2

= ℎ

[

sin 𝜃 − cos 𝜃

sin 𝜃𝑛

√1 − (sin 𝜃

𝑛 )2

]

,

and, finally,

𝑥 = ℎ sin 𝜃 [1 −cos 𝜃

√𝑛2 − (sin𝜃)2].

This formula is more elegant than the first one above and it fives the same numerical result

x = 0.329 cm.

32. Refraction in two transparent slabs

Two parallel-sided blocks of glass of refraction indices n1 = 1.4 and n2 = 1.7 and the

thicknesses d1 = 10 cm and d1 = 15 cm lie next to each other as shown in the figure. A

light ray strikes the first one at an incident angle of 40◦.

(a) Calculate the angle of refraction θ as the ray emerges from the second block to air.

(b) Calculate the distance AB, where B is the point of exit.

Solution:

First, we enhance the original drawing adding the light ray inside the system and necessary

notations.

d1 d2

0

1 1

2 3

h1 h2

(a) As all surfaces are parallel, the light ray will exit the system in the same direction as it

entered the system, the only effect being the lateral displacement of the ray. This means

= 0. This has to be explicitly demonstrated in (a). To do this, we apply the Snell’s law for all

three intercaces:

sin 𝜃0 = 𝑛1 sin 𝜃1 , 𝑛1 sin 𝜃1 = 𝑛2 sin 𝜃2 , 𝑛2 sin 𝜃2 =sin 𝜃3.

These three equations can be compacted into one:

sin 𝜃0 = sin 𝜃3

that implies = 0.

(b) The distance AB is the sum of the vertical distances in each of the two layers:

𝐴𝐵 = ℎ1 + ℎ2.

From the triangles one obtains

ℎ1 = 𝑑1 tan 𝜃1 , ℎ2 = 𝑑2 tan 𝜃2,

where the angles can be found from the Snell’s law above:

𝜃1 = arcsinsin 𝜃0

𝑛1, 𝜃2 = arcsin

sin 𝜃0

𝑛2.

The final result is

𝐴𝐵 = 𝑑1 tan arcsinsin 𝜃0

𝑛1+ 𝑑2 tan arcsin

sin 𝜃0

𝑛2.

One can produce a smarter solution expressing tan via sin with the help of the formula

tan 𝜃 =sin 𝜃

cos 𝜃=

sin 𝜃

√1 − (sin𝜃)2=

1

√1

(sin 𝜃)2 − 1

.

Finding sin1 and sin2 from the Snell’s law, one obtains

𝐴𝐵 =𝑑1

√(𝑛1

sin 𝜃0)2

− 1

+𝑑2

√(𝑛2

sin 𝜃0)2

− 1

.

or

𝐴𝐵 =𝑑1 sin 𝜃0

√𝑛12 − (sin 𝜃0)2

+𝑑2 sin 𝜃0

√𝑛22 − (sin 𝜃0)2

.

All three analytical expressions for AB are acceptable.

Substituting numbers, one obtains sin 𝜃 = sin 40° = 0.6428 and

𝐴𝐵 =10

√(1.4

0.6428)2

− 1

+15

√(1.7

0.6428)2

− 1

= 11.3 𝑐𝑚.

33. Refraction in a prism (advanced)

A light ray strikes a glass prism with the apex angle at the incidence angle 1, as shown in

the figure. (a) Find the direction-change angle . (b) Find the value of 1 in the symmetric

case, 1 = 4. Express the refraction index n via this incidence angle. Find d in the symmetric

case and express n via this . (c) Consider the thin-prism limit (small ).

Solution: (a) Refraction at the first and second interfaces obeys the Snell’s law:

sin 𝜃1 = 𝑛 sin 𝜃2 , 𝑛 sin 𝜃3 = sin 𝜃4.

In the triangle formed by the entry point, exit point, and the apex, the sum of all angles is

180°. Introducing the complementary angles between the ray and interfaces, one can write

180° = 𝛼 + 𝜃2′ + 𝜃3

′ .

Switching to the angles with respect to the normal to the interface, shown in the figure, one

can rewrite it as

180° = 𝛼 + 90° − 𝜃2 + 90° − 𝜃3,

wherefrom

𝜃3 = 𝛼 − 𝜃2. (1)

Thus, one can find 𝜃2 from the Snell’s law, then find 𝜃3 from the relation above, then find 𝜃4

from the Snell’s law. After that, to find the direction change angle , one can notice that

there are two direction changes for the light ray crossing the two interfaces. These

direction-change angles are

𝛿12 = 𝜃1 − 𝜃2, 𝛿34 = 𝜃4 − 𝜃3.

The total direction change is

𝛿 = 𝛿12 + 𝛿34 = 𝜃1 − 𝜃2 − 𝜃3 + 𝜃4 = 𝜃1 − 𝛼 + 𝜃4, (2)

where (1) was used. Now, we have to perform the whole chain of calculations to find 𝜃4, in

this formula. One obtains

𝜃2 = arcsinsin 𝜃1

𝑛, 𝜃3 = 𝛼 − 𝜃2 = 𝛼 − arcsin

sin 𝜃1

𝑛

and

𝜃4 = arcsin(𝑛 sin 𝜃3) = arcsin (𝑛 sin (𝛼 − arcsinsin 𝜃1

𝑛)). (3)

The final result is

𝛿 = 𝜃1 − 𝛼 + arcsin (𝑛 sin (𝛼 − arcsinsin 𝜃1

𝑛)). (4)

For = 0, one has

𝛿 = 𝜃1 + arcsin (𝑛 sin (−arcsinsin 𝜃1

𝑛)) = 𝜃1 − arcsin (𝑛

sin 𝜃1

𝑛) = 𝜃1 − 𝜃1 = 0,

as it should be (a glass slab does not change the direction of the light ray).

(b) In the symmetric case, 𝜃4 given by (3) is equal to 𝜃1. Taking the sine of (3) with this

substitution and dividing by n, one obtains

sin 𝜃1

𝑛= sin (𝛼 − arcsin

sin 𝜃1

𝑛).

Now taking the arcsine of this, one obtains

arcsinsin 𝜃1

𝑛= 𝛼 − arcsin

sin 𝜃1

𝑛,

wherefrom

arcsinsin 𝜃1

𝑛=

𝛼

2

and

sin 𝜃1

𝑛= sin

𝛼

2. (4)

From this equation, one can find 𝜃1 or express n via 𝜃1.

In the symmetric case, from (2) one obtains

𝜃1 =𝛼 + 𝛿

2.

Then, with the help of (4), one obtains

𝑛 =sin

𝛼 + 𝛿2

sin𝛼2

.

This formula is used in the Refraction experiment in our PHY167 lab course.

(c) We have seen that 𝛿 = 0 for 𝛼 = 0. Thus for small the result should simplify. To work

it out, we use the trigonometric formula

sin(𝜃 − 𝜃′) = sin 𝜃 cos 𝜃′ − cos 𝜃 sin 𝜃′

and the small-argument approximation

sin 𝜃 ≅ tan 𝜃 ≅ 𝜃 (in radians), cos 𝜃 ≅ 1

to simplify in (4)

sin (𝛼 − arcsinsin 𝜃1

𝑛) = sin 𝛼 cos arcsin

sin 𝜃1

𝑛− cos 𝛼 sin arcsin

sin 𝜃1

𝑛

≅ 𝛼 cos arcsinsin 𝜃1

𝑛−

sin 𝜃1

𝑛= 𝛼√1 − (

sin 𝜃1

𝑛)

2

−sin 𝜃1

𝑛

so that (4) becomes

𝛿 = 𝜃1 − 𝛼 + arcsin (𝑛𝛼√1 − (sin 𝜃1

𝑛)2

− sin 𝜃1)

= 𝜃1 − 𝛼 − arcsin(sin 𝜃1 − 𝑛𝛼√1 − (sin 𝜃1

𝑛)

2

).

Here, it seems that there is no way except of using calculus to simplify the arcsin at small .

Using

arcsin(𝑥 + 𝛿𝑥) ≅ arcsin 𝑥 +𝛿𝑥

√1 − 𝑥2, 𝛿𝑥 ≪ 1,

one obtains

𝛿 = 𝜃1 − 𝛼 − arcsin sin 𝜃1 +𝑛𝛼√1 − (

sin 𝜃1

𝑛 )2

√1 − (sin 𝜃1)2

that simplifies to

𝛿 = 𝛼 (√𝑛2 − (sin 𝜃1)2

cos 𝜃1− 1).

This expression vanishes in the limits 𝛼 = 0 (a glass slab) and 𝑛 = 0 (no glass).

If the light strikes the prism perpendicularly, 𝜃1 = 0, then the result simplifies to

𝛿 = 𝛼(𝑛 − 1). (5)

In this case, obviously, the light ray changes its direction only at the second interface. Here,

𝜃3 = 𝛼, 𝜃4 = arcsin(𝑛 sin 𝜃3) = arcsin(𝑛 sin 𝛼)

and

𝛿 = 𝛿34 = 𝜃4 − 𝜃3 = arcsin(𝑛 sin 𝛼) − 𝛼.

Using the small-argument formulas

sin 𝜃 ≅ 𝜃, arcsin 𝑥 ≅ 𝑥,

one can simplify this to (5).