SINGLE LOOP CIRCUITS (Chapter 26)
Key concepts:
A Battery is a device that maintains a potential difference between its two poles.
The poles are denoted “-“ and “+”. The potential at the “-“ pole is smaller than the potential at the “+” pole.
The important characteristic of the battery is its electromotive force (EMF) denoted as E . If the battery is ideal, the potential difference V that is maintains is equal to E . For real batteries, there is an internal resistance usually denoted "r" that acts in series with the battery.
The simplest circuit consists of a battery and resistor in a single loop circuit:
The current within the circuit flows from the region of higher potential to the region of lower potential (see picture above), i.e. from
V b
to V a
.
Due to chemical reactions, the current inside the battery flows from “-“ to “+”, i.e from the region of lower potential (V a
) to the region of the higher potential (V b
). The electromotive force is directed from “-“ to “+” i.e. it “pushes the positive charge inside the battery in the direction of lower potential to the direction of higher potential.
Real batteries have some internal resistance, therefore the potential difference they maintain is different from
E : V=E-i*r, where r is the internal resistance of the battery and i
is the current through the battery.
Real batteries can be represented by two pieces: the ideal battery which maintains
V
= E and the resistor r
. The potential difference V that is available is less than the EMF by i*r, where i is the current through the battery.
If resistors are connected in serial, the sum of resistances gives the equivalent resistance: R eq
= R
1
+ R
2
+ R
3
+…

If resistors are connected in parallel the sum of reciprocal resistances gives the reciprocal equivalent resistance
R
-1 eq
= R
-1
1
+ R
-1
2
+ R
-1
3
+…
These relations are reversed from those describing capacitors in series and parallel.
The current through a single loop circuit is the same everywhere and can be found by applying the Kirchoff loop rule : The sum of all potential differences in a complete loop through the circuit should be equal zero.
The power P is the energy (spent or gained) per unit time. In the problems related to the circuits the energy can be supplied by the batteries, the energy can dissipate in resistors and bulbs, etc.
Power P can be found as a product of the current times voltage.
For example the energy rate (power) dissipated in the resistor is the product of the current inside the resistor times the voltage across the resistor: P=i*V
=i
2
*R=V
2
/R
, where we have used the relationship i=V/R.
For ideal batteries, the battery supplies the power to the circuit P=i*E.
Typical problems related to single loop circuits:
Problem 1. A circuit consists of the real battery (E=10V, r= 2
Ω
) and the resistor R= 3
Ω
.
Calculate current and the total power released in the circuit.
A.
1 A, 10 W
B.
0.5 A, 5 W
C.
2 A, 20 W
D.
3 A, 15 W
E.
5 A, 3 W
Solution.
First, walk through the loop and write the Kirchoff equation. This will give us the equation for the current. Since the battery arrow (from “-“ to “+”) is pointed up, and there is only one battery, the direction of the current is clockwise. Walking clockwise:
E-i*r-i*R=0 i=E/(r+R)=2A.

Now, the total power released in the circuit is the sum of the power lost in the resistor r and R. Since the current is known, the useful formula for power P=i*V= i
2
*R .
Power released in r: P r
= i
2
Power released in R: P
R
= i
*r=2*2*2=8 W
2
*R=2*2*3=12 W
The total power released in the circuit is therefore 12+8=20W.
Note that the same answer can be found quicker: this is the power which ideal piece of the battery supplies to the circuit:
P=i*E=2*10=20 V
Answer C.
Problem 2. A circuit consists of the ideal battery (E=18 V) and two kinds of resistors
R
1
=3
Ω
, R
2
=2
Ω
. Calculate current through the battery and the total power released in the circuit.
A.
3 A, 64W
B.
6 A, 108 W
C.
12 A, 216 W
D.
1 A, 32 W
E.
0.5 A, 108 W
Solution.
While formally it is a multi loop circuit, the problem is simple since we need to find the current through the battery.
Therefore, first, we simplify the circuit by calculating equivalent resistance. We have first three resistors in parallel, its resistance is 1/(1/3+1/3+1/3)=1
Ω.
Second, two resistors are in parallel, therefore the equivalent resistance is 1(1/2+1/2)=1
Ω.
The third three resistors are in parallel, their resistance is 1/(1/3+1/3+1/3)=1
Ω.
After, we have three serial connections with the total resistance 1+1+1=3
Ω.
The second step is to find the current through the battery since we have a simple single loop circuit with the battery and one equivalent resistor of 3
Ω.
E-i*R eq
=0 i=E/R eq
=18/3=6 A.
The total power released in the circuit is the sum of all powers released in all resistors.
This is the same as the power which battery supplies to the circuit. Therefore
P=i*V=6*18=108 W
Answer B.

Problem 3. A fragment of the circuit is shown on the Figure. What is the potential difference between the points B and A?
A.
1 V
B.
2 V
C.
–5 V
D.
4 V
E.
–3 V
Solution.
It is easy to walk in the direction of the current from the point A to the point B. Writing the Kirchoff equation:
V
A
+E
1
-R
1
*i-E
2
-R
2
*i+E
3
=V
B
E
1
-R
1
*i-E
2
-R
2
*i+E
3
=V
B
-V
A
=2-2*3-1-3*3+15=1 V
Answer A.
Problem 4.
A bulb is connected to the battery of some EMF equal to E. Will the bulb be brighter or dimmer if it will be reconnected to a new battery of E new
=2E. By how much?
A.
Twice brighter
B.
Twice dimmer
C.
The same
D.
Four times brighter
E.
Four times dimmer
Solution.
The brightness of the bulb is proportional to its power. Therefore we have to figure out what happens to the power released in the bulb if we change the voltage across it. If the bulb has some resistance R, and it is connected to some ideal battery E, its power
P=i*V=i*E since the voltage across the bulb is equal to the EMF of the battery. The current in the circuit i=E/R.
Therefore we can express the power of the bulb via E and R as follows:
P=i*V=i*E=E*E/R=E
2
/R.
This formula tells us if we increase the applied voltage by 2, the power released in the bulb will increase quadratically, i.e. by 4.
P new
=E
2 new
/R=(2*E)
2
/R=4P old
Answer D.

MULTI LOOP CIRCUITS (Chapter 26)
Key concepts:
Multi loop circuits of batteries and resistors: loops, branches and junctions should be distinguished.
A general problem is to find currents insides every branch if the resistances and
EMF’s of the batteries are known.
Three rules apply:
Branch rule: inside every branch the current is the same. The current may be different in different branches.
Junction rule:
the sum of all currents coming into the junction is equal to the sum of all currents leaving the junction.
Loop rule: the sum of all potential differences in any complete walk through any closed loop of the circuit is equal to zero.
Strategy for solving multi-loop circuit problems:
1. Find all branches and enumerate currents.
2. Find all junctions and establish the relationships between the currents
3. Apply Kirchoff loop rules to the loops in the circuit.
The total number of equations to be written should be equal to the total number of currents, which are unknown.
Typical problems related to multi loop circuits:
Problem 5.
What is the current inside the resistor R
1
? What is the power released in the resistor R
1
?
A.
1A, 27 W
B.
2A, 13 W
C.
3 A, 27 W
D.
4 A, 12 W
E.
5 A, 5 W
Solution. While this is a multi loop circuit, the answer to that particular question can be found very easily. We see that the first loop contains the battery E
1
and only this resistor. Let us we write the Kirchoff loop rule for this particular loop. The

current is assumed to be directed down inside R
1
, therefore we walk through the loop clockwise: E
1
-i
1
*R
1
=0
From which we figure out that i
1
=E
1
/R
1
=9/3=3 A.
The power released in the resistor (energy rate dissipated in the resistor) is
P=i
1
*V=i
1
*i
1
*R=i
1
2
*R=3*3*3=27 W.
This trick works when you can find such a loop in the multi loop circuit which contains only one resistor and (possibly many) batteries. Then you can apply the Kirchoof loop rule to this particular loop and find the current inside the resistor immediately!
Answer C.
Problem 6. Order branches of bulbs by brightness, dimmest first:
A.
I, II, III
B.
II, III, I
C.
III, I, II
D.
III, II, I
E.
II, I, III
Solution. The brightness of each branch is proportional to the total power released in the bulbs of the branch. Assume that the resistance of the bulbs is the same and it is equal to
R.
Then the total power of branch I is: P
1
= i
1
*V
1
=( i
1
) the total power of branch II is: P
2
= ( i
2
)
2
*R + ( i
2
)
2
2
*R,
*R =
2*( i
2
)
2
*R, and the total power of branch III is: P
3
= ( i
3
)
2
*R + ( i
3
( i
3
)
2
*R =3*( i
3
)
2
*R.
)
2
*R +
Note that the power is expressed via the current and resistance here.
What about the currents i
1
, i
2
, i
3
?
We can find currents easily by applying the Kirchoff loop equations to the loops which accounts for the battery and the particular branch.
So, that for the branch I and the battery the equation is
E- i
1
*R=0
For the branch II and the battery, the equation is
E- i
2
*R- i
2
*R =0
For the branch III and the battery the equation is
E- i
3
*R- i
3
*R- i
3
*R =0
Therefore i
1
=E/R, i
2
=E/(2R), i
3
=E/(3R)

The resulting powers are
P
1
= ( i
1
)
2
*R=E
2
/R
P
2
= 2*( i
2
)
2
*R=2 E
2
/(4R)= E
2
/(2R)
P
3
=3*( i
3
)
2
*R=3E
2
/(9R)= E
2
/(3R)
We finally see that the brightest branch is I (has largest power), then II, and finally III.
Therefore, the dimmest is III, then II, then I.
Answer D.
Problem 7.
In the previous problem (see figure to the problem 6), if E=12 V and R= 2
Ω
, find the power released in each bulb?
A. 8W for branch I, 18W for branch II, 72W for branch III
B. 18W for branch I, 8W for branch II, 72W for branch III
C. 8W for branch I, 72W for branch II, 18W for branch III
D. 72W for branch I, 18W for branch II, 8W for branch III
E. 72W for branch I, 8W for branch II, 18W for branch III
Solution.
See derivation to the problem 15. The power released in the bulb of the branch I:
( i
1
)
2
*R=E
2
/R=12*12/2=72 W
The power released in each bulb (total two bulbs) of branch II:
( i
2
)
2
*R= E
2
/(4R)= 12*12/4/2=18 W
The power released in each bulb (total three bulbs) of branch III:
( i
3
)
2
*R= E
2
/(9R)= 12*12/9/2=8 W
Answer D.
Problem 8.
In the circuit shown in the figure find the currents through each resistor if E
1
=10 V,
E
2
=5V and all resistors are equivalent
R
1
=R
2
=R
3
=R= 5
Ω
. with

A. i
1
= 1A down, i
2
=3A up, i
3
= 2A down
B. i
1
= 1A down, i
2
=1A up, i
3
= 0
C. i
1
= 2A down, i
2
=0, i
3
= 3A down
D. i
1
= 3A up, i
2
=2A up, i
3
= 1A down
E. i
1
= 1A up, i
2
=2A down, i
3
= 0
Solution.
We have a real multi loop circuit problem now. We have three branches corresponding to three resistors. Inside each branch there is its own current. Lets enumerate currents in each resistors to be i
1
, i
2
, i
3
Assume that the direction of the current i1 in the resistor R1 is down, the direction of the current i2 in the resistor R2 is up and the direction of the current in the resistor i3 is down again.
From the junction rule we obtain (i) i2=i1+i3
From two inner loops we obtain
(ii) E1-i1*R-i2*R=0
(iii) E2-i3*R-i2*R=0
We have total three equations. Summing equations (ii) and (iii) and using Eq. (i):
E1+E2-(i1+i3)*R-2*i2*R= E1+E2-3*R*i2= 0
Therefore i2=(E1+E2)/(3R)=15/15=1A (sign is +, therefore direction chosen is OK, i.e up)
From equation (ii): i1=(E1-i2*R)/R=E1/R-i2=10/5-1=1A (sign is +, therefore direction of the current i2 is
OK, i.e. down.)
From equation (i): i3=i2-i1=1-1=0A (no current!)
We finally have i1=1A down, i2=1A up, i3=0
Answer B.

RC CIRCUITS (Chapter 26)
The RC circuit consists of the capacitor C, the resistor R, the battery E and the switch S.
When the switch S is closed in RC circuit, capacitor is charged. Application of the Kirchoff loop equation for time moment t to RC circuit leads to differential equation. The solution of the equation reads as follows:
(1.1) ( )
=
EC (1
− e
− t RC
) is the charge on the capacitor as a function of time. Current in the RC circuit:
(1.2)
= dq
=
E dt R e
− t RC =
E
R e
− t /
τ
C where time constant
= shows the characteristic time for the charging
C
RC process. The voltage across the capacitor can be found
(1.3)
= =
E (1
− e
− t RC
)
C
In the reverse situation, when the switch is opened, the capacitor with charge q
0 gets discharged. The functions for the charge, current, and voltage are the following:
(1.4) ( )
= q e
0
−
(1.5)
= dq
= − dt q
0
RC e
−
(1.6)
=
( )
= q
C C
0 e
− t RC
Problem 9.
What is the voltage across the capacitor in the RC circuit with R= 3 Ohm ,
C= 5 pF and E=9V just after the switch is closed.
A.
0 V
B.
1 V
C.
3 V
D.
5 V
E.
9 V

Solution. After the switch is closed the capacitor begins to charge. The charge on the plates as a function of time is given by
( )
=
EC (1
− e
− t RC
)
The voltage across the capacitor is thus
= =
E (1
− e
− t RC
)
C
The time constant for the circuit RC is 5pF*3 Ohms = 15 picoseconds. Even though this is a very short time, it is not zero. Therefore, just after the switch is closed, t = 0 in the expression above makes the quantity in parentheses go to zero). Thus, the voltage
V(t=0) = 0.
Answer A.
Problem 10.
How long does it take for an RC circuit with R= 3 Ohm , C= 5 pF and
E=9V to charge the capacitor to 45 pC.
A.
1 second
B.
1 hour
C.
1 day
D.
1 year
E.
infinity
Solution. After the switch is closed the capacitor is beginning to charge.
The charge on the plates as a function of time is:
( )
=
EC (1
− e
− t RC
)
We solve this for t with q known. Note that q is the limiting value = EC. t
= −
RC ln |1
−
/ |
==
3* 5*10
−
12 − =
Answer E

MAGNETIC FIELDS (Chapter 27)
PLEASE NOTE THAT IN ALL FORMULAE THE VECTOR PRODUCT (CROSS
PRODUCT) IS DENOTED WITH THE SQUARE BRACKETS: c=[ab]
Key concepts:
A moving charged particle creates a magnetic field around it. Do not confuse magnetic fields with electric fields: the electric fields are associated with the charges; the magnetic fields are associated with the moving charges. Therefore, a moving charged particle creates both an electric field and a magnetic field.
If a moving charged particle is placed inside a magnetic field it experiences magnetic force. This force is proportional to the charge of the particle, its velocity and the strength of the magnetic field. Again, it is essential that the particle is moving: the faster the particle moves, the greater the force it feels from the external magnetic field.
When a particle of charge q moves with velocity v
inside a magnetic field
B
it feels a magnetic force F given by the formula:
F
=q*[ vB
] which was found as the result of experimental measurement. A cross product
[ vB
] is a vector which is perpendicular to the plane made by vectors v
and
B
.
The direction of the cross product [ vB
] is fixed by the right hand rule: rotate vector v towards vector B using your right hand (use the smallest angle between them). The orientation of your right hand shows the direction of the cross product.
The direction of the magnetic force F can be different from the vector product
[ vB
] since the charge q may be negative. In that case the direction of the force
F is opposite to the direction of the vector product since F =-|q|*[ vB ].
Typical problems:
Problem 1.
Find the value and the direction of the force acting on a positively charged particle with q=2
µ
C moving with velocity v=5 km/s along x direction which is placed inside constant magnetic field B=1 kT oriented along y direction:
A.
10 mN, -z direction
B.
10 mN, +z direction

C.
10 N, -z direction
D.
10 N, +z direction
E.
1 kN, +x direction
Solution. The vector product
[ vB
]
is oriented along +z direction according to the right hand rule. Since the charge is positive the direction of the force is also lying along +z direction. The value for the force is (angle
φ between v and B is 90 degrees)
F=q*v*B* sin
φ
=q*v*B=2*10
-6
*5*10
3
*1*10
3
=10 N
Answer D
Problem 2. Find the value and the direction of the force acting on a positively charged particle with q=2 mC moving with v=5 mm/s along x direction which is placed inside constant magnetic field
B=1.41 kT oriented 45 degrees in xy plane
A.
10 mN, -z direction
B.
10 mN, + z direction
C.
10 N, - z direction
D.
10 N, +z direction
E.
1 kN, +x direction
Solution. The vector product [ vB ] is oriented along +z direction according to the right hand rule. Since the charge is positive the direction of the force is also lying along +z direction. The value for the force is (angle
φ between v and B is 45 degrees)
F=q*v*B*sin
φ
=q*v*B sin45=2*10
-3
*5*10
-3
*1.41*10
3
/1.41=10 mN
Answer B.
Problem 3. Find the value and the direction of the force acting on a negatively charged article with q=-0.5 C moving with v=10 m/s along x direction which is placed inside constant magnetic field
B =(-10,20,0) T
A.
10 N, -z axis
B.
10 N, +z axis
C.
100 N, -z axis
D.
100 N, +z axis
E.
1 N, x-axis

Solution. Since vector B has z-component equal zero, it is completely oriented within xy plane. Therefore, the vector product [ vB ] is oriented along +z direction according to the right-hand rule. Since the charge is negative the direction of the force is lying along -z direction. Let us find the direction and the value for the force using vector algebra.
First, let us represent the vector B in the form: B =-10* i +20* j , where i is a unit vector along x axis and j is a unit vector along y-axis. Vector v =10* i is lying completely along x axis. The vector product is:
[ vB ]= [ 10* i*( -10* i +20* j) ]=-10*10*[ ii ]+10*20*[ ij ]
Since vector product [ ii ]=0 and [ ij ]= k , the unit vector along z-axis, the vector product becomes: [ vB ]=200* k .
Now the force is the vector product time the charge. Since the charge is negative:
F =q*[ vB ]=-0.5*200* k =-100* k, N, i.e 100 N oriented along vector – k which is –z axis.
Answer C.
Problem 4.
An electron moving with the constant speed in the direction from top to bottom enters the region of a constant magnetic field. The electron trajectory deviates to the left. What is the orientation of the magnetic field?
A.
top
B.
bottom
C.
left
D.
right
E.
inside the page
Solution. If electron deviates to the left, it experiences the force directed to the left. This is the magnetic force described as F =q*[ vB ]. We need to figure out such direction of B so that to be multiplied by vector v and taking account the sign of charge for the electron, we obtain the direction of the force to the left. To do this, we first note that if the force is directed to the left, the vector product [ vB ] should be directed to the right since electron is negatively charged particle (q<0). Second, if v is directed to the bottom, vector B should be directed inside the page, in this way vector product [ vB ] is directed to the right, and the force will be directed to the left.
Answer E.
Problem 5.
An electron is circulating within the plane of the page due to magnetic field directed inside the page. The radius of its circulating motion is 9 cm and its speed is 1.6 km/s. Find the value of magnetic field and the direction of the circulating motion,

A.
0.1
µ
T, clockwise
B.
0.1
µ
T, counterclockwise
C.
0.1 mT, clockwise
D.
0.1 mT, counterclockwise
E.
0.1 T, clockwise
Solution.
Circular motion assumes that there is force acting on the electron, which is directed towards the center of the orbit. This force is of magnetic origin, i.e.
F =e*[ vB ].
At the same time, the force associated with the circular motion is equal to m*w where centrifugal acceleration w= v
2
/R. Since velocity vector v is always perpendicular to magnetic field vector B , we can write F=e*v*B=m*v
2
/R from which
B=m*v/(R*e)=9*10
-31
*1.6*10
3
/(9*10
-2
*1.6*10
-19
)=10
-7
T.
To understand whether the electron circulates clockwise or counterclockwise, we need to look at the electron at some of its position. Assume that at some moment t it is positioned at 12pm (noon, the topmost part of its trajectory). If its velocity is directed to the right, it is going clockwise, if its velocity is directed to the left, it is going counterclockwise, At the topmost point of its trajectory the electron should experience the force directed to the bottom since it is going around the circle, and the force of the circular motion is always directed to the center. Therefore F =e*[ vB ] is directed to the bottom, and we know that e<0 and B is directed inside the page. The vector product [ vB ] should be directed to the top since F is directed to the bottom, and e is less than zero. If vector product is directed to the top and B is directed inside the page, the velocity vector v should be directed to the right. In this way, vector product [ vB ] will be directed to the top, and multiplied by negative charge, the force will be directed to the bottom. We therefore found that the velocity in topmost point is directed to the right. Therefore, the electron is going around the circle clockwise. We can fix any other of its position (say leftmost) and come to the same conclusion.
Answer A.

Key concepts:
In general, the magnetic field due to a moving charged particle is given by: where constant
µ
0
B ( r )=
µ
0
/4
π
*q*[ vr ]/r
3
/4
π=10 -7
. Note that the direction of the magnetic field is tricky: if the particle is moving from bottom to top, and a point where we look at the field is on the right, the vector product [ vr ] is directed inside the page. If the charge of the particle >0, the magnetic field will be pointed inside the page at this point r. If the charge < 0 the magnetic field is pointed outside the page at this point r. All this information is encoded into the above formula.
Since current inside the wire is just an array of moving charged particles, every wire carrying current creates magnetic field around it. To find the total field at some point r we need to sum up all the contributions to this field from all moving charges inside the wire.
In general, there is an integral, which gives the answer. For simple configurations like straight wire or circular wire the answers are:
1.
Infinite straight wire carrying current creates magnetic field which varies with the distance d from the wire as follows: B(d)=
µ
0
*i/(2
π
d), where i is the value of the current. The direction of the field can be found using the right hand rule: point your large finger along with the current, the four other fingers show the direction of the magnetic field (See figure in the textbook).
2.
A coil or a circular wire creates magnetic field at the center of the circle: B=
µ
0
*i
/(2R). The direction of the field can be found from right hand rule: rotate by your right hand in the direction of the current simulating circular motion, the orientation of the right hand will show the direction of the magnetic field within the plane inside the circle. A general arc defined by angle ϕ
creates magnetic field B=
µ
0
*I* ϕ
/(4
π
R) which for ϕ
=2
π
transforms to the expression above.
3.
A solenoid, which is a cylindrically turned wire with some (usually large) number of turns per unit length, creates non-zero constant magnetic field inside itself and zero field outside itself. The magnetic field inside the solenoid is given by:
B=
µ
0
*n*i, where n is the number of turns per unit length. The direction of the field inside the solenoid can be found by applying the right hand rule: rotate by your right hand in the direction of the current simulating circular motion, the orientation of the right hand shows the direction of the field inside the solenoid.
A wire with the current placed into magnetic field experiences a force since every moving charge inside the wire experiences the magnetic force. The total force on the wire with current i depends on the orientation of the wire and on the orientation of the magnetic field. If the wire is a straight piece of some length L, the force is given by F = i*
[ LB ], where vector L has length L an is directed towards the direction of the current.
Two pieces of wire, which brought together will repel from or attract to each other if the currents exist inside these wires. This can be simply understood since one wire will create

magnetic field around it, and the moving particles inside another wire will experience the magnetic force. For example, if two parallel wires carry currents in the same direction, they will attract to each other. If two parallel wires carry currents in different direction, they will repel from each other. In case of two parallel pieces of wire of length L separated by distance d which carry out the currents i1 and i2, the force between them is given by F=
µ
0
i1*i2*L/(2
π d).
Typical problems:
Problem 1.
Find the direction and the value for the force on the wire, which carries the current of 2 mA, if it is placed inside the magnetic field of 4 kT. The length of the wire is 3 mm, it is vertically oriented with the current flowing to the bottom. The magnetic field is horizontally oriented and is pointed to the left.
A.
24
Ν
, left
B.
24 m
Ν
, right
C.
24 kN, top
D.
24 N, bottom
E.
24 mN, inside the page
Solution.
The formula to use is: F = i*[ LB ], where L is a vector which points from top to bottom (vertical orientation with the current flowing to the bottom) and B points to the left. Both vectors are perpendicular to each other. The vector product [ LB ] points inside the page. We should not care about sign of i in this case, it is always positive, and direction of vector L takes care about direction of the current. The value of the force is
F=i*L*B=2*10
-3
*3*10
-3
*4*10
3
=24 mN.
Answer E.
Problem 2.
A wire carries a current from the right to the left. What is the orientation of the magnetic field above and below the wire?
A.
Above – inside the page, Below – outside the page
B.
Above – outside the page, Below – inside the page
C.
Above – left, Below – right
D.
Above – right, Below – left
E.
Above – bottom, Below – top

Solution.
Applying the right hand rule, orient your large finger to the left, the four other fingers point to the direction of the field above the wire, they point to inside the page. If you rotate by the right hand around the wire so that your large fingers are located below the wire, the point outside the page. This gives the answer to the problem.
Answer A.
Problem 3.
Find the value of magnetic field of the long straight wire at distance d=3 cm if it carries current of 3 mA.
A.
20 pT
B.
20 nT
C.
20
µ
T
D.
20 mT
E.
20 T
Solution.
The formula is B(d)=
µ
0
*i/(2
π
d), where all numbers are given by the problem.
B(d)=
µ
0
*i/(2
π
d)= 2
µ
0
*i/(4
π
d)=2*10
-7
*3*10
-3
/(3*10
-2
)=2*10
-8
=20 * 10
-9
T=20 nT
Answer B.
Problem 4.
Two straight wires of the length 5 m each and parallel to each other carry currents of 3 A and 5
A in opposite directions. If the distance between the wires is 1 cm what is the force between them?
A.
3 mN, repulsion
B.
3 mN, attraction
C.
1.5 mN, repulsion
D.
1.5 mN, attraction
E.
3 N, repulsion
Solution.
The formula to use is: F=
µ
0
i1*i2*L/(2
π d)=
2* 10
-7
*3*5*5/0.01=150*10
-5
N. Since the directions of the currents are opposite, it is repulsion.
Answer C.

Problem 5.
A circular wire of radius 6.28 cm is oriented within the page and carries a current of 2 A. The current flows clockwise. What are the value and the direction for the magnetic field created at its center?
A.
20
µ
T, outside the page
B.
20
µ
T, inside the page
C.
20 mT, to the left
D.
20 mT to the right
E.
20 T, to the top
Solution.
The formula to use is: B=
µ
0
*i /(2R)=4*3.14 * 10
-7
* 2 / (2 * 6.28 * 10
-2
)=
2*10
-5
T. Let us apply the right hand rule: to simulate the current flow by rotating the right hand, the orientation of the right hand should be inside the page.
Answer B.
Problem 6.
A solenoid is oriented perpendicular to the page, and carries a current in the counterclockwise direction. What are the value and the direction of the magnetic field inside it?
Assume that the current is 1 A, and the number of turns per centimeter is 25.
A.
3.14 mT, outside the page
B.
3.14 mT, inside the page
C.
3.14 T, to the left
D.
3.14 T to the right
E.
3.14 kT, to the top
Solution.
The formula to use is: B=
µ
0
*i*n=4*3.14*10
-7
*1*25*10
2
=3.14 mT. The direction is outside the page, since simulation of the current flow by rotating your right hand gives us the answer.
Answer A.

Problem 7.
Find the direction of the force on each side of rectangular frame.
A.
Top side - top, bottom side - bottom, left side - left, right side – right.
B.
Top side - bottom, bottom side - top, left side - right, right side – left.
C.
Top side - left, bottom side - right, left side - bottom, right side – top.
D.
Top side - right, bottom side - left, left side - bottom, right side – top.
E.
Top side - left, bottom side - right, left side - top, right side – bottom.
Solution.
Parallel wires attract to each other if the currents flow in the same direction.
They repel from each other if the currents flow in opposite direction. For perpendicular wires the situation is more complicated. It can be figured out by noting that the wire below itself creates a magnetic field directed inside the page. Since the current flow in the left side of the frame is on top, this is the flow of positively charged particles. They will experience the force F =q*[ vB ], pointed to the left (q>0, v is on top, B is inside the page).
Therefore the entire left part of the frame will experience the force pointed to the left.
Using similar argument, the entire right part experiences the force pointed to the right.
Answer A.
Problem 8.
Find the total force on the frame in the above problem if the current inside the wire is 30 A, and inside the frame is 20 A. The horizontal dimension of the frame is 30 cm and the vertical dimension is 7 cm. The top side of the frame is located at the distance 1 cm from the wire.
A.
3.15 N, top
B.
3.15 N, bottom
C.
3.15 mN, top
D.
3.15 mN, bottom
E.
3.15 kN, left
Solution. First note that the forces on left side and right side are the same and directed oppositely. Therefore they do not contribute to the total force. The force on the top side is directed to the top and it is given by: F top
=
µ
0
*i1*i2*L/(2
π d) with d=1cm. Evaluating it:
2*10
-7
*30*20*30*10
-2
/10
-2
=36*10
-4
N. The force on the bottom part is directed to the bottom and it is given by the same formula with d = 8 cm. Therefore, the force
F bottom
=
µ
0
*i1*i2*L/(2
π d)= 2*10
-7
*30*20*30*10
-2
/(8*10
-2
) = 4.5*10
-4
N. The total force is directed on top and it is 36*10
-4
-4.5*10
-4
=31.5*10
-4
N.= 3.15 mN.

Key concepts:
If we consider a frame in a changing magnetic a field, current will be induced inside the frame. This is the Faraday law of induction. To describe the law quantitatively we consider a flux of magnetic field thru the frame:
Φ
= B * A =B*A*cos
φ where angle
φ
is the angle between the magnetic field and the normal vector to the frame area A. In order to induce the current inside the frame, the flux thru the frame should change with the time.
This can either be done by considering magnetic field as a function of time, B(t), or by changing the area A, or by changing the relative orientation between vectors B and A , for example by rotating the frame with some angular speed so that angle
φ depends on time.
Under any of these condition the flux is a function of time,
Φ
(t) which is sufficient to generate the current inside the frame. The induced EMF inside the frame is given by the derivative of the magnetic flux with respect to time:
Ε
=-d
Φ
/dt. If we know the resistance of the frame wire, the current i induced is just i=
Ε
/R.
The tricky issue is the direction of the induced current. Consider the frame, which is lying within this page. Magnetic field is directed inside the page and changes with time. Will the current induced inside the frame flow clockwise or counterclockwise? To answer the question we distinguish the case when the magnetic field increases with time and when the magnetic field is decreases with time. If magnetic field increases with time, its time derivative points in the same direction as the field. For example if B (t)= a *t, d B /dt
= a=B /t , i.e. vectors d B /dt and B are parallel to each other. If magnetic field thru the frame decreases with time, its time derivative points in the opposite direction. For example if B (t)= a /t, d B /dt =a/ t
2
=-B/ t , i.e. vectors d B /dt and B are antiparallel to each other. Once understand that, we can formulate the rule (Lenz rule) which allows us to determine the direction of the induced current: The current is induced in such direction so that the magnetic field created by the induced current, B ind
, is pointed in the direction opposite to the direction of increase of B (t), i.e., opposite to the vector d B /dt.
Example 1. Consider the frame, which is lying within the page. Magnetic field is directed inside the page and increases with the time. What is the direction of the current induced?
Since B (t) increases, d B /dt is parallel to B , therefore d B /dt is directed inside the page.
From Lenz rule it follows that the field B ind
induced by the current inside the frame should be directed oppositely to d B /dt, i.e. B ind
is outside of the page. Therefore, the current should flow counterclockwise, since only this direction will create B ind
pointed outside the page.
Example 2. Consider the frame, which is lying within the page. Magnetic field is directed inside the page and decreases with time. What is the direction of the current induced?
Since B (t) decreases, d B /dt is antiparallel to B , therefore d B /dt is directed outside the page. From Lenz rule it follows that the field B ind
induced by the current inside the frame should be directed oppositely to d B /dt, i.e. B ind
is inside the page. Therefore, the current should flow clockwise, since only this direction will create B ind
pointed inside the page.

Therefore, we can formulate three step procedure to apply the Lenz rule: (i) transfer yourself into the coordinate system of the frame. Understand if the magnetic field increases or decreases with time and at which direction relative to the orientation of the frame, (ii) if magnetic field increases, its time derivative is parallel with the direction of the magnetic field; if magnetic field decreases, its time derivative is antiparallel with the direction of the magnetic field, (iii) choose the direction of the field induced by the current opposite to the time derivative. Once found this, choose the direction of the induced current to simulate the direction of the induced field.
Typical problems:
Problem 1.
Two circular coils are arranged as on the figure. There is a counterclockwise current in the bottom coil, which decreases with time. What is the direction of the induced current in the top coil?
A.
clockwise
B.
counterclockwise
C.
fluctuating
D.
no current
E.
none of the above
Solution.
Since the current in the bottom coil is counterclockwise, the right hand rule tells us that the magnetic field in the plane of the top coil is pointed along +z axis. Now since the current in the bottom coil decreases with time, the produced magnetic field is also decreasing function of time. It follows that the flux of this field thru the top coil decreases with time, which will induce the current. To understand the direction of the induced current, we need to figure out the direction of the time derivative of the magnetic field produced by the bottom coil. Since magnetic field decreases, its time derivative, d B /dt
,
is antiparallel with the direction of the field, i.e. it points along –z axis. The induced field should be opposite to d B /dt, as given by the Lenz rule, and it will be pointed along +z axis. To arrange such direction, the current induced in the top coil should be directed counterclockwise.
Answer B.
Problem 2.
The horizontal component of magnetic field of the Earth decreases with the distance from its center. Imagine we move a metallic frame from the surface of the Earth as shown on the figure. What is the direction of the induced current inside the frame?

A.
clockwise
B.
counterclockwise
C.
fluctuating
D.
no current
E.
none of the above
Solution.
Since the magnetic field is pointed into the frame (as shown on the figure) and when we raise the frame the magnetic field decreases, the magnetic flux thru the frame decreases with the time and this will induce the current. Let us transfer ourselves into the coordinate system associated with the frame. Then, there is a magnetic field, which passes thru the frame and decreases with the time. The time derivative d B /dt is directed oppositely to B in this case, if B is directed inside the page, the time derivative d B /dt is directed outside. The current will be induced in such a way that the induced magnetic field is opposite to d B /dt. Since d B /dt is pointed outside, the B ind
is pointed inside and for this to happen the current should flow clockwise.
Answer A.
Problem 3. A coil of the radius 1 m is placed into the magnetic field. The coil is lying within the page, and the magnetic field is pointed outside the page. If the magnetic field is changing as B(t)=5*t Tesla, what is the induced EMF inside the coil and what is the direction of the current.
A.
15.7 V counterclockwise
B.
15.7 V clockwise
C.
5 V counterclockwise
D.
5 V clockwise
E.
no EMF, no current.
Solution.
Flux thru the coil is:
Φ(
t
)
=B(t)*A=5* t *
π
*R
2
. The induced EMF is the time derivative of the flux and it is equal to: E=5*
π
*R
2
=5*3.14*1=15.7 V. To find the direction, first note that we have a field, which increases with time. Therefore its time derivative d B /dt is pointed along with the field. Since the field is pointed outside the page, so does d B /dt. According to the Lenz rule, the field induced by the current should be pointed opposite to d B /dt, i.e. inside the page. For this to happen, the coil should carry clockwise current.
Answer B