Notes and Solved Problems for Common Exam 3
(Does not include Induction)
8. MULTI LOOP CIRCUITS
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.
The strategy for solving multi-loop circuit problems is the following:
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 R1? What is the power
released in the resistor R1?
A.
B.
C.
D.
E.
1A, 27 W
2A, 13 W
3 A, 27 W
4 A, 12 W
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 E1 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 R1, therefore
we walk through the loop clockwise: E1-i1*R1=0
From which we figure out that i1=E1/R1=9/3=3 A.
The power released in the resistor (energy rate dissipated in the resistor) is
P=i1*V=i1*i1*R=i12*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.
B.
C.
D.
E.
I, II, III
II, III, I
III, I, II
III, II, I
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: P1= i1*V1=( i1 )2*R,
the total power of branch II is: P2= ( i2)2*R + ( i2)2*R =
2*( i2)2*R,
and the total power of branch III is: P3= ( i3)2*R + ( i3)2*R +
( i3)2*R =3*( i3)2*R.
Note that the power is expressed via the current and resistance here.
What about the currents i1, i2, i3?
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- i1*R=0
For the branch II and the battery, the equation is
E- i2*R- i2*R =0
For the branch III and the battery the equation is
E- i3*R- i3*R- i3*R =0
Therefore i1=E/R, i2=E/(2R), i3=E/(3R)
The resulting powers are
P1= ( i1 )2*R=E2/R
P2= 2*( i2)2*R=2 E2/(4R)= E2/(2R)
P3=3*( i3)2*R=3E2/(9R)= E2/(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= 2Ω, 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:
( i1 )2*R=E2/R=12*12/2=72 W
The power released in each bulb (total two bulbs) of branch II:
( i2)2*R= E2/(4R)= 12*12/4/2=18 W
The power released in each bulb (total three bulbs) of branch III:
( i3)2*R= E2/(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 E1=10
V, E2=5V and all resistors are equivalent with R1=R2=R3=R= 5Ω.
A. i1= 1A down, i2=3A up, i3 = 2A
down
B. i1= 1A down, i2=1A up, i3 = 0
C. i1= 2A down, i2=0, i3 = 3A down
D. i1= 3A up, i2=2A up, i3 = 1A down
E. i1= 1A up, i2=2A down, i3 = 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 i1 , i2 , i3
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.
9. RC CIRCUITS
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) q(t ) = EC (1 − e − t / RC )
is the charge on the capacitor as a function of time. Current in the RC circuit:
(1.2) i (t ) =
dq E − t / RC E − t / τ C
= e
= e
dt R
R
where time constant τ C = RC shows the characteristic time for the charging
process. The voltage across the capacitor can be found
(1.3) V (t ) =
q(t )
= E (1 − e − t / RC )
C
In the reverse situation, when the switch is opened, the capacitor with charge
q0 gets discharged. The functions for the charge, current, and voltage are the
following:
(1.4) q(t ) = q0e − t / RC
(1.5) i (t ) =
dq
q
= − 0 e − t / RC
dt
RC
(1.6) V (t ) =
q(t ) q0 − t / RC
= e
C
C
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.
B.
C.
D.
E.
0V
1V
3V
5V
9V
Solution. After the switch is closed the capacitor begins to charge. The charge
on the plates as a function of time is given by
q(t ) = EC (1 − e − t / RC )
The voltage across the capacitor is thus
V (t ) =
q(t )
= 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.
B.
C.
D.
E.
1 second
1 hour
1 day
1 year
infinity
Solution. After the switch is closed the capacitor is beginning to charge.
The charge on the plates as a function of time is:
q(t ) = 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 − q / EC |== 3* 5*10 −12 * ln |1 − 45 /(9 * 5) |= inf
Answer E
10. MAGNETIC FIELDS
PLEASE NOTE THAT IN ALL FORMULAS THE VECTOR PRODUCT (CROSS
PRODUCT) IS DENOTED WITH THE SQUARE BRACKETS: c=[axb]
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*[vxB]
which was found as the result of experimental measurement. A cross product
[vxB] is a vector which is perpendicular to the plane made by vectors v and B.
The direction of the cross product [vxB] 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
[vxB] 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|*[vxB].
Typical problems:
Problem 11. Find the value and the direction of
the force acting on a positively charged particle
with q=2 µC 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.
B.
C.
D.
E.
10 mN, -z direction
10 mN, +z direction
10 N, -z direction
10 N, +z direction
1 kN, +x direction
Solution. The vector product [vxB] 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* sinφ =q*v*B=2*10-6*5*103*1*103=10 N
Answer D
Problem 12. 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.
B.
C.
D.
E.
10 mN, -z direction
10 mN, + z direction
10 N, - z direction
10 N, +z direction
1 kN, +x direction
Solution. The vector product [vxB] 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*103/1.41=10 mN
Answer B.
Problem 13. 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.
B.
C.
D.
E.
10 N, -z axis
10 N, +z axis
100 N, -z axis
100 N, +z axis
1 N, x-axis
Y
y
Z
z
Solution. Since vector B has z-component equal zero, it is completely oriented
within xy plane. Therefore, the vector product [vxB] 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:
[vxB]= [ 10*i*(-10* i +20*j)]=-10*10*[i*i]+10*20*[i*j]
Since vector product [ixi]=0 and [ixj]=k, the unit vector along z-axis, the vector
product becomes: [vxB]=200* k.
Now the force is the vector product time the charge. Since the charge is
negative:
F=q*[vxB]=-0.5*200* k=-100*k, N, i.e 100 N oriented along vector –k which is –z
axis.
Answer C.
Problem 14. 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.
B.
C.
D.
E.
top
bottom
left
right
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*[vxB]. 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
[vxB] 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 [vxB] is directed to the right, and the force
will be directed to the left.
Answer E.
Problem 15. 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.
B.
C.
D.
E.
0.1 µT, clockwise
0.1 µT, counterclockwise
0.1 mT, clockwise
0.1 mT, counterclockwise
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*[vxB]. At the same time, the force associated with the circular motion is
equal to m*w where centrifugal acceleration w= v2/R. Since velocity vector v is
always perpendicular to magnetic field vector B, we can write F=e*v*B=m*v2/R
from which
B=m*v/(R*e)=9*10-31*1.6*103/(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 (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*[vxB] is directed to the bottom, and we know that e<0 and B is directed inside
the page. The vector product [vxB] 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 [vxB] 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.
11. MAGNETIC FIELDS FROM CURRENTS
Key concepts:
In general, the magnetic field due to a moving charged particle is given by:
B(r)= µ0/4π *q*[v*r]/r3
-7
where constant µ0/4π=10 . 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 [vxr] 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)=µ0*i/(2π 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= µ0*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=µ0*I*ϕ /(4πR) which for ϕ=2π 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=µ0*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* [ L xB], 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=µ0 i1*i2*L/(2πd).
Typical problems:
Problem 16. 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.
B.
C.
D.
E.
24 Ν, left
24 mΝ, right
24 kN, top
24 N, bottom
24 mN, inside the page
Solution. The formula to use is: F= i*[ L xB], 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 [LxB] 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*103=24 mN.
Answer E.
Problem 17. 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 18. Find the value of magnetic field of
the long straight wire at distance d=3 cm if it
carries current of 3 mA.
A.
B.
C.
D.
E.
20 pT
20 nT
20 µT
20 mT
20 T
Solution. The formula is B(d)= µ0*i/(2π d),
where all numbers are given by the problem.
B(d)= µ0*i/(2π d)= 2 µ0*i/(4π d)=2*10-7*3*10-3/(3*10-2)=2*10-8 =20 * 10-9T=20 nT
Answer B.
Problem 19. 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.
B.
C.
D.
E.
3 mN, repulsion
3 mN, attraction
1.5 mN, repulsion
1.5 mN, attraction
3 N, repulsion
Solution. The formula to use is: F=µ0 i1*i2*L/(2πd)= 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 20. 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.
B.
C.
D.
E.
20 µT, outside the page
20 µT, inside the page
20 mT, to the left
20 mT to the right
20 T, to the top
Solution. The formula to use is: B=µ0*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 21. 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.
B.
C.
D.
E.
3.14 mT, outside the page
3.14 mT, inside the page
3.14 T, to the left
3.14 T to the right
3.14 kT, to the top
Solution. The formula to use is: B=µ0*i*n=4*3.14*10-7 *1*25*102=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 22. 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*[vxB], 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 23. 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.
B.
C.
D.
E.
3.15 N, top
3.15 N, bottom
3.15 mN, top
3.15 mN, bottom
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: Ftop=µ0*i1*i2*L/(2πd) 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
Fbottom=µ0*i1*i2*L/(2πd)= 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.