Rishiraj Rana, Homework 10 Due: Apr 8 2008, noon Inst: Vitaly
This print-out should have 14 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering. The due time is Central
time.
001 (part 1 of 2) 15 points
The wire is carrying a current I .
y
180
I
R
I
O
x
I
Find the magnitude of the magnetic eld B
at O due to a current-carrying wire shown in
the gure, where the semicircle has radius r,
and the straight parts to the left and to the
right extend to innity.
1. B =
2. B =
3. B =
4. B =
5. B =
6. B =
7. B =
8. B =
9. B =
10. B =
0 I
4r
0 I
2r
0 I
7r
0 I
2r
0 I
r
0 I
r
0 I
correct
4r
0 I
3r
0 I
8r
0 I
3r
By the Biot-Savart Law,
B=
0 I
4
ds
r
.
r2
Consider the left straight part of the wire.
The line element ds at this part, if we come
in from , points towards O; i.e., in the xdirection. We need to nd d s to use the
r
Biot-Savart Law. However, in this part of the
wire, is pointing towards O as well, so d s
r
and are parallel meaning d s = 0 for this
r
r
part of the wire. It is now easy to see that the
right part, having a d s antiparallel to , also
r
gives no contribution to B at O.
Let us go through the semicircle C. The
element d s, which is along the wire, will now
be perpendicular to , which is pointing along
r
the radius towards O. Therefore
| d s | = ds
r
using the fact that is a unit vector. So the
r
Biot-Savart Law gives for the magnitude B of
the magnetic eld at O
B=
0 I
4
C
ds
r2
Since the distance r to the element ds is constant everywhere on the semicircle C, we will
be able to pull it out of the integral. The
integral is
C
ds
1
=2
2
r
r
ds =
C
1
LC ,
r2
where LC = r is the length of the semicircle.
Thus the magnitude of the magnetic eld is
B=
0 I 1
0 I
r =
.
2
4 r
4r
002 (part 2 of 2) 15 points
is in x-direction, is in y-direction,
Note: i
j
and k direction is perpendicular to paper towards reader.
Determine the direction of the magnetic
eld B at O due to the current-carrying wire.
1. B = +
Explanation:
1
2. B =
Rana, Rishiraj Homework 10 Due: Apr 8 2008, noon Inst: Vitaly
2
(1.25664 106 T m/A) (0.414 A)
2 (1.34 m)
3. B = +
=
1
4. B = k
2
1
5. B = +
i
2
= 6.1791 108 T .
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6. B =
1
i
7. B =
2
8. B = +k
9. B = k correct
1
10. B = k +
2
Explanation:
We know from Part 1 that the only contribution to the magnetic eld at O comes from
the semicircle. Furthermore, we need only
consider the direction of d s for one typir
cal segment d s. If we go along the semicircle
from left to right, and we know that is pointr
ing in towards O, the right hand rule tells us
that the eld resulting from this element is
into the paper. Since this holds for every element on the semicircle, the total eld is also
pointing into the paper.
004 (part 1 of 5) 15 points
An innite wire is bent as shown in the gure.
The current is i. It consists of three segments,
AB , BCD, and DE . AB and DE are parallel
and both innite in length. BCD is a semicircular arc, with radius OB = OC = OD =
r. The following questions are concerned with
the magnetic eld vector, B , at O.
D
E
y
C
O
II
z
III
I
x
IV
B
A
What is the component of B at O which lies
in the plane of the current; i.e., the xy plane,
due to the wire segment AB ?
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1. nonzero, parallel to quadrant II
003 (part 1 of 1) 15 points
Calculate the magnitude of the magnetic eld
at a point 134 cm from a long, thin conductor
carrying a current of 0.414 A.
The permeability of free space is
1.25664 106 T m/A.
Correct answer: 6.1791 108 T.
Explanation:
2. nonzero, parallel to quadrant I
3. nonzero, parallel to the positive y direction
4. nonzero, parallel to the negative y direction
5. nonzero, parallel to quadrant III
6
Let : 0 = 1.25664 10 T m/A ,
r = 134 cm = 1.34 m , and
I = 0.414 A .
Using Biot-Savart Law, the magnetic eld
of the wire is
0 I
B=
2r
6. zero correct
7. nonzero, parallel to the negative x direction
8. nonzero, parallel to the positive x direction
Rana, Rishiraj Homework 10 Due: Apr 8 2008, noon Inst: Vitaly
Hence,
Explanation:
Recall that the Biot-Savart Law involves a
term
I dl .
r
Since both dl and lie in the xy plane, their
r
cross product is, by denition, perpendicular
to this plane. In other words, the component
of the magnetic eld B in the xy plane is zero.
The question referred only to the segment,
AB , but this is true for the entire wire in
general.
005 (part 2 of 5) 15 points
What is the magnitude of B at O due to the
semicircle BCD alone?
0 I
2r
0 2 I
2. BBCD =
r
0 I
3. BBCD =
correct
4r
0 I
4. BBCD =
r
0 I
5. BBCD =
r
0 I
6. BBCD =
8r
0 I
7. BBCD =
4r
0 I
8. BBCD =
2r
0 I
9. BBCD =
8r
0 I
10. BBCD =
r
Explanation:
The Biot-Savart Law is given by
1. BBCD =
dB =
o I d l
r
.
2
4
r
Note: The distance from a current element of
BCD to O is a constant, namely r, so we can
pull this out of the integral. Also, the current
element, I dl, is always perpendicular to , so
r
sin = 1.
3
o I
dl
4 r2
o I
r
=
4 r2
o I
=
.
4r
BBCD =
006 (part 3 of 5) 15 points
What is the magnitude of B at O due to AB
alone?
0 I
2r
0 I
correct
2. B =
4r
0 I
3. B =
4r
0 I
4. B =
r
0 I
5. B =
2r
0 I
6. B =
8r
0 2 I
7. B =
r
0 I
8. B =
r
0 I
9. B =
r
0 I
10. B =
8r
Explanation:
1. B =
C
O
r
s
B
x
x
I
A
From the Biot-Savart law, the contribution
to the magnetic eld at O due to a current
element I x at x is given by
o I
o I x
sin =
sin ,
B =
2
4 s
4 r
Rana, Rishiraj Homework 10 Due: Apr 8 2008, noon Inst: Vitaly
x sin
x r
where =
= 2.
s
s
Upon integration, the magnetic eld contributed by a current segment from 1 to 2 is
given by
B=
o I
(cos 1 cos 2 ) .
4 r
and
For wire segment AB , we set 1 =
2
2 = . Hence
o I
cos cos
4r
2
o I
=
.
4r
BAB =
Note: This result is one half the value for the
usual innite wire, which is reasonable.
007 (part 4 of 5) 15 points
What is the direction of B at O due to the
entire wire, ABCDE ?
1. in the positive x direction
4
The permeability of free space is
1.25664 106 T m/A.
Find the magnitude of B at O due to the
entire wire ABCDE .
Correct answer: 7.73012 107 T.
Explanation:
Note: The magnetic eld due to DE is
equal in magnitude and direction to the magnetic eld due to AB .
Hence, using the results from parts 2 and 3
we obtain
Btotal = BAB + BDE + BBCD
0 I
0 I
+
=2
4r
4r
0 I 2
=
+1
4r
(1.25664 106 T m/A) (1.09 A)
=
4 (0.725 m)
2
+1
= 7.73012 107 T .
2. out of the xy plane
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3. in the negative y direction
009 (part 1 of 2) 15 points
A long, straight wire carries a current of
19.9 A. An electron travels at 96200 m/s
parallel to the wire, 38.8 cm from the wire.
The permeability of free space is
1.25664 106 N/A2 and the charge on an
electron is 1.6 1019 C.
What force does the magnetic eld of the
current exert on the moving electron?
Correct answer: 1.57887 1019 N.
Explanation:
4. in the negative x direction
5. in quadrant II
6. in quadrant I
7. in quadrant III
8. in the positive y direction
9. into the xy plane correct
10. in quadrant IV
Explanation:
From the right hand rule, we can see that
the direction of B from all three segments,
AB , BCD, and DE is directed into the page.
008 (part 5 of 5) 15 points
Let r = 0.725 m and I = 1.09 A.
Let : 0
v
q
I
r
= 106 1.25664 N/A2 ,
= 96200 m/s ,
= 1.6 1019 C ,
= 19.9 A , and
= 38.8 cm = 0.388 m .
Magnetic eld produced by a straight,
current-carrying wire is
0 I
B=
2r
Rana, Rishiraj Homework 10 Due: Apr 8 2008, noon Inst: Vitaly
(4 107 ) (19.9 A)
2 (0.388 m)
= 1.02577 105 T .
=
The electron moves in the direction which is
perpendicular to the direction of the magnetic
eld caused by the long wire. So the force on
the electron in a magnetic eld is
F = Bvq
= (1.02577 105 T) (96200 m/s)
(1.6 1019 C)
= 1.57887 1019 N .
5
Explanation:
As the electron moves, its current is opposite to the direction of its motion, and if this
electron is in the same direction as the wires
current, the force on the electron is directed
perpendicular to the electron motion and towards the wire, and if the electrons current is
in the opposite direction as the wires current,
the force is directed oppositely from the rst
situation. But the problem does not give the
direction of the current and we do not know
if the electron is repelled or attracted to the
wire.
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010 (part 2 of 2) 15 points
Which of the following statements is correct ?
1. The force on the electron is directed perpendicular to the electrons motion and directed away from the wire.
2. The force on the electron is directed opposite to the electrons motion.
3. The force on the electron is perpendicular
to the electrons motion and perpendicular to
the plane in which both the electron position
and the wire lie, with the direction determined
by the right hand rule.
4. The force on the electron is directed along
the electrons motion.
5. The force on the electron is directed perpendicular to the electrons motion but with
the limited information given we cannot determine if this force is directed towards or
away from the wire. correct
6. The force on the electron is directed perpendicular to the electrons motion and must
be directed towards the wire.
7. The force on the electron is perpendicular
to the electrons motion and perpendicular to
the plane in which both the electron position
and the wire lie, with not enough information
given to determine the direction of the force.
011 (part 1 of 3) 10 points
Refer to a long, straight wire carrying constant current I .
What can be concluded about the magnitude of the magnetic eld at distance a from
the wire? ( is read is proportional to)
I2
1. B
a
2. B I a
3. B
I
correct
a
4. B a
5. B I
Explanation:
Amperes Law tells us that therefore
B 2 a = 0 I
0 I
1
B=
.
2a
a
012 (part 2 of 3) 10 points
If a stationary charge Q were located a distance a from the wire, the force on that charge
due to the magnetic eld of the wire would be
1. innite.
2. dependent on current I.
Rana, Rishiraj Homework 10 Due: Apr 8 2008, noon Inst: Vitaly
6
keywords:
3. dependent on distance a.
014 (part 1 of 1) 10 points
Consider a long wire and a rectangular current
loop.
I1
4. 0. correct
5. in the direction of the wire.
Explanation:
Magnetic force on a charge is proportional to
the velocity of the charge (FB = q v B ), so the
force is zero when the velocity is zero.
013 (part 3 of 3) 10 points
If an identical wire were parallel to the rst
wire with a distance a between them and the
second wire also carried the same current I
(in the same direction), then the force on the
second wire would be
2. F =
3. F =
4. F =
5. dependent on the value of a only.
5. F =
Explanation:
The magnetic force on a segment of wire of
length L carrying current I is given by
6. F =
FB = L I B ,
7. F =
I1
F12
I2
B1
The right hand rule for cross-products determine that the force F12 on the second wire
due to the rst will point toward the rst wire.
C
Determine the magnitude and direction of
the net magnetic force exerted on the rectangular current loop due to the current I1 in the
long straight wire above the loop.
2. attractive to the rst wire. correct
where B in this case is the eld due to the
other wire and I points in the direction of
the current. This eld B1 due to the rst
wire points perpendicular to the second wire
(where currents I1 and I2 point into the page).
b
D
1. F =
4. 0.
a
I2
1. repulsive from the rst wire.
3. time-varying.
B
A
8. F =
0 I 1 I 2
b
, up correct
2
a (a + b)
0 I 1 I 2
(a b),left
2
a
0 I 1 I 2
, down
2
a+b
0 I 1 I 2
a b, down
2
0 I 1 I 2 a
(b a), up
2
0 I 1 I 2
(a b), right
2
ab
0 I 1 I 2
, down
2
a+b
0 I 1 I 2
(a + b), up
2
Explanation:
To compute the net force on the loop, we
need to consider the forces on segments AB ,
BC , CD, and DA. The net force on the loop
is the vector sum of the forces on the pieces of
the loop. The magnetic force on AB due to
the straight wire can be calculated by using
B
FAB = I2
A
ds B .
In order to use this, we need to know the
magnitude and direction of the magnetic eld
Rana, Rishiraj Homework 10 Due: Apr 8 2008, noon Inst: Vitaly
at each point on the wire loop. We can apply
the Biot-Savart Law. The result of this is that
the magnitude of the magnetic eld due to the
straight wire is
B=
0 I 1
,
2r
and the direction of the magnetic eld is given
by the right hand rule; the eld curls around
the straight wire with the eld coming out of
the page above the wire and the eld going
into the page below the wire. We can now
nd the force on the segment AB ; applying
the right hand rule to nd the direction of the
cross product, dsB , we see that the force will
be in the up direction. Since the wire along
the segment AB is straight and always at a
right angle to B , the cross product simplies
to B ds. Since the magnitude of the magnetic
eld is constant along segment AB , it can
come out of the integral which simplies to
give us the result,
FAB = I2 B1
0 I 1
= I2
2a
0 I 1
2 ( a + b)
Floop = FAB FCD
0 I 1
0 I 1
= I2
I2
2 a
2 (a + b)
0 I 1 I 2
b
=
.
2
a (a + b)
I1
A
I2 B 1
,
and its direction is down. This is because
the direction of the current is now in in the
opposite direction along segment CD!
We can do the use the same procedure for
segments BC and DA, but because the magnetic eld decreases with distance from the
straight wire, B is changing along these segments. This means that the integrals are not
as simple. Using the right hand rule, we see
that the force on segment BC is directed towards the right and the force on segment DA
is directed towards the left. Because the two
segments of wire are symmetrically placed,
their magnitudes will be equal. Since these
forces on the loop have equal magnitudes but
opposite directions, they will cancel. Looking more closely at segments BC and DA,
FAB
I2
B
I2
I2
C
I2 B1 FCD
Here, a positive force is in the up direction,
and a negative force is in the down direction.
So, the direction of the net force is up.
D
Following the same argument, we see that the
force on the segment CD is
FCD = I2
we see that for each small portion of the segment BC , we can nd a small portion on the
segment DA such that the forces on these
two portions are the same magnitude (The
small portions are the same distance from the
straight wire.); because the currents are in opposite directions, the forces on them will be
in opposite directions. Their contributions to
the net force cancel.
The net force on the loop is then
I2
.
7
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