Download AP Physics C - Problem Drill 23: Magnetic Fields Question 01 1

AP Physics C - Problem Drill 23: Magnetic Fields
1. Which of these statements about magnetism is incorrect?
Question 01
(A) Iron, Cobalt and Nickel are examples of ferromagnetic materials.
(B) Most materials are only slightly magnetic.
(C) There are no magnetic monopoles.
(D) A compass points to geographical north.
(E) A permanent magnet contains magnetic domains that are similarly aligned.
A. Incorrect!
These are ferromagnetic materials.
B. Incorrect!
Most materials are magnetic.
C. Incorrect!
Unlike electric charges which can exist on their own, there are no magnetic
monopoles; you will never find a north pole without a south pole.
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Choice
D. Correct!
The point of a compass is made of a ferromagnetic material that will point to
magnetic north, which is not the same as geographical north.
E. Incorrect!
A magnetic domain is a region inside a ferromagnetic material that acts like a tiny
magnet. A permanent magnet is created by applying a magnetic field to the
material; the fields in the domains become more aligned with the applied field, and
remain in alignment along time after the applied field is removed.
Solution
Iron, Cobalt and Nickel are ferromagnetic materials. All materials are magnetic but
most are only very slightly magnetic.
Unlike electric charges which can exist on their own, there are no magnetic
monopoles; you will never find a north pole without a south pole.
A magnetic domain is a region inside a ferromagnetic material that acts like a tiny
magnet. A permanent magnet is created by applying a magnetic field to the
material; the fields in the domains become more aligned with the applied field, and
remain in alignment along time after the applied field is removed.
The point of a compass is made of a ferromagnetic material that will point to
magnetic north, which is not the same as geographical north.
The correct answer is (D).
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Question No. 2 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
2. Five students are discussing how to use the right hand rule, which one needs to
do some more studying?
Question 02
(A) Student A says the thumb points in the direction of conventional current.
(B) Student B says the fingers point in the direction of the B field.
(C) Student C says the palm of the hand lies in the plane of the force on a
current carrying wire.
(D) Student D says that the Right Hand Rule can be used to find force on a
charge in a B field.
(E) Student E says that when using the RHR for a negative charge the thumb
points in the opposite direction to the way the charge is moving.
A. Incorrect!
The thumb does point in the direction of conventional current.
B. Incorrect!
The fingers do point in the direction of the magnetic field.
C. Correct!
To find the force on a current carrying wire in a magnetic field, the thumb points in
the direction of the conventional current, the fingers point in the direction of the
magnetic field and force on the wire will be in a direction perpendicular to the palm
of the hand.
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Choice
D. Incorrect!
The Right Hand Rule can be used to find the force on a charge in a B-field.
Remember the direction of conventional current is opposite to the direction that a
negative charge moves, so the direction of the thumb is reversed when using it
with negative charges.
E. Incorrect!
The Right Hand Rule can be used to find the force on a charge in a B-field.
Remember the direction of conventional current is opposite to the direction that a
negative charge moves, so the direction of the thumb is reversed when using it
with negative charges.
The thumb does point in the direction of conventional current.
The fingers do point in the direction of the magnetic field.
The Right Hand Rule can be used to find the force on a charge in a B-field.
Remember the direction of conventional current is opposite to the direction that a
negative charge moves, so the direction of the thumb is reversed when using it
with negative charges.
Solution
To find the force on a current carrying wire in a magnetic field, the thumb points in
the direction of the conventional current, the fingers point in the direction of the
magnetic field and force on the wire will be in a direction perpendicular to the palm
of the hand.
The correct answer is (C).
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Question No. 3 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
3. A wire in the plane of the page carries a conventional current as shown. Which
direction is the resulting force on the wire:
X
X
X
X
Question 03
(A) Directed into the page.
(B) Directed out of the page.
(C) Directed to the right.
(D) Directed to the left.
(E) None of the above
X
X
X
X
X
X
X
X
X
X
X
X
A. Incorrect!
Use the right hand rule. The thumb points in the direction of the conventional
current. The extended fingers point in the direction of the magnetic, B, field. The
palm points in the direction of the magnetic force exerted. You could also imagine a
vector perpendicular to the palm that shows the direction of the force.
B. Incorrect!
Use the right hand rule. The thumb points in the direction of the conventional
current. The extended fingers point in the direction of the magnetic, B, field. The
palm points in the direction of the magnetic force exerted. You could also imagine a
vector perpendicular to the palm that shows the direction of the force.
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Each Answer
Choice
C. Correct!
Use the right hand rule. The thumb points in the direction of the conventional
current. The extended fingers point in the direction of the magnetic, B, field. The
palm points in the direction of the magnetic force exerted. You could also imagine a
vector perpendicular to the palm that shows the direction of the force.
D. Incorrect!
Use the right hand rule. The thumb points in the direction of the conventional
current. The extended fingers point in the direction of the magnetic, B, field. The
palm points in the direction of the magnetic force exerted. You could also imagine a
vector perpendicular to the palm that shows the direction of the force.
E. Incorrect!
Use the right hand rule. The thumb points in the direction of the conventional
current. The extended fingers point in the direction of the magnetic, B, field. The
palm points in the direction of the magnetic force exerted. You could also imagine a
vector perpendicular to the palm that shows the direction of the force.
Use the right hand rule. The thumb points in the direction of the conventional
current. The extended fingers point in the direction of the magnetic, B, field. The
palm points in the direction of the magnetic force exerted. You could also imagine a
vector perpendicular to the palm that shows the direction of the force.
Note the Xs for the magnetic field. This means the field points into the plane of the
page.
Solution
The arrow indicates the direction of current. There is no need to reverse any
direction since the conventional current direction is given.
Point the fingers of your right hand into the page. The thumb then points
downward. Your palm should be facing to the right. This is the direction of the force
on the wire.
The correct answer is (C).
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Question No. 4 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
4. A wire carrying a 30 A current has a length of 12 cm between the pole faces of
a magnet at an angle of 60 ° to the field. The magnetic field is approximately
uniform at 0.90 T. What is the force on the wire?
Question 04
(A) 3.2 N
(B) 280 N
(C) 0 N
(D) 2.8 N
(E) 1.6 N
A. Incorrect!
Don’t forget to account for the angle between the current and the magnetic field.
In this case its 60° degrees.
B. Incorrect!
Be sure to convert the distance given into meters so that it matches the units of the
other quantities.
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Choice
C. Incorrect!
This would be true if the current and field were parallel. However, in this case the
angle is 60 °.
D. Correct!
Use the formula for the magnetic force on a current carrying wire. FB= BiL sinθ.
Substitute all the given values and calculate carefully. Note the angle between the
current and B field. You may also need to convert the distance into meters.
E. Incorrect!
Take the sine of the angle between the wire and the B field not the cosine of the
angle.
Known:
Length of wire, L = 12 cm
Current, i = 30 A
Angle with field, θ = 60°
Magnetic Field, B = 0.90 T
Unknown:
Define:
Solution
N
S
Force on wire, F =? N
Draw a diagram to help visualize the problem.
Use the formula for the magnetic force on a current carrying wire.
F = BiLsinθ
Also, be sure to note that in this case the angel between the current and
B field is given as 60 degrees
Output:
First convert the distance in centimeters to meters.
12 cm ×
1m
= 0.12 m
100 cm
FB = (0.9 T)(30 A)(.12 m)(sin60o) = 2.8 N
The correct answer is (D).
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Question No. 5 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
5. If a 0.20 m wire carrying a current of 10 A experiences a force of 0.05 N when
placed in a field of 2.5 x 10-3 T, what is the angle between the wire and the field?
Question 05
(A) 1°
(B) 90°
(C) 0
(D) 0.14°
(E) 45°
A. Incorrect!
This is the sin of the angle.
B. Correct!
Use the formula F= BiLsin θ, and rearrange to find θ.
Feedback on
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Choice
C. Incorrect!
Make sure you are using sin-1 θ, to find the angle.
D. Incorrect!
Rearrange the formula F= BiLsin θ, to find θ.
E. Incorrect!
Use the formula F= BiLsin θ.
Known:
Force, F = 0.05 N
B-field, B = 2.5 x 10-4 T
Current, I = 10 A
Length, L = 0.2 m
Unknown: Angle, θ =?
Define:
F = BiL sin θ
Rearrange:
Solution
Output:
Sin θ =
Sin θ =
Sin θ =
F
BiL
F
BiL
0.05 N
2.5 × 10-4 T × 10 A × 0.2 m
=1
θ = sin-1(1)
θ = 90D
Substantiate: Units are correct, sig figs are correct, Magnitude is reasonable.
The correct answer is (B).
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Question No. 6 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
6. Look at the diagram, the arrow indicates
the direction of travel, of an electron
introduced in to a B field. In which direction
will the electron move?
Question 06
X
X
X
X
X
X
X
X
X
X
X
X
-
X
X
X
(A) Into the paper/screen.
(B) Out of the paper/screen.
X
(C) It will continue to move in the same
direction.
(D) Clockwise, in the plane of the
paper/screen.
(E) Anticlockwise, in the plane of the paper/screen.
A. Incorrect!
Use the right hand rule, remember the thumb points in direction of conventional
current and the force on the electron is in the direction perpendicular to the palm.
B. Incorrect!
Use the right hand rule, remember the thumb points in direction of conventional
current and the force on the electron is in the direction perpendicular to the palm.
Feedback on
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Choice
C. Incorrect!
There will be a force on the electron the direction of the force can be found using
the right hand rule.
D. Correct!
Use the right hand rule, remember the thumb points in direction of conventional
current and the force on the electron is in the direction perpendicular to the palm.
The force is always perpendicular to the direction of travel and pointed inwards
which results in the electron moving in a clockwise circle.
E. Incorrect!
Use the right hand rule, remember the thumb points in direction of conventional
current and the force on the electron is in the direction perpendicular to the palm.
Use the right hand rule, remember the thumb points in direction of conventional
current and the force on the electron is in the direction perpendicular to the palm.
The force is always perpendicular to the direction of travel and pointed inwards
which results in the electron moving in a clockwise circle.
Solution
The correct answer is (D).
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Question No. 7 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
7. An electron is traveling perpendicular to a B-field of 0.75 T, if the force on the
electron is 4.0 x 10-17 N, what is the speed of the electron?
Question 07
(A) Insufficient information
(B) 5.3 x 10-17 m/s
(C) 0 m/s
(D) 250 m/s
(E) 330 m/s
A. Incorrect!
There is sufficient information to calculate the velocity, identify a formula that ties
all the information together and remember that the charge on an electron is 1.6 x
10-19 C.
B. Incorrect!
Remember that the charge on an electron is 1.6 x 10-19 C.
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Choice
C. Incorrect!
The electron will be move in the field.
D. Incorrect!
Remember you need to include the value for B-field in the calculation.
E. Correct!
Use the formula FB = qv sin θ and rearrange to find velocity. Remember that
electron is travelling perpendicular to the B-field so, θ is 90 degrees.
Known:
Charge on an electron, q = 1.6 x 10-19 C
Force, F = 4 x 10-17 N
Angle between electron and field = 90°
Unknown: Speed of electron, v = ? m/s
Define:
FB = qv sin θ
Rearrange:
Output:
Solution
v=
v=
FB
q sinθ
4.0 × 10−17N
1.6 × 10
−19
C × 0.75 T × sin 90
=
4.0 × 10−17N
1.2 × 10−19 C
= 330 m/s
Substantiate: Units are correct, sig figs are correct, magnitude is reasonable.
The correct answer is (E).
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Question No. 8 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
8. A charged particle travels at 140 m/s perpendicular to a 0.5 T magnetic field.
The radius of the path is 6.0 m. Determine the mass of the particle if its charge is
known to be 8.2 x 10-4 C.
Question 08
(A) 3.2 x 10-5 kg
(B) 10.2 g
(C) 1.8 x 105 kg
(D) 1.8 x 10-5 kg
(E) None of the above
A. Incorrect!
The centripetal force on the particle is equal to mass times velocity squared divided
by the radius.
B. Incorrect!
The magnetic force on the particle is equal to the charge of the particle times the
velocity of the particle, times the magnetic field strength.
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Choice
C. Incorrect!
The magnetic field is perpendicular to the velocity of the particle and the sin of 90
degrees is one.
D. Correct!
The magnetic force exerted on the particle acts as a centripetal force. These two
quantities can be equated. Rearrange the equality to solve for our unknown value,
the mass. m= qBr/v. Then substitute carefully and calculate.
E. Incorrect!
The magnetic force exerted on the particle acts as a centripetal force. These two
quantities can be equated.
Known:
Charge, q = 0.20 x 10-4 C
Velocity, v = 140 m/s
B field, B = 0.50 T
Radius of the path, r = 6.0 m
Unknown: Mass of the particle, m = ? kg
Define:
The magnetic force exerted on the particle acts as a centripetal force.
These two quantities can be equated.
q v B sinθ =
mv2
r
The mass is the unknown quantity, solve for that variable. Since the field
and velocity are perpendicular, the sin θ = 1 and drops out.
Solution
m=
qvBr
2
v
=
qBr
v
Next, substitute the known values.
Output:
m =
(8.2 x 10-4 C)(.5 T)(6.0 m)
140 m/s
= 1.8 x 10−5 kg
Substantiate: Units are correct, sig figs are correct, magnitude is reasonable.
The correct answer is (D).
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Question No. 9 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
9. A vertical electric wire in the wall of a building carries a DC current of 25 A
upward. What is the magnitude of the magnetic field at a point 10 cm to the side
of this wire?
Question 09
(A) 5 x 10-5 T
(B) 5 x 10-7 T
(C) 5 x 10-9 T
(D) 5 x 10-11 T
(E) None of the above
A. Correct!
Use the formula to find the B field at a distance from a current carrying wire. The
magnetic field equal to the permeability of free space times the current, divided by
2 pi times the distance away from the wire.
B. Incorrect!
Look at the information given and choose a suitable formula that fits the known and
unknown values.
Feedback on
Each Answer
Choice
C. Incorrect!
The current, and distance from the wire are given values and the unknown is the
field, identify a formula that includes all these factors.
D. Incorrect!
Be sure to convert the centimeter distance given into meters so that it is in SI
units.
E. Incorrect!
We are looking for the magnetic field strength, B. The current, and distance from
the wire are the given values. This should narrow your selection of formulas to
choose from.
Known: Distance to wire, r = 10 cm
Current, i = 25 A
Unknown: B field, B = ? T
Define:
B=
μoi
2πr
μo is the permeability of free space, a constant analogous to εo in
Coulombs law.
μo = 4∏x10-7 Tm/A.
Output: First convert the 10 cm into meters.
Solution
10 cm x
1m
= 0.10 m
100 cm
Substitute values into the equation.
B=
4π x10−7 Tm/A × 25 A
= 5 x 10−5 T
2π (0.10 m)
Substantiate: Units are correct, sig figs are correct, magnitude is reasonable.
Note: This relatively small magnetic field is comparable in size to the Earth’s
magnetic field. It seems reasonable that currents such as this could disrupt
compasses nearby.
The correct answer is (A).
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Question No. 10 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
10. Two long parallel wires separated by a distance d, carry equal current as
shown. What is the direction of the net magnetic field due to the wires at a point
halfway between them?
Question 10
(A) Directed into the page.
(B) Directed out of the page.
(C) Directed to the right.
(D) Directed to the left.
(E) Zero in magnitude.
A. Correct!
Use the right hand rule. The thumb points in the direction of the conventional
current. The fingers extend, or curl in this case, in the direction of the magnetic
field. Do this for each of the wires to get a magnetic field direction. Those two
fields add to yield a net magnetic field that is into the page at that location.
B. Incorrect!
Use the right hand rule. The thumb points in the direction of the conventional
current. The fingers extend, or curl in this case, in the direction of the magnetic
field.
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Each Answer
Choice
C. Incorrect!
Use the right hand rule. The thumb points in the direction of the conventional
current. The fingers extend, or curl in this case, in the direction of the magnetic
field.
D. Incorrect!
Use the right hand rule. The thumb points in the direction of the conventional
current. The fingers extend, or curl in this case, in the direction of the magnetic
field.
E. Incorrect!
Use the right hand rule. The thumb points in the direction of the conventional
current. The fingers extend, or curl in this case, in the direction of the magnetic
field.
Use the right hand rule. The thumb points in the direction of the conventional
current. The fingers extend, or curl in this case, in the direction of the magnetic
field.
For the wire on the left, this procedure gives a magnetic field that goes into the
page at the point that is halfway between the wires.
Solution
For the wire on the right, this procedure gives a magnetic field that also goes into
the page at the point that is halfway between the wires.
Thus, those two fields add to yield a net magnetic field that is into the page at that
location.
The correct answer is (A).
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