Physics 202 Problems - Week 5 Worked Problems

Physics 202 Problems - Week 5
Worked Problems
Chapter 22: 27, 36, 43, 82
Problem 22.27)
A 0.45-m copper rod with a mass of 0.17 kg carries a current of 11 A in the positive x direction.
What are the magnitude and direction of the minimum magnetic field needed to levitate the rod?
We are given the length, L = 0.45m, the mass, m = 0.17kg, and the current running through
the wire, I = 11A. If we want to find the magnetic field needed to levitate the wire we should first
draw the free body diagram.
~
We know that the force on the wire from the magnetic field is given by F~B = IL B
sinθ,
where θ is the angle between the current and the magnetic field, and the direction is given by the
right-hand rule. In this case, we want the minimum magnetic field, so we will look at the case
when θ = 90 or sinθ = 1. We know that the net force on the wire should be zero for it to levitate.
Therefore, we see from the free body diagram above that
X
~ = F~B yˆ − W
~ yˆ = F~B yˆ − mg yˆ = 0
F~ = F~B + W
Thus,
~ FB = mg
Or
~
IL B
= mg
~
Solving for B
, we find that
mg
~
= 0.34T
B =
IL
We also need the direction. Clearly, in order for the wire to levitate, the magnetic force would
need to be in the positive y direction as shown in the free body diagram. By the right-hand rule,
this means that the magnetic field would have to point in the negative z direction. Therefore, we
~ = −0.34T zˆ
obtain the final answer B
Problem 22.36)
Two current loops , one square and the other circular, have one turn made from wires of the
same length.
(a) If these loops carry the same current and are placed in magnetic fields of equal magnitude, is
the maximum torque of the square loop greater than, less, than, or the same as the maximum
torque of the circular loop? Explain.
(b) Calculate the ratio of the maximum torques, τsquare /τcircle .
(a) If both loops are made of a wire of length L, then the square will have sides of length x =
L
and the circle will have a radius of R = 2π
as shown in the figure below.
2
L
4
Then the area of the square would be Asquare = L16 and the area of the circle would be
2
Acircle = L
4π . Since the torque on the loop from the magnetic field is directly proportional to
the area of the loop, the torque will be greater on the circular loop, since it has a larger area.
~
(b) The maximum torque on a single loop carrying a current I, is given by |τmax
~ | = IA B
.
Therefore, since the current and magnetic field are the same the ratio of the maximum
torques on the loops is given by
τsquare
τcircle
~
IAsquare B
A
4πL2
π
= square =
=
=
2
~
A
16L
4
circle
IAcircle B Finally, we find that the ratio of the maximum torques is
τsquare
π
= = 0.79
τcircle
4
Problem 22.43)
Consider the long, straight, current-carrying wires shown in the figure below. One wire carries
a current of 6.2A in the positive y direction; the other wire carries a current of 4.5A in the positive
x direction.
(a) At which of the two points, A or B, do you expect the magnitude of the net magnetic field
to be greater? Explain.
(b) Calculate the magnitude of the net magnetic field at points A and B.
(a) The magnetic field will be greater at point A. At point B, the magnetic field from the wire
along the x axis will be pointing out of the page and the magnetic field from the wire along
the y axis will be pointing into the page, so they will work against one another. On the other
hand, at point A, both wires create a field that points out of the page so they work together
to create a stronger magnetic field.
(b) Now let’s calculate the magnetic field at these two points. At point A, both of the fields
are
P~
~ ~ out of the page. and the total field will be the vector sum at that point B = (B1 + B2 )ˆ
z.
We also know that the magnetic field created by a single wire is given by
µ I
~
0
B =
2πr
,
where r is the distance away from the wire, I is the current, and µ0 = 4π × 10− 7T m/A.
Therefore, the total field at point A is given by
B~A =
µ0 I 1
µ0 I2
+
2πr1 2πr2
zˆ =
µ0 (4.5A)
µ0 (6.2A)
+
2π(0.16m) 2π(0.16m)
zˆ
Finally, the magnetic field at point A is B~A = 1.33 × 10−5 T zˆ
At point B, we can
find the total field by the vector sum of the individual fields. Now,
P~
~ ~ z
B = (B1 − B2 )ˆ
Which we can rewrite,
B~B =
µ0 I1
µ0 I 2
−
2πr1 2πr2
zˆ =
µ0 (4.5A)
µ0 (6.2A)
−
2π(0.16m) 2π(0.16m)
Finally, the magnetic field at point B is B~B = −2.13 × 10−6 T zˆ
zˆ
Problem 22.82)
A single current-carrying loop of radius R is placed along next to a long, straight wire as shown
in the figure below. The current in the wire points to the right and is of magnitude I.
(a) In which direction must the current flow in the loop to produce zero magnetic field at its
center? Explain.
(b) Calculate the current in part (a).
(a) We know that the long, straight wire produces a magnetic field that points out of the page in
the center of the loop by the right-hand rule. So, if we want the total field to be zero at the
center of the loop, then the loop must create a magnetic field that points out of the page. If
this is the case, then the current must be going clockwise around the loop. This too comes
from the right-hand rule.
(b) We know that the total magnetic field at the center of the loop is given by
X
~ = Bwire
~ + Bloop
~ = Bwire
~ − Bloop
~ zˆ = 0
B
~ ~ From this, we see that Bloop
= Bwire We know that the magnitude of the magnetic field produced by a single loop at its center is
given by
~ µ0 Iloop
Bloop =
2R
and the magnetic field from the wire is given by
µ I
µ0 Iwire
~ µ0 Iwire
= 0 wire
=
Bwire =
3R
2πr
π3R
2π 2
We can set the two equal and solve for Iloop .
µ0 Iloop
µ0 Iwire
=
2R
π3R
Iloop =
2Iwire
2I
=
3π
3π
Physics 202 Problems - Week 5
Worked Problems
Chapter 23: 70, 10, 22, 30
Problem 23.70)
A cubical box 22cm on a side is placed into a uniform 0.35-T magnetic field. Find the net
magnetic flux through the box.
There are two ways that this question could be answered. The first is conceptual and requires
no calculation; the second is slightly more involved.
Method 1:
Whenever you are given a closed surface, the magnetic flux through it will always be zero.
This is true even if there are currents inside the the surface. This comes from the fact that magnetic
field lines start and end at the same point. This essentially comes from the fact that there is no
such thing as magnetic charge, or point sources of magnetic fields. One could also argue that since
the magnetic field is uniform, there is equal flux going into and coming out of the closed surface.
Therefore the net flux is zero.
Method 2:
~ = Bx x
Consider an arbitrary magnetic field B
ˆ + By yˆ + Bz zˆ and some cube with sides of length
a as shown below.
This cube has six surfaces that will contribute to the net flux. By convention, we chose our
normal vector to point out of the surface on all sides. We can calculate each of these fluxes using
~ ·A
~ = BAcosθ. As shown in the following figure.
Φ=B
where A = a2 is the area of each side.
Then the net flux through the surface will be the sum of the flux through the above surfaces.
Φnet = Φ1 + Φ2 + Φ3 + Φ4 + Φ5 + Φ6
Putting everything together, we see that
Φnet = −By A − Bx A + By A + Bx A + Bz A − Bz A
Therefore, the total flux through the cube is Φnet = 0 . This result is true no matter how the
magnetic field is oriented since we used an arbitrary field. The result is also independent of the
field magnitude of the field.
Problem 23.10)
The figure below shows the magnetic flux through a single coil as a function of time.
(a) At what times shown in the plot does the magnetic flux have the greatest magnitude?
(b) What is the value of the magnitude of the magnetic flux at these times?
(c) At what times does the induced emf have its greatest magnitude?
(d) What are the maximum values of the induced emf?
(a) From the graph, we see that the magnetic flux has its greatest magnitude at the times
t = 0.0s, 0.2s, 0.4s and 0.6s . These are the times when the absolute value of the magnetic
flux is greatest.
(b) From the graph, we see that the magnitude of the magnetic flux at these times is |Φ| = 4Wb .
(c) The induced emf in a single coil is given by
∆Φ || = ∆t This is just the slope of our plot. It has its maximum value when the magnetic flux is equal
to zero. This occurs at the times t = 0.1s, 0.3s and 0.5s .
(d) The flux as a function of time appears to follow the form
2πt
Φ(t) = A cos(ωt) = A cos
T
This is similar to the position from equilibrium of a mass attached to a spring as a function
of time. Recall, x(t) = A cos(ωt). For this system the speed as a function of time was given
by x(t) = −Aω sin(ωt). In this case the speed was the instantaneous slope of the position
versus time plot. Likewise, the slope of our plot of magnetic flux versus time will follow the
form
2πA
∆Φ
= −Aω sin(ωt) = −
sin
∆t
T
2πt
T
From the plot was identify that A = 4Wb and T = 0.4s. Therefore, at t = 0.1s, 0.3s and 0.5s,
|| = 62.8V .
Problem 23.22)
The figure below shows a current-carrying wire and a circuit containing a resistor R.
(a) If the current in the wire is constant, is the induced current in the circuit clockwise, counterclockwise, or zero? Explain.
(b) If the current in the wire increases, is the induced current in the circuit clockwise, counterclockwise, or zero? Explain.
(a) We know that the magnetic field from the current-carrying wire is points out of the page
through the circuit. We also know that the induced emf in the loop created by the circuit
has a magnitude given by
∆Φ || = ∆t In this case, the normal to the area of the loop is parallel to the field so the magnetic flux
through the circuit area, A, is given by
~ ·A
~ = B
~ Acosθ = B
~ A
Φ=B
Therefore the induced emf through the circuit is given by
∆ B
~ A || = ∆t
We know that the area of the circuit loop is fixed. So the only thing that will create a changing flux is a changing magnetic field. That being said, we know that the magnetic field is
not changing when the current in the long, straight wire is constant. Therefore, there is no
induced emf in the circuit, and by Ohm’s Law (|| = IR), there is no induced current.
~ · A.
~
Note: We should be a little cautious about our definition of the magnetic flux, Φ = B
This is only true when the magnetic field is constant over the area; however, it does give us
some conceptual insight to this problem.
The induced current in the circuit is zero.
(b) If on the other hand the current increases in the wire, we would expect the magnetic field to
increase through the wire. This will create an induced emf in the circuit loop. If we care only
about the direction of the induced current, then all we need to know is the induced current
will try to create a field that opposes the changing magnetic field through the loop. In this
case the field is increasing out of the page. Therefore, the induced current would create a
magnetic field that is into the page. By the right hand rule, we know that this current must
be clockwise around the loop.
The induced current in the circuit is clockwise around the loop.
Problem 23.30)
The figure below shows a zero-resistance rod sliding to the right on two zero-resistance rails
separated by the distance L = 0.45m. The rails are connected by a 12.5Ω resistor, and the entire
system is in a uniform magnetic field with a magnitude of 0.750T.
(a) Find the speed at which the bar must be moved to produce a current of 0.125A in the resistor.
(b) Would your answer to part (a) change if the bar was moving to the left instead of the right?
Explain.
(a) We know that the magnitude of the induced emf in the loop created by the rods and resistor
is given by
∆ B
~ A || = ∆t
since the magnetic field is parallel to the normal direction of the loop’s enclosed area. In this
case, the magnetic field is constant over the area and is not changing. What is changing here
is the area of the loop. So the induced emf becomes
∆A ~
|| = B
∆t We also know that the rod is moving at a constant speed, v = ∆x
∆t , and that the height of
the loop is fixed at, L. Then we can say that
∆A ∆(Lx) ∆x ∆t = ∆t = L ∆t = Lv
Finally, we see that the induced emf is
~
|| = B
Lv
We can then use Ohm’s Law to relate the induced emf to the induced current.
~
B Lv
I = || /R =
R
Now we can simply solve for the speed.
IR
v = ~L
B
Therefore, we require that the rod have a speed of v = 4.63m/s .
(b) The answer to part (a) is independent of the direction that the rod is moving on the rails.
The only difference between the two cases would be that the current would go in the opposite
direction through the resistor. The current would still have the same magnitude.