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PhET Tutorial: Faraday's Electromagnetic Lab
Faraday's law of induction deals with how a magnetic flux induces an emf in a circuit. Recall that magnetic flux depends on magnetic field strength and the effective area the field is passing through. We'll start our investigation by looking at the field strength around a bar magnet.
Select the Bar Magnet tab. Deselect Show Compass, and drag the bar magnet to the center of the screen. Then, select Show Field Meter, which allows you to measure the strength of the magnetic field at any location.

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The magnetic field is strongest where the field lines are close to each other or we can say the dense area of field lines.
C

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Select the Pickup Coil tab, and place the bar magnet inside the coil containing two loops.
Try to find a location where the stationary magnet induces a current in the coil and causes the light bulb to shine. Which of the following is correct?

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There is no induced current in the coil, so the light bulb does not shine, if the magnet is stationary (for any location of the magnet). b/c: Apparently, a magnetic field that isn't changing in time doesn't induce a current in the coil.

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Now, let's look at a situation with changing flux. Starting from the far left of the screen, move the magnet to the right so it goes through the middle of the current loop at a constant speed and out to the right of the loop.
Roughly where is the magnet when the light bulb is the brightest? (The brightness of the light bulb is depicted by the length of the rays emanating from it.)

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The light bulb is brightest when either end of the magnet is in the middle of the coil.
This is consistent with Faraday's law of induction, which states that the induced emf in the coil is equal to the time rate of change of the magnetic flux through the coil. Since the magnetic field is strongest near the poles of the magnet (as we saw in Part A) and is weaker near the middle of the magnet, the magnetic flux changes the quickest when the poles are going through the coil loop. This causes the greatest induced emf, which causes the greatest current through the light bulb.

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How does the brightness of the light bulb change if the magnet is moved through the coil more quickly?

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The bulb shines more brightly.
As the magnet is approaching the coil, the magnetic flux is increasing since the field strength is stronger closer to the magnet. The faster the magnet approaches, the quicker the magnetic flux increases, which causes a greater induced emf and a greater current.

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Select the Transformer tab, which contains an electromagnet at left (containing a DC battery) and a pickup coil attached to a light bulb at right. The electromagnet produces a magnetic field very similar to that of the bar magnet.
Try to find a location where the stationary electromagnet (with a DC current) induces a current in the pickup coil and causes the light bulb to shine. Which of the following is correct?

answer

There is no induced current in the coil, so the light bulb does not shine, if the electromagnet is stationary (for any location of the electromagnet).
Just as we saw earlier (in Part B), a constant magnetic flux does not cause an induced emf through the pickup coil. According to Faraday's law of induction, an emf is induced only when the magnetic flux changes with time.

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Now, select an AC current source at top right (instead of the DC battery). You can adjust the frequency of oscillation of the current using the horizontal slider bar (in the blue AC Current Supply window), and you can adjust the peak current with the vertical slider bar.
Place the electromagnet to the left of the pickup coil (as shown below).
How does the average brightness of the light bulb depend on the AC frequency?

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The brightness increases when the frequency increases.
A higher frequency causes the magnetic field produced by the current to change more rapidly, which causes a greater rate of change of the magnetic flux through the pickup coil. This greater emf causes a greater current.

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With the electromagnet still situated just to the left of the pickup coil, turn the frequency all the way down to its lowest setting (5%). Watch how the brightness changes in time and how it correlates with the instantaneous current (which is shown by the position of the red vertical line in the AC current supply meter graph). When is the brightness the greatest?

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When the current is zero.
The induced emf that drives the current through the light bulb is proportional to the rate of change of the flux. The magnetic flux is proportional to the current in the electromagnet, so the emf is largest when the current is varying the most quickly. This occurs when the current is zero. Notice that the instantaneous rate of change of the current is zero when the current is at a maximum or minimum value, which explains why the brightness of the bulb is zero at those moments.

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Adjust the frequency to 100% so the bulb is always "on."
With the electromagnet still just to the left of the pickup coil, how does the average brightness of the light bulb depend on the number of loops of the pickup coil? (You can change the number of loops in the Pickup Coil properties box at bottom right.)

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The brightness increases when the number of loops increases.
The total magnetic flux through the pickup coil is equal to the magnetic flux through each loop times the number of loops. So when you increase the number of loops, you increase the flux, which makes the flux change more quickly in time as the current oscillates. This greater rate of change of the magnetic flux causes a greater induced emf in the pickup coil.

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With the electromagnet still just to the left of the pickup coil, how does the average brightness of the light bulb depend on the loop area of the pickup coil? (You can change the loop area (as a percentage) using the slider bar at bottom right.)

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The brightness increases when the loop area of the pickup coil increases.
The brightness increases when the loop area of the pickup coil increases.

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Now, place the electromagnet centered inside the pickup coil, as shown below.
How does the average brightness of the light bulb depend on the loop area of the pickup coil?

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The brightness decreases when the loop area of the pickup coil increases.
This somewhat surprising result is due to the magnetic flux actually decreasing with an increasing loop area (for a particular instantaneous current in the electromagnet) since the magnetic field from the electromagnet reverses its direction outside its coil. This field (outside the coil) produces a flux that in part cancels out the magnetic flux due to the field inside the coil. The smaller magnetic flux causes a slower rate of change of the magnetic flux as the current oscillates.
You can see this by looking at the direction of the electromagnet's field in the following two situations.

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Â± Motional EMF in a Conducting Rod
In the figure, a conducting rod with length L = 25.0 cm moves in a magnetic field Bâƒ— of magnitude 0.500 T directed into the plane of the figure. The rod moves with speed v = 7.00 m/s in the direction shown. (Figure 1)

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A-FU

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When the charges in the rod are in equilibrium, which point, a or b, has an excess of positive charge?

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F=qv x B
Using Right-Hand-Rule, the forces are acting on the rod in the right side. Charges are forced to move right.
So therefore the posive excess charge moves toward b.

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In what direction does the electric field then point?

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Electric potential moves from areas of high potential to low potiential, so therefore
b->a.

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When the charges in the rod are in equilibrium, what is the magnitude E of the electric field within the rod?
Express your answer in volts per meter to at least three significant digits.

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If the charge on the rod is in equilibrium, then the electric force and the magnetic force acting on the charge on the rod are equal and opposite in direction.
Fe=Fm
qE=qv x B
E=vB
=(7.00 m/s)(0.500 T)
=3.50 V/m

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Which point, a or b, is at higher potential?

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b
Some work must have been done in order to create this potential difference, i.e., to separate the charges in the rod. This work was done by the initial force required to pull the rod in opposition to the force on it due to the interaction of the transient current in it with the magnetic field. Note that once the charges are in equilibrium, no force is required to keep the rod moving with constant velocity.

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What is the magnitude Vba of the potential difference between the ends of the rod?
Express your answer in volts to at least three significant digits.

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emf is equal to the potential difference
0.875V
(the work is done in the next problem)

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What is the magnitude E of the motional emf induced in the rod?
Express your answer in volts to at least three significant digits.

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Vba=- integral from b to a of E x dL
=E integral of dL
=EL
=(3.5V/m)[(25.0cm)(1m/100cm)]
=0.875 V

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Induced Current in a Metal Loop Conceptual Question
For each of the actions depicted below, a magnet and/or metal loop moves with velocity vâƒ— (vâƒ— is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop.

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A-E

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For the action depicted in the figure, (Figure 1) indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop).

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The loop will act as a magnet and there will be force repulsion.
Counterclockwise

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For the action depicted in the figure, (Figure 2) indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop).

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The face of the south pole will become the north pole, so there will be a force of attraction between them, which makes the motion of the coil opposed. Lenz's Law shows that:
counter clockwise

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For the action depicted in the figure, (Figure 3) indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop).

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If stationary, no change in magnetic flux, so there will be no induced emf; 0.
Zero

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For the action depicted in the figure, (Figure 4) indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop).

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No magnetic flux, magnet isn't getting closer or farther away. So no induced emf and no current flow; 0.
Zero

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For the action depicted in the figure, (Figure 5) indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop).

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Clockwise

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Faraday's Law and Induced Emf
Learning Goal:
To understand the terms in Faraday's law and to be able to identify the magnitude and direction of induced emf.
Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux. Mathematically, it can be written as
E=âˆ’Î”Î¦BÎ”t,
where E is the emf induced in a closed loop, and
Î”Î¦BÎ”t
is the rate of change of the magnetic flux through a surface bounded by the loop. For uniform magnetic fields the magnetic flux is given by Î¦B=Bâƒ— â‹…Aâƒ— =BAcos(Î¸), where Î¸ is the angle between the magnetic field Bâƒ— and the normal to the surface of area A.
To find the direction of the induced emf, one can use Lenz's law:
The induced current's magnetic field opposes the change in the magnetic flux that induced the current.
For example, if the magnetic flux through a loop increases, the induced magnetic field is directed opposite to the "parent" magnetic field, thus countering the increase in flux. If the flux decreases, the induced current's magnetic field has the same direction as the parent magnetic field, thus countering the decrease in flux.
Recall that to relate the direction of the electric current and its magnetic field, you can use the right-hand rule: When the fingers on your right hand are curled in the direction of the current in a loop, your thumb gives the direction of the magnetic field generated by this current.

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A-F

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Which of the following changes would induce an electromotive force (emf) in the loop? When you consider each option, assume that no other changes occur.
Check all that apply.

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The magnitude of Bâƒ— increases.
The magnitude of Bâƒ— decreases.
The loop rotates about the vertical axis (vertical dotted line) shown in the diagram.
The loop rotates about the horizontal axis (horizontal dotted line) shown in the diagram.

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Find the flux Î¦B through the loop.
Express your answer in terms of x, y, and B.

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Î¦B = xyB

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If the magnetic field steadily decreases from B to zero during a time interval t, what is the magnitude E of the induced emf?
Express your answer in terms of x, y, B, and t.

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E = xyB/t

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If the magnetic field steadily decreases from B to zero during a time interval t, what is the magnitude I of the induced current?
Express your answer in terms of x, y, B, t, and the resistance R of the wire.

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I = xyB/Rt

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If the magnetic field steadily decreases from B to zero during a time interval t, what is the direction of the induced current?

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counterclockwise
The flux decreases, so the induced magnetic field must be in the same direction as the original (parent) magnetic field. Therefore, the induced magnetic field is out of the page. Using the right-hand rule, we deduce that the direction of the current is counterclockwise.

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Which of the following changes would result in a clockwise emf in the loop? When you consider each option, assume that no other changes occur.
Check all that apply.

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The magnitude of Bâƒ— increases.
Clockwise emf implies that the induced magnetic field is directed into the page. Therefore, the magnetic flux of the original field must be increasing. Only the first option corresponds to increasing flux.

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Induced EMF and Current in a Shrinking Loop
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 163 cm , but its circumference is decreasing at a constant rate of 13.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.

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A and B

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Find the magnitude of the emf E induced in the loop after exactly time 3.00 s has passed since the circumference of the loop started to decrease.
Express your answer numerically in volts to three significant digits.

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2.05Ã—10âˆ’2 V
Radius Initially = 163 / 2 x 3.14 = 19.73 cm/s
rate of decrease of radius = 13 / 2 x 3.14 = 2.07 cm/s = dr/dt
E = dÏ†/dt
= B x dA/dt
A = Ï€r2
dA/dt = 2 Ï€ r (dr/dt)
r at five seconds = 19.73 - 3 x 2.07 = 15.44
E = 0.8 x 2 x 3.14 x (19.73x 10-2) x (2.07 x 10-2)

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Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

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clockwise

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Â± Magnetic Flux and Induced EMF in a Coil
In a physics laboratory experiment, a coil with 160 turns enclosing an area of 13.3 cm2 is rotated during the time interval 4.90Ã—10âˆ’2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.30Ã—10âˆ’5 T .

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A-C

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What is the total magnitude of the magnetic flux ( Î¦initial) through the coil before it is rotated?
Express your answer numerically, in webers, to at least three significant digits.

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phi=nBAcos(theta)
phi=(160)(5.3e-5)cos0
phi=1.13e-5 Wb

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What is the magnitude of the total magnetic flux Î¦final through the coil after it is rotated?
Express your answer numerically, in webers, to at least three significant digits.

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phi1=nBAcos(theta)
phi2=nBAcos(90)=0

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What is the magnitude of the average emf induced in the coil?
Express your answer numerically (in volts) to at least three significant digits.

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emf=derivative of phi
phi2-phi1/dt
(0-1.3e-5)/(5e-2)=
2.3e-4V

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Faraday's Law and Electric Fields
Learning Goal:
To understand the terms in Faraday's law for magnetic induction of electric fields, and contrast these fields with those produced by static charges.
Faraday's law describes how electric fields and electromotive forces are generated from changing magnetic fields. It relates the line integral of the electric field around a closed loop to the change in the total magnetic field integral across a surface bounded by that loop:
E=âˆ’dÎ¦Mdt=âˆ’ddtâˆ«Bâƒ— (râƒ— ,t)â‹…dAâƒ— ,
where E=âˆ®Eâƒ— (râƒ— ,t)â‹…dlâƒ— is the line integral of the electric field, and the magnetic flux is given by
Î¦M=âˆ«Bâƒ— (râƒ— ,t)â‹…dAâƒ— =âˆ«B|dA|cos(Î¸),
where Î¸ is the angle between the magnetic field Bâƒ— and the local normal to the surface bounded by the closed loop.
(Figure 1)
Direction: The line integral and surface integral reverse their signs if the reference direction of dlâƒ— or dAâƒ— is reversed. The right-hand rule applies here: If the thumb of your right hand points along dAâƒ— , then the fingers point along dlâƒ— . You are free to take the loop anywhere you choose, although usually it makes sense to choose it to lie along the path of the circuit you are considering.

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A-F

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Consider the direction of the electric field in the figure. Assume that the magnetic field points upward, as shown. Under what circumstances is the direction of the electric field shown in the figure correct?

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There are two approaches:
One approach is to realize that the electric field would produce a current in the same direction as the field if there were a wire present. Find the magnetic field due to this (imaginary) current using the
right-hand rule. The magnetic field produced by this current must oppose the change in the original magnetic field. You should be able to tell what the change is.
The other approach is more mathematical. Choose a direction for and . Suppose is in the direction of the electric field shown. Then has the same direction as that of the magnetic field
shown. The left-hand sid
if B(t) decreases with time

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Now consider the magnetic flux through a surface bounded by the loop. Which of the following statements about this surface must be true if you want to use Faraday's law to relate the magnetic flux to the line integral of the electric field around the loop?

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The surface can be any surface whose edge is the loop.
You are free to take any surface bounded by the loop as the surface over which to evaluate the integral. The result will always be the same, owing to the continuity of magnetic field lines (they never start or end anywhere, since there are no magnetic charges).

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When can an electric field be measured at any point from the force on a stationary test charge at that point?

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No matter how the field is generated
In fact, this operation defines an electric field. Similarly, if the test charge is moving, it will measure magnetic fields.

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When can an electric field that does not vary in time arise?

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Only if the field is generated by a coulomb field of static charges
Only if the field is generated by a changing magnetic field
*In either of the above two cases

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When will the integral âˆ®Eâƒ— â‹…dlâƒ— around any closed loop be zero?

answer

Only if the field is generated by the coulomb field of static charges or a constant current
The electric field generated by a static charge or a constant current always has zero loop integral. Even when the constant current takes the form of a continuous line of evenly-spaced charges moving with constant velocity, the resulting electric field will have a zero loop integral. An electric field generated by any other configuration of moving charges (moving through the loop) would have a non-zero loop integral.

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A cylindrical iron rod of infinite length with cross-sectional area A is oriented with its axis of symmetry coincident with the z axis of a cylindrical coordinate system as shown in the figure. It has a magnetic field inside that varies according to Bz(t)=B0+B1t. Find the theta component EÎ¸(R,t) of the electric field at distance R from the z axis, where R is larger than the radius of the rod. (Figure 2)
Express your answer in terms of A, B0, B1, R, and any needed constants such as Ïµ0, Ï€, and Î¼0.

answer

EÎ¸(R,t) = âˆ’AB12Ï€R

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Inductance of a Coaxial Cable
A coaxial cable consists of alternating coaxial cylinders of conducting and insulating material. Coaxial cabling is the primary type of cabling used by the cable television industry and is also widely used for computer networks such as Ethernet, on account of its superior ability to transmit large volumes of electrical signal with minimum distortion. Like all other kinds of cables, however, coaxial cables also have some inductance that has undesirable effects, such as producing some distortion and heating.

answer

Just Part A

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Consider a long coaxial cable made of two coaxial cylindrical conductors that carry equal currents I in opposite directions (see figure). The inner cylinder is a small solid conductor of radius a. The outer cylinder is a thin walled conductor of outer radius b, electrically insulated from the inner conductor. Calculate the inductance per unit length Ll of this coaxial cable. (Figure 1) ( L is the inductance of part of the cable and l is the length of that part.) Due to what is known as the "skin effect", the current I flows down the (outer) surface of the inner conducting cylinder and back along the outer surface of the outer conducting cylinder. However, you may ignore the thickness of the outer cylinder.
Express your answer in terms of some or all the variables I, a, b, and Î¼0, the permeability of free space.

answer

Ll = Î¼0ln(ba)/2Ï€
The capacitance per unit length Cl of such a coaxial (c/a)ble is 2Ï€Ïµ0/lnba. So the product L/lâ‹…C/l=Î¼oâ‹…Ïµ0=1/c^2, where c is the speed of light in vacuum! As you might have guessed, this is not a coincidence, but a result that is quite generally true for such systems.

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Â± Basic Properties of Inductors
Learning Goal:
To understand the units of inductance, the potential energy stored in an inductor, and some of the consequences of having inductance in a circuit.
After batteries, resistors, and capacitors, the most common elements in circuits are inductors. Inductors usually look like tightly wound coils of fine wire. Unlike capacitors, which produce a physical break in the circuit between the capacitor plates, the wire of an inductor provides an unbroken continuous path in which current can flow. When the current in a circuit is constant, an inductor acts essentially like a short circuit (i.e., a zero-resistance path). In reality, there is always at least a small amount of resistance in the windings of an inductor, a fact that is usually neglected in introductory discussions.
Recall that current flowing through a wire generates a magnetic field in the vicinity of the wire. If the wire is coiled , such as in a solenoid or an inductor, the magnetic field is strongest within the coil parallel to its axis. The magnetic field associated with current flowing through an inductor takes time to create, and time to eliminate when the current is turned off. When the current changes, an EMF is generated in the inductor, according to Faraday's law, that opposes the change in current flow. Thus inductors provide electrical inertia to a circuit by reducing the rapidity of change in the current flow.

answer

A-D

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Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?

answer

1H=1(Vs/A)

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What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so the current drops to essentially zero in 0.015 seconds?
Express your answer in volts to two significant figures.

answer

(560mH)(2.5A-0A)/(0.015s)=93V

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Which of the following changes would increase the potential energy stored in an inductor by a factor of 5?
Check all that apply.

answer

Increasing the inductance by a factor of 5; leaving the current unchanged
Leaving the inductance unchanged; increasing the current by a factor of 5âˆš
Reducing the inductance by a factor of 5; increasing the current by a factor of 5

question

Which of the graphs illustrate how the current through an inductor might possibly change over time?(Figure 1)
Type the numbers corresponding to the right answers in alphabetical order. Do not use commas. For instance, if you think that only graphs C and D are correct, type CD.

answer

ABC
All real circuits, even those that do not specifically have inductors in them, have at least a small amount of inductance, just as real inductors have a small amount of resistance in their windings. Circuit analysis in textbooks often assumes ideal batteries, resistors, capacitors, and inductors and hence neglects such subtle details of real circuits.

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Â± Energy Stored in an Inductor
The electric-power industry is interested in finding a way to store electric energy during times of low demand for use during peak-demand times. One way of achieving this goal is to use large inductors.
What inductance L would be needed to store energy E=3.0kWh (kilowatt-hours) in a coil carrying current I=300A?

answer

E=(L*I^2 )/2
where
L = value of the inductance in Henry
I = Value of current in amperes
to store E=3kWh = 3000watts for one hour
we need to convert this to joules
so, E = 3000*60*60 (it, 3000 watts = 3000 joules per second. 3000Watt hours = 3000 watts continuously for one hour = 3000 joules per 60*60 seconds)
E=10800000
then, just use the value of I=300, and get
L=2*E/I^2 = 240 Henry
This is probably not the best way to store energy: unless the coil is a superconductor, the amount of heat dissipated in the coil would be enormous. At this point, there is no way to produce large superconducting coils. Think of this problem as a practice exercise rather than a realistic example.