Answers to questions posted by students
11. A block of mass 2.0 kg is released from rest at point P as shown in the figure below. It is found that the block finally comes to rest at point Q.
The length of the track PQ is 2.5 m What us the average value of frictional force acting on the block when it is moving from P to Q?
All the initial energy of the block was lost to work done against friction.
The average frictional force is 4.0 N.
7. A bullet of mass 0.010 kg travelling horizontally at 100 m/s is stopped after penetrating through 0.20 m of wood. What is the average retarding force applied to the bullet by the wood?
All the kinetic energy of the bullet was lost to work done against friction.
The frictional force is 250 N.
15) A magnetic field B is applied through a coil in a direction perpendicular to the plane of the coil as shown.
Use Faraday's law and Lenz's law to state and explain the direction of the induced current through resistor R when the magnetic field B is decreasing. [4]
When the magnetic flux density B decreases, the magnetic flux linkage of the coil into the page, given by the formula ϕ = B A, decreases. By Faraday's Law of electromagnetic induction, there will be an induced e.m.f. produced in the coil.
This induced e.m.f. in the coil will cause an induced current to flow.
By Lenz's law, since the magnetic flux linkage into the page has deceased, the induced current will produce magnetic flux into the page. According to the right hand grip rule, the induced current will flow in the clockwise direction as shown below.
Therefore, the current will flow through the resistor from the right to the left.
12) A long bar magnet is suspended from a helical spring so that one pole of the magnet lies within a long cylindrical coil as shown.
The magnet is given a small vertical displacement so that one pole of the magnet oscillates within the coil.
a) An induced e.m.f. is measured across the terminals of the coil.
i) Sketch the graph of e.m.f. against time to show how the induced e.m.f. varies with time for two periods of oscillation. [2]
ii) Using the principles of EM induction, explain the variation of the induced e.m.f. shown in your graph. [3]
b) State and explain a change that occurs in the amplitude of the oscillations of the magnet when a resistor is connected across XY. [3]
a) i) Assuming that the magnet is displacement upwards and then released, the induced e.m.f. is as shown:
a) ii) At the top-most position, the magnetic flux linkage is the least. This because the magnetic field lines of the bar magnet spread out so the fewer magnetic field lines pass through the bottom-most coil.
At the bottom-most position, the magnetic flux linkage is the greatest as most of the magnetic field lines pass through the bottom-most coil.
Assuming that the magnet is displaced upwards and released, the magnetic flux linkage varies as shown:
By using the relationship \( \varepsilon = -{{d \phi } \over {d t}}\) , the graph of the induced e.m.f. against time is obtained.
b) When a resistor is connected across XY, the induced e.m.f. will cause an induced current to flow.
According to Lenz's law, the induced current flowing in the coil will produce a magnetic field to oppose the change producing it. In this case, the magnetic field produced by the induced current will repel the magnet as it falls into the coil and attract the magnet as it rises away from the coil.
This causes the amplitude of the oscillation of magnetic to decrease with time. The kinetic energy of the magnet is converted into electrical energy which is then dissipated as thermal energy by the resistor.
The figure below shows the cross section of a long cylindrical magnet, as well as the plan view of the radial magnetic field. The cylindrical magnets are placed vertically. A thin circular aluminum ring of radius r and resistance R falls through the magnetic field as shown in the diagram. In the vicinity of the aluminum ring, themagnetic flux density is 2.0 T.
[Resistivity of aluminium = 3.1 x 10-8 Ω m, density of aluminium = 2.7 x 103 kg m-3]
a) By considering the plan view of the magnet, indicate the polarity of the magnets and the direction of the induced current.
b) Describe the motion of the aluminum ring when it is released in the radial magnetic field.
c) Show that the current in the aluminum ring when it is falling at a speed v is given by
\( I= {{2 \pi r B v} \over {R}}\)
d) Calculate the terminal velocity of the ring.
a) Magnetic field lines are from the North pole to the South pole, hence the outer ring is a North pole and the inner cylinder is a South pole.
Based on Fleming's Right hand rule, since the ring is moving into the page and the magnetic field lines are radially inwards, the induced current flows in the anti-clockwise direction.
b) When it is first released, it falls with an acceleration of 9.81 ms-2. As velocity of the the aluminium ring increases, the acceleration of the ring decreases until it eventually reaches zero. The ring has reached terminal velocity.
c)
d)
A long, straight wire, which carries a current, is placed in the vicinity of a circular coil. Using the definition of magnetic flux, explain the following statements.
a) When the wire is placed along the diameter of the coil, the net flux through the coil is zero.
b) When the wire is placed along the axis to the coil, the net flux through the coil is zero.
a) The perpendicular component of the magnetic flux density with respect to the area of the coil on the left side of the coil is into the page.
The perpendicular component of the magnetic flux density with respect to the area of the coil on the right side of the coil is out of the page.
By symmetry, the magnitude of the magnetic flux density on both sides are the same.
Hence, the resultant perpendicular component of the magnetic flux density is zero. Therefore, the magnetic flux through the coil is zero.
b) The magnetic field lines are along the surface of the coil. This means that the perpendicular component of the magnetic flux density with respect to the area of the coil is zero.
Hence, the magnetic flux through the coil is zero.
The photo on the right shows a refrigerator. Explain why the freezer compartment is often located at the top of a refrigerator.
The air that is cooled in the freezer becomes denser and sinks. Warmer air flows in to be cooled. This creates a convection current which allows the entire refrigerator to be cooled more effectively.
*Note that this is an old design of a refrigerator. It is still commonly used in low-end refrigerators. The shelves are usually made of wires and allow air to circulate freely in the fridge. Most modern refrigerator have separate compartments and separate cooling systems for each compartments.
This is not a post about questions by students, but about something which they might be interested in.
The solar eclipse occurred on 9th March 2016 and I was lucky enough to be able to view it from Marina Barrage with my significant other, Meng Choo. The weather in the morning was good and there were a few clouds obstructing the view of the sun only occasionally during the entire 2 hour period of the eclipse. In addition, since the eclipse occurred in the morning, it was much cooler.
I used a Celestron Travelscope 70 and a sun funnel to project the image of the sun onto a screen. This set-up uses a series of lenses to project an image on the screen and causes the image of the sun to be inverted on the screen. There was also the hassle of constantly adjusting the telescope to track the sun as it moved across the sky. I might consider getting a tracking mount in the future.
The barrage was not particularly crowded. There were some people at the barrage, and almost everyone could have a view of the eclipse on the sun funnel. While most brought out their camera, it was not easy to take pictures of the sun as the sun was extremely bright and caused the moon to be washed out. However, viewing through solar eclipse glasses made the progress of the eclipse clearly visible. I would recommend that everyone viewing solar eclipses to always bring a pair. The difference between viewing the sun through a camera and through the glasses is a very dramatic difference.
Typical image through a camera during the solar eclipse
Eclipse through solar eclipse glasses (Simulated Image)
The compiled photo of the projected image on the sun funnel is shown below. The eclipse started at around 7.20am, reached its peak at 8.24am and ended around 9.30am. The images have been flipped to show the correct orientation of the sun.
The next solar eclipse visible in Singapore will be on Dec 26 2019 around 1230pm. Since the sun will almost be directly overhead, it should be visible from any open space in Singapore. One thing to make sure is to order a pair solar eclipse glasses a few months ahead of time!
Below is a gallery of photos taken during the solar eclipse. A big thanks goes out to Meng Choo for taking most of the pictures. Click on images for a larger version.
Please contact me if you would like any of the full-res jpegs for non-commercial use.
On Fig, 8.1, sketch on the card the pattern of the magnetic field produced by the current in the wire. The detail of your sketch should suggest the variation in the strength of the field. Show the direction of the field with arrows.
A straight current carrying conductor will produce a circular magnetic field. The direction of the magnetic field can be determined by the right hand grip rule.
The direction of current is indicated by the thumb and the direction of the magnetic field is given by the fingers.
The magnetic field lines are closer together nearer the wire because the magnetic field is stronger nearer the wire.
Two charged sole metal spheres A and B are situated in a vacuum. Their centres separated by a distance of 12.0 cm as illustrated in Fig. 2.1. The diagram is not to scale.
Point P is a point on the line joining the centres of the two spheres. Point P is a distance x from the centre of sphere A.
The variation with distance X of the electric field strength E at point P is shown in Fig 2.2. A positive value of E on the graph at a point corresponds to an electric field vector pointing horizontally to the right.
State and explain:
1. the signs of charges of A and B.
2. the ratio of the radius of sphere A to that of sphere B.
3. how the electric potential varies inside spheres A and B.
1. Both A and B are positive.
Near A, the electric field strength is positive. This means that the electric field strength is in the direction towards the right and away from A. This means that a positive point test charge will experience a force away from A. Hence A is positive.
Near B, the electric field strength negative. This means that the electric field strength is in the direction towards the left and away from B. This means that a positive point test charge will experience a force away from B. Hence B is positive.
2. From the graph, we can see that the radius of A given by the region from x = 0.0 cm to x = 1.4 cm where the electric field strength is zero. Hence, the radius of sphere A is 1.4 cm.
The radius of sphere B is given by the the region from x = 11.4 cm to x = 12.0 cm where the electric field strength is zero. Hence, the radius of sphere B is 0.6 cm.
Therefore, the ratio is 7 : 3.
3. The electric potential remains constant inside sphere A and sphere B. This is because the of the relationship \(E = -{dV \over dx}\). Since E is 0 inside the spheres, V must be constant in order for \( {{dV} \over {dx}} \) to be 0.
(Since the spheres are conductors, the electric charges in the form of mobile electrons inside are free to move. If there is a potential difference, there will be an electric force on the electrons. These electrons will move and redistribute themselves until there is a constant electric potential inside the sphere and no longer any electric force on the electrons.)
