A coil of thin wire, consisting of many meters of wire, is subjected to the voltage of a hand-cranked generator. A strip of aluminum about 30 cm long is subjected to the same voltage. Which generator should be harder to crank? Should the long thin wire or the aluminum foil strip have the greater resistance? How does the answer to one question answer the other?
We can predict which will have the greater electrical resistance, and we can also interpret the feeling of mechanical resistance in the crank of a generator as it 'pumps' current across both.
Cranking at the same speed for both we feel more mechanical resistance with the aluminum strip, indicating that more energy is being dissipated. This indicates a greater current and therefore lesser resistance.
Reasoning in terms of resistivity and current-flow models we recall that for a given potential difference greater length implies a lesser potential gradient, which is the same thing as a lesser electrical field, resulting in less acceleration of charges between collisions and hence less current. Less cross-sectional area implies fewer available charge carriers per unit of length so that fewer charges will drift past a given point in a given time interval. So for a given material greater length and lesser cross-sectional area both imply less current for a given voltage. Less current for a given voltage implies greater electrical resistance.
There is a moderate difference in resistivity of the two materials, but the much greater length and possibly lesser cross-sectional area of the wire in the coil will more than overcome differences in the two conducting materials and the coil will be expected to have much greater resistance than the aluminum strip.
If we wish to use a single source to run a large current through an aluminum strip at the same time as we run as much current as possible through the wire loop, how should we connect the source to the two?
We could connect the source in series with the strip and the coil, or in parallel. If we connect a small and a large resistance in series, the total resistance will be just a little greater than that of the larger resistance. We will therefore get about the same current as if we had connected just to the coil, and this same current will pass through both the strip and the coil. The voltage of the source will drop mainly through the coil; there will be little voltage drop through the strip.
If we connect the strip and the coil in parallel, each will be subject to the total voltage of the source. This voltage will result in a much greater current through the strip, as compared to the current of the series combination. The voltage drop of the coil will be a little greater than in the series combination, where the coil didn't experience quite all the voltage of the source. So the aluminum strip will carry much more current, while the coil will carry slightly more than in the series combination.
It is therefore completely to our advantage to use a parallel combination.
When an aluminum strip is run horizontally just above a coil of wire, with the central axis of the wire vertical, and a current is run through both the coil and the strip, is there any effect on the strip? If so what do you think is the source of this effect?
We see that when charge is stored on a capacitor, which we then discharge in brief intervals through the circuit, the aluminum strip does experience a slight deflection. The effect is much smaller than when the same current was passed through the strip in the presence of a ceramic magnet.
It is difficult to determine the exact direction of the force on the strip, due to its greater mechanical resistance to deflection along its width.
Use the compass to investigate the magnetic field created when a current is run through the coil. In what direction(s) is the field strongest?
The compass aligns most strongly when placed along the axis of the coil, and aligns in the direction of the axis.
How is the direction of the current through the coil related to the direction of the magnetic field?
If the axis of the coil points West and current flows counterclockwise around the coil, as viewed from the West, then the compass indicates that the field points West. If the current (viewed from the West) flows clockwise around the coil then the field points to the East. The direction of the field is consistent at all points along the axis.
How is the direction of the current in a thin aluminum strip related to the direction of the magnetic field directly beneath the strip? What about the direction of the magnetic field directly above the strip?
When current travels from South (positive) to North (negative) is it claimed that the North-facing pole of a compass located directly beneath the strip points West, while the North-facing pole of a compass located directly above the strip points East.
Again, when the direction of the current is reversed the directions of the magnetic field are reversed.
Consider the aluminum strip to consist of a number of segments each 1 cm long. Suppose that the strip is carrying a current of 2 amps, directed to the North. If we consider the magnetic field source to be the product of the current and the length of the segment, then what are the magnitude and direction field source due to a 1 cm segment?
The current is 2 amps toward the North, and the length of the segment is 1 cm. So the source is 2 amps * 1 cm = 2 amps * .01 m = .02 amp meters, directed toward the North.
If the compass is placed 10 cm from the aluminum strip, positioned at the same height as the strip and so that a vector from the center of the 1 cm segment to the compass is perpendicular to the strip, then the magnitude of the magnetic field created by the strip at the position of the compass is governed by an inverse-square law
B = k ' * ( I `dL) / r^2 (if vector r is perpendicular to vector I `dL, and if r is much greater than `dL).
k ' is a constant equal to 10^-7 Tesla m / amp. Tesla is the unit of magnetic field. What is the magnitude of the field due to the 1 cm segment?
We have seen that for this segment I `dL = .02 amp meter. With the 10 cm separation we find that the field due to the segment is
B = k ' * ( I `dL) / r^2 = 10^-7 Tesla / (amp m) * (.02 amp m) / (.10 m)^2 = 2 * 10^-7 Tesla.
Note that the magnetic field of the Earth is on the order of 10^-4 Tesla, so this field won't significantly deflect a compass from the direction of the Earth's field.
At the position of the compass the direction of the magnetic field due to the segment is perpendicular to the vector r from the center of the segment to the compass and also to the direction of the current. Given only this much information there are two possible directions for the direction of the field. What are they?
The current flows horizontally toward the North and the r vector is perpendicular to the segment, so must be either East or West (this has not yet been specified). A vector perpendicular to North and to the East-West direction must be vertical. So the field points either straight up or straight down.
The decision between up and down for the direction of the magnetic field in the preceding question is decided by the right-hand rule. If your fingers point in the direction of the current and are positioned so that when your fingers are curled they move toward the direction of the r vector, then your thumb will point in the direction of the field. If the current moves toward the North and the compass is placed to the East of the strip, then what will be the direction of the field?
If you point the fingers of your right hand to the North, then in order for the fingers to curl in the Eastward direction the palm of your hand must fact to the East. Try this; you might have to twist your hand at an awkward angle.
If you do this your thumb will end up pointing downward.
The figure below depicts this situation, with the current running along the y axis to the right (regarded as North) and the compass positioned along the positive x axis.
If the compass was positioned below directly below the segment, with current running North, what would be the direction of the field at the position of the compass?
The r vector would be directed downward. Positioning the right hand with the fingers pointing North and ready to curl to a downward orientation, the thumb would be pointing to the West. So the field is directed to the West.
Answer the same question if the compass is positioned directly East of the segment, and again if the compass is vertically above the segment.
The r vector would point East in the first case, which would require the thumb to point vertically upward. For a vertical r vector the thumb would point to the East. So the field vectors would be, respectively, vertically upward and to the East.
The figure below depicts the field vectors as they would be seen looking at the segment 'end-on', looking in the direction of the current (i.e., for the present situation looking toward the North). Note how the vectors decrease rapidly in magnitude.
If the r vector is not perpendicular to the segment then it makes an angle theta, not 90 deg (which would make it perpendicular) with the segment. In this case the field is given by
Suppose that we want to find the field due to a 1 cm segment of the same current as in previous example, but that the segment is located 5 cm 'behind' the previous segment as measured along the aluminum strip. How far is this segment from the compass? What angle does the vector r from the middle of the segment to the compass make with the current segment? What is the sine of this angle?
The vector r from the midpoint of the segment to the compass is the hypotenuse of a right triangle whose legs are 5 cm (along the aluminum strip) and 10 cm (perpendicular to the strip). The Pythagorean Theorem tells us that the distance r is therefore sqrt(.05^2 + .10^2) m = sqrt(.0125) m = .112 m (note that we have converted distances to meters for this calculation, anticipating that we will need MKS units in our final calculation).
The angle is arctan ( 10 / 5) = arctan(2) = 63 deg, approx.
The sine of the angle is the ratio 10 / sqrt(125) = 0.894 approx.
What therefore is the magnitude of the magnetic field at the position of the compass?
The magnitude of the field is B = k ' (I L) / r^2 * sin(theta), with I L still equal to .02 amp m, r = sqrt(125) cm and sin(theta) = 10 / sqrt(125). Thus we have
What is the direction of this field?
The direction is perpendicular to the r vector and the direction of the segment. Both of these vectors lie in a horizontal plane so the direction of the field vector is still vertical; the right-hand rule still has the fingers in the direction of the current and positioned so that the fingers will curl toward the r vector, which will still require the thumb to point downward. So the field is vertically downward.
The figure below shows the situation for a short segment of length L lying x units 'behind' the position of the compass, in the sense that if we move directly from the compass to the aluminum strip along the shortest possible path (which will be perpendicular to the strip) we will then have to travel distance x along the strip to arrive at the center of the segment. The compass is at distance a from the strip, with a much greater than the length of the segment.
