Current strength
( If ).
Current Density
Where S- square cross section conductor.
Current density in the conductor
where is the speed of ordered movement of charges in a conductor, n- charge concentration, e- elementary charge.
Dependence of resistance on conductor parameters
Where l- length of the conductor, S- cross-sectional area of the conductor, - resistivity, - specific conductivity.
Dependence of resistivity on temperature
,
where is the temperature coefficient of resistance, is the resistivity at .
Resistance for series (a) and parallel (b) connection of conductors
where is the resistance of the th conductor, n– number of conductors.
Ohm's law:
for a homogeneous chain section
,
for a non-uniform section of the chain
,
for closed circuit
Where U- voltage on a homogeneous section of the circuit, - potential difference at the ends of the circuit section, - emf of the source, r- internal resistance of the current source.
Short circuit current
Current work in time t
Current power
Joule-Lenz law (the amount of heat released when current passes through a conductor)
Current source power
Current source efficiency
.
Kirchhoff's rules
1) - for nodes;
2) 
- for contours,
where is the algebraic sum of the current strengths converging at the node, is the algebraic sum of the EMF in the circuit.
2.1.
A voltage of 1 V is maintained at the ends of a copper wire 5 m long. Determine the current density in the wire (copper resistivity 
).
A. 
B. 
S. D. 
2.2. A 5 Ohm resistor, a voltmeter and a current source are connected in parallel. The voltmeter shows a voltage of 10 V. If you replace the resistor with another with a resistance of 12 Ohms, the voltmeter will show a voltage of 12 V. Determine the emf and internal resistance of the current source. Neglect the current through the voltmeter.
A. B.
S. D.
2.3. Determine the current strength in a circuit consisting of two elements with an emf equal to 1.6 V and 1.2 V and internal resistances of 0.6 Ohm and 0.4 Ohm, respectively, connected by poles of the same name.
A. B. C. D.
2.4. The galvanic element gives a current of 0.2 A to an external resistance of 0.5 Ohm. If the external resistance is replaced by 0.8 Ohm, then the current in the circuit is 0.15 A. Determine the short circuit current.
A. B. C. D.
2.5. A load is connected to a current source with an emf of 12 V. The voltage at the source terminals is 8 V. Determine the efficiency of the current source.
A. B. C. D.
2.6. The external current source circuit consumes 0.75 W. Determine the current strength in the circuit if the source emf is 2V and the internal resistance is 1 ohm.
A. B. C. D.
2.7. A current source with an emf of 12 V and an internal resistance of 1 Ohm is connected to a load with a resistance of 9 Ohms. Find: 1) the current strength in the circuit, 2) the power released in the external part of the circuit, 3) the power lost in the current source, 4) the total power of the current source, 5) the efficiency of the current source.
2.8. The winding of an electric boiler has two sections. If one section is turned on, the water boils after 10 minutes, if the other, then after 20 minutes. How many minutes will it take for water to boil if both sections are turned on: a) sequentially; b) in parallel? The voltage at the boiler terminals and the efficiency of the installation are considered the same in all cases.
A. [a) 30 minutes, b) 6.67 minutes] B. [a) 6.67 minutes; b) 30 min]
C. [a) 10 min; b) 20 min] D. [a) 20 min; b) 10 min]
2.9. An ammeter with a resistance of 0.18 Ohm is designed to measure current up to 10 A. What resistance should be taken and how to turn it on so that this ammeter can measure current up to 100 A?
A.V.
S. D.
2.10. A voltmeter with a resistance of 2000 Ohms is designed to measure voltages up to 30 V. What resistance should be taken and how to turn it on so that this voltmeter can measure voltages up to 75 V?
A.V.
S. D.
2.11 .* The current in a conductor with a resistance of 100 Ohms increases uniformly from 0 to 10 A over 30 s. What is the amount of heat released in the conductor during this time?
A. B. C. D.
2.12.* The current in a conductor with a resistance of 12 Ohms decreases uniformly from 5 A to 0 within 10 s. How much heat is released in the conductor during this time?
A. B. C. D.
2.13.* A uniformly increasing current flows through a conductor with a resistance of 3 ohms. The amount of heat released in the conductor in 8 s is equal to 200 J. Determine the charge flowing through the conductor during this time. IN starting moment time the current was zero.
A. B. C. D.
2.14.* The current in a conductor with a resistance of 15 Ohms increases uniformly from 0 to a certain maximum within 5 s. During this time, an amount of heat of 10 kJ was released in the conductor. Find the average current in the conductor over this period of time.
A. B. C. D.
2.15.* The current in the conductor increases uniformly from 0 to a certain maximum value within 10 s. During this time, an amount of heat of 1 kJ was released in the conductor. Determine the rate of increase of current in the conductor if its resistance is 3 ohms.
A. B. C. D.
2.16. In Fig. 2.1 = = , R 1 = 48 Ohm, R 2 = 24 Ohm, voltage drop U 2 across resistance R 2 is 12 V. Neglecting the internal resistance of the elements, determine the current strength in all sections of the circuit and the resistance R 3 .
 |
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Rice. 2.1 Fig. 2.2 Fig. 2.3
2.17. In Fig. 2.2 = 2V, R 1 = 60 Ohm, R 2 = 40 Ohm, R 3 = R 4 = 20 Ohm, R G = 100 Ohm. Determine the current I G through the galvanometer.
2.18. Find the current strength in the individual branches of the Wheatstone bridge (Fig. 2.2) provided that the current strength passing through the galvanometer is zero. EMF source 2V, R 1 = 30 Ohm, R 2 = 45 Ohm, R 3 = 200 Ohm. Neglect the internal resistance of the source.
2.19. In Fig. 2.3 = 10 V, = 20 V, = 40 V, and resistance R 1 = R 2 = R 3 = 10 Ohm. Determine the strength of currents through resistances ( I) and through sources (). Neglect the internal resistance of the sources. [ I 1 =1A, I 2 =3A, I 3 =2A, =2A, =0, =3A]
2.20. In Fig. 2.4 = 2.1 V, = 1.9 V, R 1 = 45 Ohm, R 2 = 10 Ohm, R 3 = 10 Ohm. Find the current strength in all sections of the circuit. Neglect the internal resistance of the elements.
Rice. 2.4 Fig. 2.5 Fig. 2.6
2.21. In Fig. 2.5 the resistances of the voltmeters are equal to R 1 =3000 Ohms and R 2 =2000 Ohms; R 3 =3000 Ohm, R 4 =2000 Ohm; =200 V. Find the voltmeter readings in the following cases: a) key TO open, b) key TO closed. Neglect the internal resistance of the source. [a)U 1 =120 V, U 2 =80 V, b)U 1 =U 2 =100 V]
2.22. In Fig. 2.6 = =1.5 V, internal resistances of the sources r 1 =r 2 =0.5 Ohm, R 1 =R 2 = 2 Ohm, R 3 = 1 Ohm. The resistance of the milliammeter is 3 ohms. Find the milliammeter reading.
2.23.
In Fig. 2.7 = = 110 V, R 1 = R 2 = 200 Ohm, voltmeter resistance 1000 V. Find the voltmeter reading. Neglect the internal resistance of the sources.
Rice. 2.7 Fig. 2.8 Fig. 2.9
2.24. In Fig. 2.8 = 2V, internal resistances of the sources are 0.5 Ohm, R 1 = 0.5 Ohm, R 2 = 1.5 Ohm. Find the current strength in all sections of the circuit.
2.25. In Fig. 2.9 = = 100 V, R 1 = 20 Ohm, R 2 = 10 Ohm, R 3 = 40 Ohm, R 4 = 30 Ohm. Find the ammeter reading. Neglect the internal resistance of the sources and the ammeter.
2.26. What current strength is shown by the ammeter in Fig. 2.10, the resistance of which is R A = 500 Ohm, if = 1 V, = 2 V, R 3 = 1500 Ohm and the voltage drop across the resistance R 2 is 1 V. Neglect the internal resistance of the sources.
2.27. In Fig. 2.11 =1.5 V, =1.6 V, R 1 =1 kOhm, R 2 =2 kOhm. Determine the voltmeter readings if its resistance R V = 2 kOhm. Neglect source resistance.
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Rice. 2.10 Fig. 2.11 Fig. 2.12
2.28. In Fig. 2.12 resistance R 1 = 5 Ohm, R 2 = 6 Ohm, R 3 = 3 Ohm. Find the ammeter reading if the voltmeter shows 2.1 V. Neglect the resistance of the source and the ammeter.
2.29 . Determine the emf of the source in the circuit in Fig. 2.13, if the current flowing through it is 0.9 A, the internal resistance of the source is 0.4 Ohm. R 1 =30 Ohm, R 2 =24 Ohm, R 3 =50 Ohm, R 4 =40 Ohm, R 5 =60 Ohm.
2.30. Find the ammeter readings in the circuit in Fig. 2.14, if the EMF is 19.8 V, the internal resistance is 0.4 Ohm, R 1 = 30 Ohm, R 2 = 24 Ohm, R 3 = 50 Ohm, R 4 = 40 Ohm, R 5 = 60 Ohm.
 |
Rice. 2.13 Fig. 2.14 Fig. 2.15
2.31 . Find the values of all resistances in the circuit in Fig. 2.15, if a current of 0.4 μA flows through resistance R 1, a current of 0.7 μA flows through resistance R 2, 1.1 μA through resistance R 3, and no current flows through resistance R 4. Neglect the internal resistance of the elements. E 1 =1.5 V; E 2 =1.8 V.
Rice. 2.16 Fig. 2.17 Fig. 2.18
2.32. Determine E 1 and E 2 in the diagram in Fig. 2.16, if R 1 = R 4 = 2 Ohm, R 2 = R 3 = 4 Ohm. The current flowing through resistance R 3 is 1A, but no current flows through resistance R 2. Internal resistance of elements r 1 =r 2 =0.5 Ohm.
2.33. Determine the current strength in all sections of the circuit in the circuit in Fig. 2.17, if E 1 =11 V, E 2 =4 V, E 3 =6 V, R 1 =5 Ohm, R 2 =10 Ohm, R 3 =2 Ohm. Internal resistance of sources r 1 =r 2 =r 3 =0.5 Ohm.
2.34. In the diagram in Fig. 2.18 R 1 =1 Ohm, R 2 =2 Ohm, R 3 =3 Ohm, current through the source is 2A, potential difference between points 1 And 2 is equal to 2 V. Find the resistance R 4.
Electromagnetism
Basic formulas
Magnetic induction is related to tension magnetic field ratio
Where 
- magnetic constant,
Magnetic permeability of an isotropic medium.
The principle of superposition of magnetic fields
where is the magnetic induction created by each current or moving charge separately.
Magnetic field induction created by an infinitely long straight conductor carrying current,
where is the distance from the current-carrying conductor to the point at which magnetic induction is determined.
Magnetic induction of a field created by a straight conductor carrying a current of finite length
,
where are the angles between the current element and the radius vector drawn from the point in question to the ends of the conductor.
Magnetic field induction in the center of a circular conductor with current
where is the radius of the circular turn.
Magnetic field induction on the axis of a circular conductor with current
,
where is the radius of the circular coil, is the distance from the center of the coil to the point at which the magnetic induction is determined.
Magnetic field induction inside a toroid and an infinitely long solenoid
where is the number of turns per unit length of the solenoid (toroid).
Magnetic field induction on the axis of a solenoid of finite length
,
where are the angles between the coil axis and the radius vector drawn from a given point to the ends of the coil.
The Ampere force acting on a current-carrying conductor element in a magnetic field is
where is the angle between the directions of current and magnetic field induction.
Magnetic moment of a current-carrying circuit
where is the contour area,
Unit vector normal (positive) to the contour plane.
The torque acting on a current-carrying circuit placed in a uniform magnetic field is
,
where is the angle between the direction of the normal to the contour plane and the magnetic induction of the field.
The force of interaction between two straight parallel conductors carrying currents and
,
where is the length of the conductor, is the distance between them.
Magnetic flux through the pad
where , is the angle between the direction of the magnetic induction vector and the normal to the site.
Magnetic flux of a non-uniform field through an arbitrary surface
where integration is carried out over the entire surface.
Magnetic flux of a uniform field through a flat surface
Work done by moving a current-carrying conductor in a magnetic field
where is the flux of magnetic induction crossed by the conductor as it moves.
The Lorentz force acting on a moving charged particle in a magnetic field is
where is the charge of the particle, is the speed of the particle, is the angle between the directions of the particle speed and the magnetic induction of the field.
E.D.S. induction
The potential difference at the ends of a conductor moving in a magnetic field is
where is the speed of movement of the conductor, is the length of the conductor, is the angle between the directions of the speed of movement of the conductor and the magnetic induction of the field.
E.D.S. self-induction
where is the inductance of the circuit.
Solenoid inductance
,
where is the cross-sectional area of the solenoid, is the length of the solenoid, is the total number of turns.
Magnetic field energy of a current-carrying circuit
Volumetric magnetic field energy density
.
3.1.
In Fig. Figure 3.1 shows a cross section of two straight infinitely long conductors carrying current. The AC distance between the conductors is 10 cm, I 1 = 20 A, I 2 = 30 A. Find the magnetic induction of the field caused by the currents I 1 and I 2 at points M 1, M 2 and M 3. Distances M 1 A = 2 cm, AM 2 = 4 cm and CM 3 = 3 cm.
A.V.
S. D.
3.2. Solve the previous problem provided that currents flow in one
direction.
A.V.
S. D.
3.3. Two straight infinitely long conductors are located perpendicular to each other and are in the same plane (Fig. 3.2). Find the magnetic field induction at points M 1 and M 2 if I 1 = 2 A and I 2 = 3 A. Distances AM 1 = AM 2 = 1 cm, DM 1 = CM 2 = 2 cm.
 |  |
||
Rice. 3.2 Fig. 3.3
A.V.
S. D.
3.4. Two straight infinitely long conductors are located perpendicular to each other and are in mutually perpendicular planes (Fig. 3.3). Find the magnetic field induction at points M 1 and M 2 if I 1 = 2 A and I 2 = 3 A. Distances AM 1 = AM 2 = 1 cm and AC = 2 cm.
A.V.
S. D.
3.5. In Fig. Figure 3.4 shows a cross section of three straight infinitely long conductors carrying current. Distances AC=CD=5 cm; I 1 =I 2 =I; I 3 =2I. Find the point on the straight line AD at which the magnetic field induction caused by the currents I 1, I 2, I 3 is zero.
 |
A. B.
S. D.
3.6. Solve the previous problem provided that all currents flow in the same direction.
A. B.
C.D.
3.7. Two circular turns with a radius of 4 cm each are located in parallel planes at a distance of 0.1 m from each other. Currents flow through the turns I 1 = I 2 = 2 A. Find the magnetic induction of the field on the axis of the turns at a point located at an equal distance from them. Currents in turns flow in one direction.
A. B. C. D.
3.8. Solve the previous problem provided that the currents flow in opposite directions.
A. B. C. D.
3.9. A current of 2A flows through a long conductor bent at an angle. Find the magnetic induction of the field at a point lying on the bisector of this angle and located at a distance of 10 cm from the vertex of the angle.
A. B. C. D.
3.10. Along a conductor bent into a rectangle with sides A= 8 cm and V= 12 cm, current flows I= 50 A. Determine the strength and magnetic induction of the field at the intersection point of the diagonals of the rectangle.
A.V.
S. D.
3.11. A current of force I = 2 A flows through a wire frame shaped like a regular hexagon. In this case, a magnetic field B = 41.4 μT is formed in the center of the frame. Find the length of the wire from which the frame is made.
A. B. C. D.
3.12. A current flows through a conductor bent in the shape of a circle. Magnetic field at the center of the circle B = 6.28 µT. Without changing the current strength in the conductor, it was given the shape of a square. Determine the magnetic induction of the field at the point of intersection of the diagonals of this square.
A. B. D.
3.13. The solenoid winding contains two layers of wire turns tightly adjacent to each other with a diameter of d = 0.2 mm. Determine the magnetic induction of the field on the axis of the solenoid if a current I = 0.5 A flows through the wire.
A. B. C. D.
3.14. A thin ring with a mass of 15 g and a radius of 12 cm carries a charge uniformly distributed with a linear density of 10 nC/m. The ring rotates uniformly with a frequency of 8 s -1 relative to an axis perpendicular to the plane of the ring and passing through its center. Determine the ratio of the magnetic moment of the circular current created by the ring to its angular momentum.
A. B. C. D.
3.15. Two infinitely long straight parallel conductors, the distance between which is 25 cm, carry currents of 20 and 30 A in opposite directions. Determine the magnetic induction of the field at a point located at a distance of 30 cm from the first and 40 cm from the second conductor.
A. B. C. D. [27.0 µT]
3.16. Determine the magnetic induction of the field on the axis of a thin wire ring with a radius of 10 cm, through which a current of 10 A flows, at a point located at a distance of 15 cm from the center of the ring.
A. B. C. D.
3.17. A direct current of 3 A flows through a wire bent into a square with a side of 60 cm. Determine the magnetic induction of the field in the center of the square.
A. B. C. D.
3.18. A current flowing through a wire ring made of copper wire with a cross-section of 1.0 mm 2 creates a magnetic field induction of 0.224 mT in the center of the ring. The potential difference applied to the ends of the wire forming the ring is 0.12 V. What current flows through the ring?
A. B. C. [ 2 A] D.
3.19. A current of 2 A, flowing through a coil 30 cm long, creates a magnetic field induction of 8.38 mT inside it. How many turns does the coil contain? The diameter of the coil is considered small compared to its length.
A. B. C. D.
3.20. An infinitely long wire forms a circular loop tangent to the wire. The radius of the loop is 8 cm. A current of 5A flows through the wire. Find the magnetic field induction at the center of the loop.
A. B. C. D.
3.21*.
Find the distribution of magnetic field induction along the axis of a circular coil with a diameter of 10 cm, through which a current of 10 A flows. Make a table of values for values in the interval 0-10 cm every 2 cm and draw a graph with a scale. [ 
] .
3.22*. Determine, using the vector circulation theorem, the magnetic induction of the field on the axis of a toroid without a core, through the winding of which, containing 300 turns, a current of 1A flows. The outer diameter of the toroid is 60 cm, the inner diameter is 40 cm.
3.23. Two infinite rectilinear parallel conductors with identical currents flowing in the same direction are located at a distance R from each other. To move them apart to a distance of 3R, 220 nJ of work is spent on each centimeter of the length of the conductor. Determine the current strength in the conductors.
A. B. C. D.
3.24. A straight conductor 20 cm long, through which a current of 40 A flows, is in a uniform magnetic field with an induction of 0.5 Tesla. How much work does the field force do to move the conductor 20 cm if the direction of movement is perpendicular to the lines of magnetic induction and the conductor.
A. B. C. D.
3.25. In a uniform magnetic field, the induction of which is 0.5 T, a conductor moves uniformly at a speed of 20 cm/s perpendicular to the field. The length of the conductor is 10 cm. A current of 2A flows through the conductor. Find the power expended to move the conductor.
A. B. C. D.
3.26. Magnetic induction of a uniform field 0.4 Tesla. In this field, a conductor 1 m long moves uniformly at a speed of 15 cm/s so that the angle between the conductor and the field induction is equal to . A current of 1A flows through the conductor. Find the work done by moving the conductor during 10 s of movement.
A. B. C. D.
3.27. A conductor 1 m long is located perpendicular to a uniform magnetic field with an induction of 1.3 Tesla. Determine the current in a conductor if, when moving at a speed of 10 cm/s in a direction perpendicular to
field and the conductor, in 4 s the energy of 10 J is consumed to move the conductor.
A. B. C. D.
3.28. In a uniform magnetic field with an induction of 18 μT in a plane perpendicular to the induction lines, there is a flat circular frame consisting of 10 turns with an area of 100 cm 2 each. A current of 3A flows in the frame winding. What should be the direction of the current in the frame so that when it is rotated around one of the diameters, the field forces do positive work? What is the magnitude of this work?
A. B. C. D.
3.29. A square circuit with a side of 20 cm, through which a current of 20 A flows, is freely established in a uniform magnetic field with an induction of 10 mT. Determine the change in the potential energy of the contour when rotating around an axis lying in the plane of the contour through an angle .
A. B. C. D.
3.30. A current of 10A flows through a circular coil with a radius of 15 cm. The coil is located in a uniform magnetic field with an induction of 40 mT so that the normal to the plane of the circuit makes an angle with the magnetic induction vector. Determine the change in the potential energy of the contour when it is rotated through an angle in the direction of increasing the angle.
A. B. C. D.
3.31. A circular frame with a current of area 20 cm 2 is fixed parallel to a magnetic field with an induction of 0.2 T, and a torque of 0.6 mN m is applied to it. When the frame was released, it turned and its angular velocity became 20 s -1. Determine the strength of the current flowing in the frame.
A. B. C. D. [15 A]
3.32. Two long horizontal conductors are parallel to each other at a distance of 8 mm. The upper conductor is fixed motionless, and the lower one hangs freely below it. How much current must be passed through the upper wire so that the lower one can hang without falling? A current of 1A flows through the lower one and the mass of each centimeter of conductor length is 2.55 mg.
A. B. C. D.
3.33 . The magnetic flux through the cross-sectional area of the solenoid (without core) is 5 μWb. The length of the solenoid is 35 cm. Determine the magnetic moment of this solenoid.
A. B. C. D.
3.34. A circular contour is placed in a uniform magnetic field so that the plane of the contour is perpendicular to the field lines. Magnetic field induction 0.2 Tesla. A current of 2A flows through the circuit. The radius of the contour is 2 cm. What work is done when the contour is rotated by ?
A. B. C. D.
3.35*. Next to a long straight wire carrying a current of 30A, there is a square frame with a current of 2A. The frame and wire lie in the same plane. The axis of the frame passing through the middles of the opposite sides is parallel to the wire and spaced from it at a distance of 30 mm. Frame side 20 mm. Find the work that needs to be done to rotate the frame around its axis by . .
3.36*. Two straight long conductors are located at a distance of 10 cm from each other. Currents of 20A and 30A flow through the conductors. How much work per unit length of conductors must be done to move these conductors apart to a distance of 20 cm? .
3.37. A proton, accelerated by a potential difference of 0.5 kV, flying into a uniform magnetic field with an induction of 0.1 T, moves in a circle. Determine the radius of this circle.
A. B. C. D.
3.38. An alpha particle flies into a magnetic field with an induction of 1 Tesla at an angle of 2 mm/s. Determine the radius of the helix that the alpha particle will describe?
A. B. C. D.
3.39. A magnetic field with an induction of 126 μT is directed perpendicular to an electric field whose strength is 10 V/m. An ion flying at a certain speed flies into these crossed fields. At what speed will it move in a straight line?
A. B. C. D.
3.40. An electron, accelerated by a potential difference of 6 kV, flies into a uniform magnetic field at an angle to the direction of the field and begins to move along a helical line. The magnetic induction of the field is 130 mT. Find the pitch of the helix.
A. B. C. [ 1.1 cm] D.
3.41. A proton flew into a uniform magnetic field at an angle to the direction of the field lines and moves in a spiral, the radius of which is 2.5 cm. The magnetic induction of the field is 0.05 Tesla. Find the kinetic energy of the proton.
A.V.
S. D.
3.42. Determine the frequency of revolution of an electron in a circular orbit in a magnetic field with an induction of 1 Tesla. How will the rotation frequency change if an alpha particle rotates instead of an electron?
3.43. A proton and an alpha particle, accelerated by the same potential difference, fly into a uniform magnetic field. How many times is the radius of curvature of the proton trajectory smaller than the radius of curvature of the alpha particle trajectory?
A. B. C. D.
3.44. A particle carrying one elementary charge flew into a uniform magnetic field with an induction of 0.05 Tesla. Determine the angular momentum that the particle had when moving in a magnetic field if its trajectory was an arc of a circle with a radius of 0.2 mm.
A.V.
S. D.
3.45. An electron moves in a circle in a uniform magnetic field with an induction of 31.4 mT. Determine the electron's orbital period.
A. B. C. D.
3.46. Find the ratio q/m for a charged particle if it, flying at a speed of 10 8 cm/s into a uniform magnetic field with a strength of 2 10 5 A/m, moves along a circular arc with a radius of 8.3 cm. The direction of the speed of the particle is perpendicular to the direction magnetic field.
A. B. C. D.
3.47. An electron, accelerated by a potential difference of 3 kV, flies into the magnetic field of the solenoid at an angle to its axis. The number of ampere-turns of the solenoid is 5000. The length of the solenoid is 26 cm. Find the pitch of the helical trajectory of the electron in the magnetic field of the solenoid.
A. B. C. D.
3.48. A charged particle moves in a magnetic field in a circle at a speed of 1 Mm/s. The magnetic induction of the field is 0.3 Tesla. The radius of the circle is 4 cm. Find the charge of the particle if it is known that its kinetic energy is 12 keV.
A.V.
S. D.
3.49*. The Serpukhov proton accelerator accelerates these particles to an energy of 76 GeV. If we ignore the presence of accelerating gaps, we can assume that accelerated protons move along a circle of radius 236 m and are held there by a magnetic field perpendicular to the orbital plane. Find the magnetic field required for this. .
3.50*. The charged particle passed through an accelerating potential difference of 104 V and flew into the electric (E = 100 V/m) and magnetic (B = 0.1 T) fields crossed at right angles. Determine the ratio of the particle's charge to its mass if, moving perpendicular to both fields, the particle does not experience deviations from a rectilinear trajectory. .
3.51. In a uniform magnetic field with an induction of 0.1 Tesla, a frame containing 1000 turns rotates uniformly. Frame area 150 cm2. The frame makes 10 rps. Determine the maximum emf. induction frame. The axis of rotation lies in the plane of the frame and is perpendicular to the direction of the field.
A. B. C. D.
3.52. The wire coil is located perpendicular to the magnetic field, the induction of which varies according to the law B = B o (1 + e to t), where B o = 0.5 T, k = 1 s -1. Find the magnitude of the emf induced in the coil at a time equal to 2.3 s. The coil area is 0.04 m2.
A. B. C. D.
3.53. A square frame made of copper wire is placed in a magnetic field with an induction of 0.1 T. The cross-sectional area of the wire is 1 mm 2, the frame area is 25 cm 2. The normal to the plane of the frame is parallel to the field lines. What charge will pass through the frame when the magnetic field disappears? The resistivity of copper is 17 nOhm m.
A. B. C. D.
3.54. A ring of aluminum wire is placed in a magnetic field perpendicular to the magnetic induction lines. Ring diameter 20 cm, wire diameter 1 mm. Determine the rate of change of the magnetic field if the strength of the induction current in the ring is 0.5A. The resistivity of aluminum is 26 nOhm m.
A. B. C. D.
3.55. In a magnetic field, the induction of which is 0.25 T, a rod 1 m long rotates with a constant angular velocity 20 rad/s. The axis of rotation passes through the end of the rod parallel to the field lines. Find the e.m.f. induction occurring at the ends of the rod.
A. B. C. D.
3.56. A ring of wire with a resistance of 1 mOhm is placed in a uniform magnetic field with an induction of 0.4 Tesla. The plane of the ring makes an angle with the induction lines. Determine the charge that will flow through the ring if it is pulled out of the field. The area of the ring is 10 cm2.
A. B. C. D.
3.57. A coil containing 10 turns, each with an area of 4 cm 2, is placed in a uniform magnetic field. The coil axis is parallel to the field induction lines. The coil is connected to a ballistic galvanometer with a resistance of 1000 ohms, the resistance of the coil can be neglected. When the coil was pulled out of the field, 2 µC flowed through the galvanometer. Determine the field induction.
A. B. C. D.
3.58. A wire is wound in one layer on a rod of non-magnetic material with a length of 50 cm and a cross-section of 2 cm 2 so that for every centimeter of the length of the rod there are 20 turns. Determine the energy of the magnetic field of the solenoid if the current strength in the winding is 0.5A.
A. B. C. D.
3.59. Find the potential difference at the ends of the car's axle that occurs when it moves horizontally at a speed of 120 km/h, if the length of the axle is 1.5 m and the vertical component of the earth's magnetic field strength is 40 A/m.
A. B. C. D.
3.60. A coil of wire is placed on a solenoid with a length of 20 cm and a cross-sectional area of 30 cm2. The solenoid winding has 320 turns and carries a current of 3A. What is the e.m.f. is induced in a coil placed on the solenoid when the current in the solenoid disappears within 0.001 s?
A. B. C. [0.18 V] D.
3.61. A coil with a diameter of 10 cm and 500 turns is placed in a magnetic field. The axis of the coil is parallel to the lines of magnetic induction field. What is the average value of emf? induction in the coil if the magnetic induction of the field increases within 0.1 s from zero to 2 Tesla?
A. B. C. D.
3.62*. A flywheel with a diameter of 3 m rotates around horizontal axis at a speed of 3000 rpm. Determine the emf induced between the rim and the axis of the wheel if the plane of the wheel makes an angle with the plane of the magnetic meridian. The horizontal component of the earth's magnetic field is 20 µT. .
3.63*. A copper hoop with a mass of 5 kg is located in the plane of the magnetic meridian. What charge is induced in it if it is rotated about the vertical axis by ? The horizontal component of the earth's magnetic field is 20 µT. The density of copper is 8900 kg/m 3, the resistivity of copper is 17 nOhm m. .
3.64*. In a uniform magnetic field, the induction of which is 0.5 T, a coil containing 200 turns, tightly adjacent to each other, rotates uniformly with a frequency of 300 min -1. The cross-sectional area of the coil is 100 cm2. The axis of rotation is perpendicular to the axis of the coil and the direction of the magnetic field. Determine the maximum emf induced in the coil. .
The classification of any electrical wire includes the main parameters represented by conductivity, cross-sectional area or diameter, materials from which the conductor is made, typical insulation protection features, level of flexibility, and thermal resistance indicators.
The area or cross-section of the conductor is one of the most important criteria for selecting a wire.
Most wide application find wire brands PUNP and PUGNP, as well as VPP, PHCB and PKGM, which have the following basic technical characteristics that are very important for obtaining a safe connection:
- PUNP- a flat wire product of the installation or so-called installation type, with single-wire copper cores in PVC insulation. This type is distinguished by the number of cores, as well as a rated voltage within 250 V with a frequency of 50 Hz and an operating temperature from minus 15 °C to plus 50 °C;
- PUGNP- flexible variety with stranded cores. The main indicators, which are represented by the nominal voltage level, frequency and temperature operating conditions, do not differ from similar data from PUNP;
- APB- aluminum single-core variety, round wire with protective PVC insulation and a single-wire or multi-wire core. The difference between this type is its resistance to damage. mechanical type, vibrations and chemical compounds. The operating temperature ranges from minus 50 °C to plus 70 °C;
- PBC- multi-core copper variety with PBX insulation, which gives the wire high density and a traditional round shape. The heat-resistant core is designed for a nominal level of 380 V at a frequency of 50 Hz;
- PKGM- a power installation type, represented by a single-core copper wire with silicone rubber or fiberglass insulation impregnated with a heat-resistant compound. The operating temperature ranges from minus 60 °C to plus 180 °C;
- PHCB- heating single-core variety in the form of a single-wire wire based on galvanized or blued steel. The operating temperature ranges from minus 50 °C to plus 80 °C;
- runway- single-core copper variety with stranded core and PBX or polyethylene insulation. The operating temperature ranges from minus 40 °C to plus 80 °C.
In conditions of low power, copper wire ШВП with protective external PBX insulation is used. The stranded core has excellent flexibility, and the wire product itself is designed for a maximum of 380 V, with a frequency within 50 Hz.
Wire products of the most common types are sold in coils, and most often have white insulation.
Conductor cross-sectional area
IN last years there is a noticeable decrease in the quality characteristics of manufactured cable products, as a result of which the resistance indicators - the cross-section of the wires - suffer. The diameter of any conductor in mandatory must comply with all parameters declared by the manufacturer.
Any deviation of even 15-20% can cause significant overheating of the electrical wiring or melting of the insulating material, therefore, the choice of conductor area or thickness must be given increased attention not only in practice, but also from a theoretical point of view.
Conductor cross-section
Parameters most important for the right choice conductor cross-sections are reflected in the following recommendations:
- the thickness of the conductor is sufficient for the unhindered passage of electric current, with the maximum possible heating of the wire within 60 °C;
- conductor cross-section is sufficient for sharp decline voltage not exceeding permissible values, which is especially important for very long electrical wiring and significant currents.
Special attention needs to be paid maximum performance worker temperature regime, above which the conductor and protective insulation become unusable.
The cross-section of the conductor used and its protective insulation must necessarily ensure full mechanical strength and reliability of the electrical wiring.
Conductor cross section formula
As a rule, wires have a circular cross-section, but the permissible current ratings must be calculated according to the cross-sectional area. In order to independently determine the cross-sectional area in a single-core or stranded wire, the sheath, which is the insulation, is carefully opened, after which the diameter in the single-core conductor is measured.
The area is determined in accordance with the physical formula well known even to schoolchildren:
S = π x D²/4 or S = 0.8 x D², where:
- S is the cross-sectional area in mm2;
- π - number π, standard value equal to 3.14;
- D is the diameter in mm.
Conductor
Measurements of a stranded wire will require its preliminary fluffing, as well as subsequent counting of the number of all veins inside the bundle. The diameter of one component element is then measured and the cross-sectional area is calculated according to the standard formula given above. On final stage measurements, the areas of the veins are summed up in order to determine the indicators of their overall cross-section.
To determine the diameter of the wire core, a micrometer or caliper is used, but if necessary, you can use a standard student ruler or centimeter. The core of the wire to be measured must be wound as tightly as possible onto the stick with two dozen turns. Using a ruler or centimeter, you need to measure the winding distance in mm, after which the indicators are used in the formula:
D = l/n,
- l is represented by the core winding distance in mm;
- n is the number of turns.
It should be noted that a larger cross-section of the wire allows for a margin of current indicators, as a result of which the level of load on the electrical wiring can be slightly exceeded.
To independently determine the wire cross-section of a monolithic core, you need to use a conventional caliper or micrometer to measure the diameter of the inner part of the cable without protective insulation.
Table of correspondence between wire diameters and cross-sectional area
Determining a cable or wire cross-section using a standard physical formula is one of the rather labor-intensive and complex processes that does not guarantee the most accurate results, so it is advisable to use special, ready-made tabular data for this purpose.
| Cable core diameter | Section indicators | Conductors with copper core type | ||
| Power in 220 V network conditions | Current | Power in 380 V network conditions | ||
| 1.12 mm | 1.0 mm 2 | 3.0 kW | 14 A | 5.3 kW |
| 1.38 mm | 1.5 mm 2 | 3.3 kW | 15 A | 5.7 kW |
| 1.59 mm | 2.0 mm 2 | 4.1 kW | 19 A | 7.2 kW |
| 1.78 mm | 2.5 mm 2 | 4.6 kW | 21 A | 7.9 kW |
| 2.26 mm | 4.0 mm 2 | 5.9 kW | 27 A | 10.0 kW |
| 2.76 mm | 6.0 mm 2 | 7.7 kW | 34 A | 12.0 kW |
| 3.57 mm | 10.0 mm 2 | 11.0 kW | 50 A | 19.0 kW |
| 4.51 mm | 16.0 mm 2 | 17.0 kW | 80 A | 30.0 kW |
| 5.64 mm | 25.0 mm 2 | 22.0 kW | 100 A | 38.0 kW |
| 6.68 mm | 35.0 mm 2 | 29.0 kW | 135 A | 51.0 kW |
How to determine the cross-section of a stranded wire?
Stranded wires are also known as stranded or flexible cables, which are single-core wires tightly wound into one bundle.In order to independently correctly calculate the cross-section or area of stranded wires, you must initially calculate the cross-section of each wire in the bundle, and then multiply the result by their total number.
When charged particles move, electric charge is transferred from one place to another. However, if charged particles undergo random thermal motion, such as free electrons in a metal, then there will be no charge transfer (Fig. 143). An electric charge moves through the cross section of a conductor only if, along with chaotic movement, electrons participate in ordered movement (Fig. 144). In this case, they say that an electric current is established in the conductor.
From the VII class physics course you know that electric current is the ordered (directed) movement of charged particles. Electric current arises from the ordered movement of free electrons in a metal or ions in electrolytes.
However, if you move a generally neutral body, then, despite the ordered movement of a huge number of electrons and atomic nuclei, no electric current occurs. The total charge transferred through any cross section of the conductor will be equal to zero, since charges of different signs move with the same average speed. A current in a conductor will arise only in the case when, when charges move in one direction, the positive charge transferred through the cross section is not equal in magnitude to the negative one.
Electric current has a certain direction. The direction of current is taken to be the direction of movement of positively charged particles. If the current is formed by the movement of negatively charged particles, then the direction of the current is considered opposite to the direction of movement of the particles.
Actions of current. We do not directly observe the movement of particles in a conductor. However, the presence of electric current can be judged by the actions or phenomena that accompany it.
First, the conductor through which the current flows heats up.
Secondly, electric current can change the chemical composition of the conductor, for example, releasing its chemical components (copper from a solution of copper sulfate, etc.). Of such kind
processes are not observed in all conductors, but only in solutions (or melts) of electrolytes.
Thirdly, the current has a magnetic effect. Thus, a magnetic needle near a current-carrying conductor rotates. The magnetic effect of current, in contrast to chemical and thermal, is the main one, since it manifests itself in all conductors without exception. The chemical effect of current is observed only in electrolytes, and heating is absent in superconductors (see § 60).
Current strength. If an electric current is established in a circuit, this means that an electric charge is constantly transferred through the cross-section of the conductor. The charge transferred per unit time serves as the main quantitative characteristic of current, called current strength. If a charge is transferred through the cross-section of a conductor over time, then the current strength is equal to:
Thus, the current strength is equal to the ratio of the charge transferred through the cross-section of the conductor over a time interval to this time interval. If the current strength does not change over time, then the current is called constant.
Current strength, like charge, is a scalar quantity. It can be both positive and negative. The sign of the current depends on which direction along the conductor is taken as positive. Current strength if the direction of the current coincides with the conventionally selected positive direction along the conductor. Otherwise
The strength of the current depends on the charge carried by each particle, the concentration of the particles, the speed of their directional movement and the cross-sectional area of the conductor. Let's show it.
Let the conductor have a cross section with an area of 5. Let us take the direction from left to right as the positive direction in the conductor. The charge of each particle is equal. In the volume of the conductor, limited by sections and 2, contains particles, where is the concentration of particles (Fig. 145). Their total charge If particles move from left to right with an average speed, then during the time all particles contained in the volume under consideration will pass through section 2. Therefore, the current strength is equal.
Does electric current have force? Yes, imagine... What is strength needed for? Well, what for, in order to do useful work, or maybe not useful :-), The main thing is to do something. Our body also has power. Some people have such strength that they can smash a brick to smithereens with one blow, while others cannot even lift a spoon :-). So, my dear readers, electric current also has power.
Imagine a hose with which you water your garden.
Let the hose be a wire, and the water in it be an electric current. We opened the faucet slightly and water ran through the hose. Slowly, but still she ran. The jet strength is very weak. We can't even spray someone with a hose with that kind of stream. Now let's open the faucet to the fullest! And our flow is such that it’s even enough to water the neighbor’s plot :-).
Now imagine that you are filling a bucket. Will you fill it faster with pressure from a hose or from a faucet? The diameter of the hose and faucet are equal
Of course, with pressure from the yellow hose! But why does this happen? The thing is that the volume of water coming out of the faucet and the yellow hose over an equal period of time is also different. Or in other words, The number of water molecules running out of a hose is much greater than out of a faucet in the same time.
It's exactly the same story with wires). That is, over an equal period of time, the number of electrons passing through the wire can be completely different. Now we can define the current strength.
So, current is the number of electrons passing through the cross-sectional area of a conductor per unit of time, say, per second. Below in the figure, this same cross-sectional area of the wire through which the electric current runs is shaded with green lines.
- for direct current -
where I is direct current strength;
- for intermittent current - in two ways:
1) according to the formula -
Q = 〈 I 〉 Δ t ,
where 〈 I 〉 is the average current strength;
2) graphically - as the area of a curvilinear trapezoid (Fig. 8.1).
IN International system Charge units are measured in coulombs (1 C).
The strength of the current is determined by the speed, concentration and charge of the current carriers, as well as the cross-sectional area of the conductor:
where q is the charge modulus of the current carrier (if the current carriers are electrons, then q = 1.6 ⋅ 10 −19 C); n is the concentration of current carriers, n = = N /V ; N is the number of current carriers passing through the cross section of the conductor (located perpendicular to the speed of current carriers) during time Δt, or the number of current carriers in the volume V = Sv Δt (Fig. 8.2); S is the cross-sectional area of the conductor; v is the modulus of the speed of movement of current carriers.
Current density is determined by the strength of the current passing through a unit cross-sectional area of a conductor located perpendicular to the direction of the current:
where I is the current strength; S is the cross-sectional area of the conductor (located perpendicular to the speed of current carriers).
The current density is vector quantity.
The direction of the current density j → coincides with the direction of the velocity of positive current carriers:
j → = q n v → ,
where q is the charge modulus of the current carrier (if the current carriers are electrons, then q = 1.6 ⋅ 10 −19 C); v → - speed of movement of current carriers; n is the concentration of current carriers, n = N /V; N is the number of current carriers passing through the cross section of the conductor (located perpendicular to the speed of current carriers) during time Δt, or the number of current carriers in the volume V = Sv Δt (Fig. 8.2); v is the modulus of the current carrier velocity; S is the cross-sectional area of the conductor.
In the International System of Units, current density is measured in amperes divided by square meter (1 A/m2).
The current strength in gases (electric current in gases is caused by the movement of ions) is determined by the formula
I = N t ⋅ | q | ,
where N /t is the number of ions that pass through the cross section of the vessel every second (every second); |q | - ion charge modulus:
- for a singly charged ion -
|q | = 1.6 ⋅ 10 −19 C,
- for a doubly charged ion -
|q | = 3.2 ⋅ 10 −19 C
Example 1. The number of free electrons in 1.0 m 3 of copper is 1.0 ⋅ 10 28. Find the speed of directional movement of electrons in a copper wire with a cross-sectional area of 4.0 mm 2, through which a current of 32 A flows.
Solution. The speed of directional movement of current carriers (electrons) is related to the current strength in the conductor by the formula
where q is the charge modulus of the current carrier (electron); n is the concentration of current carriers; S is the cross-sectional area of the conductor; v is the modulus of the speed of directional movement of current carriers in the conductor.
Let us express from this formula the desired quantity - the speed of current carriers -
v = I q n S .
To calculate the speed, we will use the following values of the quantities included in the formula:
- the magnitude of the current and the cross-sectional area of the conductor are specified in the problem statement: I = 32 A, S = 4.0 mm 2 = 4.0 ⋅ 10 −6 m 2 ;
- the value of the elementary charge (equal to the modulus of the electron charge) is a fundamental constant (constant value): q = 1.6 ⋅ 10 −19 C;
- current carrier concentration - the number of current carriers per unit volume of a conductor -
n = N V = 1.0 ⋅ 10 28 1 = 1.0 ⋅ 10 28 m −3 .
Let's do the calculation:
v = 32 1.6 ⋅ 10 − 19 ⋅ 1.0 ⋅ 10 28 ⋅ 4.0 ⋅ 10 − 6 = 5.0 ⋅ 10 − 3 m/s = 5.0 mm/s.
The speed of directional movement of electrons in the specified conductor is 5.0 mm/s.
Example 2. The current strength in the conductor increases uniformly from 10 to 12 A in 12 s. What charge passes through the cross section of the conductor during the specified time interval?
Solution. The strength of current in a conductor changes over time. Therefore, the charge transferred by current carriers through a cross-section of a conductor located perpendicular to the speed of current carriers over a certain period of time can be calculated in two ways.
1. The required charge can be calculated using the formula
Q = 〈 I 〉 Δ t ,
where 〈 I 〉 is the average current strength; ∆t - time interval, ∆t = 12 s.
The current strength increases uniformly in the conductor; therefore, the average current strength is given by
〈 I 〉 = I 1 + I 2 2 ,
where I 1 is the current value at the initial moment of time, I 1 = 10 A; I 2 - current value at the final moment of time, I 2 = 12 A.
Substituting the expression for the average current strength into the formula for calculating the charge, we get
Q = (I 1 + I 2) Δ t 2.
The calculation gives the value
Q = (10 + 12) ⋅ 12 2 = 132 C = 0.13 kC.
The figure shows the dependence I (t) specified in the problem conditions.
The charge transferred by current carriers through the cross section of a conductor located perpendicular to the speed of current carriers during a specified period of time, numerically equal to area trapezoid bounded by four lines:
- straight line I (t);
- perpendicular to the time axis, restored from point t 1;
- perpendicular to the time axis, restored from point t 2 ;
- time axis t.
We perform the calculation using the formula for the area of a trapezoid:
Q = 12 + 10 2 ⋅ 12 = 132 C = 0.13 kC.
Both methods of calculating the charge transferred by current carriers over a specified period of time give the same result.
The idea of electric current can be approached from different positions. One of them is macroscopic, the other is based on an analysis of the conductivity mechanism. For example, the flow of fluid through pipes can be viewed as a continuous movement of matter, but it can also be analyzed in terms of the movement of fluid particles.
The first idea of electric current arose at that stage of development of physics when the mechanism of conduction was not yet known. It was then that the physical quantity arose - current strength, which shows which electric charge passes through the cross section of the conductor per unit time. Current strength. The unit of current is ampere (A): .
From the definition of current strength, two features of this quantity follow. One of them is the independence of the current strength from the cross-section of the conductor through which the current flows. The second is the independence of the current strength from the spatial arrangement of the circuit elements, which you can see more than once: no matter how the conductors are moved, this does not affect the current strength. The current is called permanent, if the current does not change over time.
Thus, the idea of electric current and its strength arose when it was not yet clear what it was.
A study of the electrical conductivity of various substances has shown that in different substances various charged free particles move under the influence of an electric field during the flow of current. For example, in metals these are electrons, in liquids these are positive and negative ions, in semiconductors these are electrons and “holes”. Not only the types of particles are different, but also the nature of their interaction with the substance in which the current flows. Thus, free electrons in metals move freely for some time between the nodes of the crystal lattice, then collide with ions located at the nodes. In electrolytes, ions interact with each other and with the atoms of the liquid.
But for all substances there is: particles in the absence of a field move chaotically; when a field appears, a very small amount of speed is added to the speed of chaotic movement either in the direction of the field (for positive particles) or in the direction opposite to the field (for negative particles). This added speed is called drift speed. The average speed of chaotic movement is hundreds of meters per second, the drift speed is several millimeters per second. However, it is this small addition that explains all the effects of the current.
For any substances, you can obtain a formula for calculating the current strength: 
, where is the concentration of charged particles, is the charge of one particle, and is the cross-sectional area.
Thus, electricity is the ordered movement of charged particles.
It may seem that this formula contradicts the statement that the current strength is independent of the cross-sectional area of the conductor. But this independence is an experimental fact. It can be explained by the fact that the drift speed is greater where the cross section is smaller, and through a larger cross section the particles drift more slowly.
It is an empirical fact that when applied to a conductor constant potential difference goes through it D.C.. This fact contradicts, at first glance, the formula . Indeed, with a constant potential difference in a substance, a field with a constant field strength is created. Consequently, a constant force acts on free particles and their speed should increase. It turns out that at a constant voltage the current strength should increase in proportion to time. This does not happen because when current flows in a substance, electrical resistance. It is this that ensures constant current strength at a constant potential difference.
To measure resistance, it is necessary to study the dependence of current on voltage. A graph of such a dependence is called Volt-ampere characteristic. Three types of current-voltage characteristics are possible (Fig. 40).
“Conductors and dielectrics” - The electrical characteristics of a medium are determined by the mobility of charged particles in it. Dielectrics. Free charges are charged particles of the same sign that can move under the influence of an electric field. Dielectrics - gases, distilled water, benzene, oils, porcelain, glass, mica, etc. External electric field.
"Golden Section" - Intercession Cathedral (St. Basil's Cathedral). Admiralty. Intercession of the Virgin Mary on the Nerl. Painting in the second floor foyer. Research objectives: Golden ratio– proportion. St Basil's Church. Purpose of the study: To derive the law of beauty of the world from the point of view of mathematics. Golden ratio in architecture. Completed by 10th grade student Yulia Smetanina.
“Sections of a parallelepiped” - 1. Introductory speech by the teacher - 3 min 2. Activation of students’ knowledge. Rectangle CKK’C’ - section ABCDA’B’C’D’. Homework. The cutting plane intersects the faces along segments. ? MNK - section of parallelepiped ABCDA’B’C’D’. Task: construct a section through the edge of a parallelepiped and point K. Independent work students.
“Proportions of the golden section” - Division of a segment by the “golden section”. "Golden Pentagon". Euclid, Leonardo da Vinci, Luca Pacioli. "Golden Rectangle". Inanimate nature. For example, the ratios of land and water on the Earth's surface are in the golden ratio. The harmony of the Universe is based on numbers. "Golden ratio" in nature, art and architecture.
“Construction of sections” - If the section is taken out, then draw an open line, two thick strokes. Designation of sections. It is more convenient to show some dimensions of part elements on sections. Sections in the drawings are divided into extended and superimposed. Sections are made to the same scale as the image to which it relates.
“Conductor in an electrical circuit” - Solve the problem. Connection of conductors. Electric light bulbs in a Christmas tree garland are connected in series. Determine the resistance of the circuit. The resistance of each resistor is 3 ohms. 1. Two conductors with a resistance of 4 Ohms and 2 Ohms are connected in series. Series connection I = I1 = I2 U = U1 + U2 R = R1 + R2 For identical conductors R = nR1.
The copper conductor has a length of 500 m and a cross-sectional area of 0.5 mm2. A) what is the current strength in the conductor when the voltage at its ends is 12V? The resistivity of copper is 1.7 times 10 -8 powers of Ohm times m b) Determine the speed of the ordered movement of electrons. The concentration of free movement for copper is equal to 8.5 multiplied by 10 to the 28th degree of meters to the minus 3 degrees, and the modulus of the electron charge is equal to 1.6 multiplied by 10 to the minus 19 degrees of C c) A second copper conductor of twice the diameter is connected in series to the first conductor . What will be the speed of ordered movement of electrons in the second conductor?
Solution for question a)
What do we know about current, voltage and resistance?
I=U/R, U=I*R
I - current in Amperes,
U - voltage in Volts
R - resistance in Ohms
What is a current of 1 Ampere?
This is a current at which a charge of 1 Coulomb passes through a conductor in 1 second.
1 A = 1 C/s(1 Ampere equals 1 Coulomb per second)
What do we know from the conditions?
U = 12 V - voltage
p = 1.7*10e-8 Ohm*m - resistivity "rho" (the resistance value of a conductor with a cross-section of 1 square meter and a length of 1 meter).
Our conductor has a cross-section S=0.5 mm^2 or 0.0000005 m^2 or 0.5*10e-6 m^2 (in one square meter 1000000 sq. Millimeters - 1000*1000) and length L=500m
We get the conductor resistance
R=p*L/S=1.7*10e-8 * 500 / 0.5*10e-6 = 0.000000017*500/0.0000005 = 17 Ohm
The current will then be:
I=U/R=12/17 A (0.706. Ampere)
Solution for question b)
The current strength I is also expressed through the following quantities:
I=e*n*S*Vav
e - electron charge, C
n - electron concentration, pcs/m^3 (pcs per cubic meter)
S - cross-sectional area, m^2
Vav - average speed of ordered movement of electrons, m/s
That's why
Vav=I/(e*n*S)= (12/17) / (1.6*10e-19 * 8.5*10e+28 * 0.5*10e-6) = 11.657*10e-3 m/s (or 11.657 mm/s)
Solution for question c)
We reason similarly to solutions a) and b)
First you need to find the total current (total resistance).
T.K. Condition c) talks about diameter, we conclude that all the wires are round.
The length of the second wire is not specified. Let's say it's also 500m.
The area of a circle is determined by the ratio:
S=(pi*D^2)/4,
where D is the diameter of the circle,
pi = 3.1415926.
Thus, when the diameter is doubled, the cross-sectional area of the wire quadruples,
when the diameter triples, the cross-sectional area of the wire increases nine times, etc.
Total S2 = S1*4= 0.5*10e-6 * 4 = 2*10e-6 M^2
If the cross-sectional area of the wire is quadrupled, then, with the same length, its resistance will decrease fourfold.
Total R2=R1/4= 17/4 Ohm = 4.25 Ohm
The total resistance in a series connection adds up, so
I=U/R=U/(R1+R2)=12/(17+17/4)= 48/85 = 0.5647. A
The ordered speed of electrons for the second conductor will then be:
Vav=I/(e*n*S2)= (48/85)/(1.6*10e-19 * 8.5*10e+28 * 2*10e-6) = 0.02076*10e-3 m/s (or 0.02076 mm/s)
