The magnetic field at a distance r from a very long straight wire, carrying a steady current I, has a magnitude equal to
" The integral of B around any closed mathematical path equals u0 times the current intercepted by the area spanning the path "
Example: Problem 31.5Six parallel aluminum wires of small, but finite, radius lie in the same plane. The wires are separated by equal distances d, and they carry equal currents I in the same direction. Find the magnetic field at the center of the first wire. Assume that the currents in each wire is uniformly distributed over its cross section.
A schematic layout of the problem is shown in Figure 31.3. The magnetic field generated by a single wire is equal to
31.2. The solenoidA solenoid is a device used to generate a homogeneous magnetic field. It can be made of a thin conducting wire wound in a tight helical coil of many turns. The magnetic field inside a solenoid can be determined by summing the magnetic fields generated by N individual rings (where N is the number of turns of the solenoid). We will limit our discussion of the magnetic field generated by a solenoid to that generated by an ideal solenoid which is infinitely long, and has very tightly wound coils.
The ideal solenoid has translational and rotational symmetry. However, since magnetic field lines must form closed loops, the magnetic field can not be directed along a radial direction (otherwise field lines would be created or destroyed on the central axis of the solenoid). Therefore we conclude that the field lines in a solenoid must be parallel to the solenoid axis. The magnitude of the magnetic field can be obtained by applying Ampere's law.
Example: Problem 31.14A long solenoid of n turns per unit length carries a current I, and a long straight wire lying along the axis of this solenoid carries a current I'. Find the net magnetic field within the solenoid, at a distance r from the axis. Describe the shape of the magnetic field lines.
The magnetic field generated by the solenoid is uniform, directed parallel to the solenoid axis, and has a magnitude equal to
Example: Problem 31.15A coaxial cable consists of a long cylindrical copper wire of radius r1 surrounded by a cylindrical shell of inner radius r2 and outer radius r3 (see Figure 31.5). The wire and the shell carry equal and opposite currents I uniformly distributed over their volumes. Find formulas for the magnetic field in each of the regions r < r1, r1 < r < r2, r2 < r < r3, and r > r3.
The magnetic field lines are circles, centered on the symmetry axis of the coaxial cable. First consider an integration path with r < r1. The path integral of B along this path is equal to
31.3. Motion of charges in electric and magnetic fieldsThe magnetic force acting on particle with charge q moving with velocity v is equal to
The effect of a magnetic field on the motion of a charged particle can be used to determine some of its properties. One example is a measurement of the charge of the electron. An electron moving in a uniform magnetic field will described a circular motion with a radius given by eq.(31.25). Suppose the electron is accelerated by a potential V0. The final kinetic energy of the electron is given by
Another application of the effect of a magnetic field on the motion of a charged particle is the cyclotron. A cyclotron consists of an evacuated cavity placed between the poles of a large electromagnet. The cavity is cut into two D-shaped pieces (called dees) with a gap between them. An oscillating high voltage is connected to the plates, generating an oscillating electric field in the region between the two dees. A charged particle, injected in the center of the cyclotron, will carry out a uniform circular motion for the first half of one turn. The frequency of the motion of the particle depends on its mass, its charge and the magnetic field strength. The frequency of the oscillator is chosen such that each time the particle crosses the gap between the dees, it will be accelerated by the electric field. As the energy of the ion increases, its radius of curvature will increase until it reaches the edge of the cyclotron and is extracted. During its motion in the cyclotron, the ion will cross the gap between the dees many times, and it will be accelerated to high energies.
Up to now we have assumed that the direction of the motion of the charged particle is perpendicular to the direction of the magnetic field. If this is the case, uniform circular motion will result. If the direction of motion of the ion is not perpendicular to the magnetic field, spiral motion will result. The velocity of the charged particle can be decomposed into two components: one parallel and one perpendicular to the magnetic field. The magnetic force acting on the particle will be determined by the component of its velocity perpendicular to the magnetic field. The projection of the motion of the particle on the x-y plane (assumed to be perpendicular to the magnetic field) will be circular. The magnetic field will not effect the component of the motion parallel to the field, and this component of the velocity will remain constant. The net result will be spiral motion.
31.4. Crossed electric and magnetic fieldsA charged particle moving in a region with an electric and magnetic field will experience a total force equal to
The drift velocity of the electrons depend on the current I in the wire, its cross sectional area A and the density n of electrons (see Chapter 28):
31.5. Forces on a wireA current I flowing through a wire is equivalent to a collection of charges moving with a certain velocity vd along the wire. The amount of charge dq present in a segment dL of the wire is equal to
Example: Problem 31.33A balance can be used to measure the strength of the magnetic field. Consider a loop of wire, carrying a precisely known current, shown in Figure 31.9 which is partially immersed in the magnetic field. The force that the magnetic field exerts on the loop can be measured with the balance, and this permits the calculation of the strength of the magnetic field. Suppose that the short side of the loop measured 10.0 cm, the current in the wire is 0.225 A, and the magnetic force is 5.35 x 10-2 N. What is the strength of the magnetic field ?
Consider the three segments of the current loop shown in Figure 31.9 which are immersed in the magnetic field. The magnetic force acting on segment 1 and 3 have equal magnitude, but are directed in an opposite direction, and therefore cancel. The magnitude of the magnetic force acting on segment 2 can be calculated using eq.(31.41) and is equal to
31.6. Torque on a current loopIf a current loop is immersed in a magnetic field, the net magnetic force will be equal to zero. However, the torque on this loop will in general not be equal to zero. Suppose a rectangular current loop is placed in a uniform magnetic field (see Figure 31.10). The angle between the normal of the current loop and the magnetic field is equal to [theta]. The magnetic forces acting on the top and the bottom sections of the current loop are equal to
The work that must be done against the magnetic field to rotate the current loop by an angle d[theta] is equal to - [tau] d[theta]. The change in potential energy of the current loop when it rotates between [theta]0 and [theta]1 is given by