Sources of the мagnetic field/ презентация

Physics 1 Voronkov Vladimir Vasilyevich

Lecture 12 Sources of the Magnetic Field The Biot-Savart Law Ampere’s Law The effects of magnetic fields. The production and properties of magnetic fields.

Current Produces Magnetic Field The magnetic field dB at a point P due to the current I through a length element ds is given by the Biot–Savart law. The direction of the field is out of the page at P and into the page at P´.

The Biot-Savart Law The experimental observations for the magnetic field dB at a point P associated with a length element ds of a wire carrying a steady current I: The vector dB is perpendicular both to ds (which points in the direction of the current) and to the unit vector directed from ds toward P. The magnitude of dB is inversely proportional to r2, where r is the distance from ds to P. The magnitude of dB is proportional to the current and to the magnitude ds of the length element ds. The magnitude of dB is proportional to sin, where is the angle between the vectors ds and .

The foregoing experimental observations can be expressed in one formula: The foregoing experimental observations can be expressed in one formula:

Note that the field dB in the Biot-Savart law is the field created by the current in only a small length element ds of the conductor. To find the total magnetic field B created at some point by a current of finite size, we must sum up contributions from all current elements Ids that make up the current. That is, we must evaluate B by integrating over the entire current distribution: Note that the field dB in the Biot-Savart law is the field created by the current in only a small length element ds of the conductor. To find the total magnetic field B created at some point by a current of finite size, we must sum up contributions from all current elements Ids that make up the current. That is, we must evaluate B by integrating over the entire current distribution:

Magnetic Field of a Thin Straight Wire Using the Biot-Savart law we can find the magnetic field at point P, created by a thin straight wire with current in it: a is the distance from the wire to P 1, 2 are the angles shown in the picture.

Magnetic Field of an Infinitely Long Wire For a very long thin straight wire we can consider 1=0, 2=, then: a is the distance from the wire to P I is the current in the wire This expression shows that the magnitude of the magnetic field is proportional to the current and decreases with increasing distance from the wire.

Magnetic Field around a Wire Because of the symmetry of the wire, the magnetic field lines are circles concentric with the wire and lie in planes perpendicular to the wire. The magnitude of B is constant on any circle of radius a and is given by the expression on the previous slide:

Magnetic Force Between Two Parallel Conductors Two long, straight, parallel wires separated by distance a and carrying currents I1 and I2 in the same direction. The force exerted on one wire due to the magnetic field set up by the other wire is:

Ampere’s Law The line integral of B*ds around any closed path equals 0I, where I is the total steady current passing through any surface bounded by the closed path.

Example for the Ampere’s Law

Magnetic Field of a Solenoid A solenoid is a long wire wound in the form of a helix. Magnetic field lines for a tightly wound solenoid of finite length, carrying a steady current. The field in the interior space is strong and nearly uniform.

Cross-sectional view of an ideal solenoid, where the interior magnetic field is uniform and the exterior field is close to zero. Cross-sectional view of an ideal solenoid, where the interior magnetic field is uniform and the exterior field is close to zero.

Magnetic Flux The magnetic flux through an area element dA is B·dA = BdA cos where dA is a vector perpendicular to the surface and has a magnitude equal to the area dA. Therefore, the total magnetic flux ФB through the surface is

Magnetic flux through a plane lying in a magnetic field The flux through the plane is zero when the magnetic field is parallel to the plane surface. The flux through the plane is a maximum when the magnetic field is perpendicular to the plane.

Gauss’s Law in Magnetism The net magnetic flux through any closed surface is always zero: Here is a scalar multiplication of two vectors. Zero net magnetic flux through any closed surface means that magnetic field lines has no source. It is based on the fact that there exist no magnetic monopoles.

The magnetic field lines of a bar magnet form closed loops. Note that the net magnetic flux through a closed surface surrounding one of the poles (or any other closed surface) is zero. (The dashed line represents the intersection of the surface with the page.) The magnetic field lines of a bar magnet form closed loops. Note that the net magnetic flux through a closed surface surrounding one of the poles (or any other closed surface) is zero. (The dashed line represents the intersection of the surface with the page.) The number of lines entering the surface equals the number of lines leaving it.

Displacement Current There is a charging capacitor, with current I two imaginary surfaces S1 and S2, and path P, bounding to S1 and S2. When the path P is considered as bounding S1, then because the conduction current passes through S1. When the path is considered as bounding S2, then because no conduction current passes through S2. Thus, we have a contradiction.

This contradiction is resolved by introducing a new quantity – the displacement current: This contradiction is resolved by introducing a new quantity – the displacement current: Є0 is a free space permittivity, a constant ФЕ is the electric flux: As the capacitor is being charged (or discharged), the changing electric field between the plates may be considered equivalent to a current that acts as a continuation of the conduction current in the wire.

General form of Ampere’s Law So considering the displacement current, we can write the General form of Ampere’s Law (or Ampere-Maxwell law):

So the electric flux through S2 is So the electric flux through S2 is Where E is the electric field between the plates, A is the area of the plates, then So the electric flux through S2 is

Then the displacement current through S2 is Then the displacement current through S2 is That is, the displacement current Id through S2 is precisely equal to the conduction current I through S1!

Units in Si Magnetic field B T= N*s/(C*m) T= N/(A*m) Electric Field E V/m=N/C Magnetic permeability of free space: