*Notes to a video lecture on http://www.unizor.com*

__Two Currents - Problems 1__

IMPORTANT NOTES

As we know from a previous lecture, the magnetic field force lines

around a long thin straight line wire with electric current running

through it are circular in planes perpendicular to a wire and centered

at the points of the wire.

These lines are attributed a direction that can be determined using the

*rule of the right hand*or Maxwell's

*corkscrew rule*.

The

*rule of the right hand*states that, if you wrap your right

hand around a wire such that your thumb points to a direction of an

electric current in the wire, which is, by definition, from positive to

negative, then your fingers will point to a direction of the magnetic

field lines. This direction is the direction of any tangential to a line

vector of force.

The

*corkscrew rule*states that, if we imagine a corked bottle

along the wire such that the direction of an electric current enters the

bottle through a cork, to open the bottle we need to rotate the regular

cork opener in the direction of the magnetic lines around the wire.

The magnitude of the intensity of a magnetic field (a vector of force) depends on the electric current's amperage

*and the distance*

**I***to the wire as follows*

**R**

**B = μ**_{0}I/(2π·R)where

*is a constant called*

**μ**_{0}*permeability*of vacuum.

The direction of this force vector at any point of space around a wire

is always in the same plane as a circular magnetic line going through

the same point and is tangential to this magnetic line.

The direction of the vector corresponds to the direction of the magnetic line.

*Problem 1A*

Two ideally long and thin wires are positioned in space with Cartesian coordinates.

One goes through point

*on Z-axis (*

**A(0,0,a)***) parallel to X-axis and carries an electric current of amperage*

**a > 0***in the positive direction of X-axis.*

**I**Another wire goes through point

*on Z-axis parallel to Y-axis and carries an electric current of the same amperage*

**B(0,0,−a)***in the positive direction of Y-axis.*

**I**Determine the X-, Y- and Z-components and magnitude of the vector of magnetic field intensity at the origin of coordinates

*.*

**(0,0,0)***Solution*

The magnetic field at point

*is a combination of two fields - one with intensity vector*

**(0,0,0)***, the source in the first wire that carries electric charge*

**B**_{1}*parallel to X-axis, from which the origin of coordinates is at distance*

**I***, and another with intensity vector*

**a***, the source in the second wire that carries electric current*

**B**_{2}*parallel to Y-axis, from which the origin of coordinates is at distance*

**I***.*

**a**The resulting field intensity vector is a vector sum of vectors

*and*

**B**_{1}*.*

**B**_{2}The first magnetic field has its force lines forming a cylindrical

surface with an axis being the first wire parallel to X-axis. At point

*the direction of this magnetic field intensity vector is perpendicular to a radius from the first wire towards point*

**(0,0,0)***, that is along the Y-axis towards its positive direction.*

**(0,0,0)**Considering the values of electric current

*and the distance to the source (the first wire) equaled to*

**I***, the magnitude of this vector is*

**a**

**|B**_{1}| = μ_{0}·I/(2π·a)So, this vector in Cartesian coordinates is

{

*}*

**0; μ**_{0}·I/(2π·a); 0The second magnetic field has its force lines forming a cylindrical

surface with an axis being the second wire parallel to Y-axis. At point

*the direction of this magnetic field intensity vector is perpendicular to a radius from the second wire towards point*

**(0,0,0)***, that is along the X-axis towards its positive direction.*

**(0,0,0)**Considering the values of electric current

*and the distance to the source (the second wire) equaled to*

**I***, the magnitude of this vector is*

**a**

**|B**_{2}| = μ_{0}·I/(2π·a)So, this vector in Cartesian coordinates is

{

*}*

**μ**_{0}·I/(2π·a); 0; 0Therefore, the combined vector of magnetic field intensity has its three coordinates

{

*}*

**μ**_{0}·I/(2π·a); μ_{0}·I/(2π·a); 0The magnitude of this vector is

**μ**_{0}·I·√2/(2π·a)*Problem 1B*

Two ideally long and thin wires are positioned in space parallel to Z-axis.

One goes through point

*(*

**A(a,0,0)***) and carries an electric current of amperage*

**a > 0***in the negative direction of the Z-axis.*

**I**_{1}Another wire goes through point

*(*

**B(0,b,0)***) and carries an electric current of amperage*

**b > 0***in the positive direction of the Z-axis.*

**I**_{2}Determine the X-, Y- and Z-components and magnitude of the vector of magnetic field intensity at the origin of coordinates

*.*

**(0,0,0)***Solution*

The magnetic field at point

*is a combination of two fields - one with intensity vector*

**(0,0,0)***, the source in the first wire that carries electric charge*

**B**_{1}*, from which the origin of coordinates is at distance*

**I**_{1}*, and another with intensity vector*

**a***, the source in the second wire that carries electric current*

**B**_{2}*, from which the origin of coordinates is at distance*

**I**_{2}*.*

**b**The resulting field intensity vector is a vector sum of vectors

*and*

**B**_{1}*.*

**B**_{2}The Z-coordinate of both vectors is zero.

The first magnetic field has its force lines forming a cylindrical surface with an axis being the first wire. At point

*the direction of this magnetic field intensity vector is perpendicular to a radius from the first wire towards point*

**(0,0,0)***, that is along the Y-axis towards its positive direction.*

**(0,0,0)**Considering the values of electric current

*and the distance to the source (the first wire) equaled to*

**I**_{1}*, the magnitude of this vector is*

**a**

**|B**_{1}| = μ_{0}·I_{1}/(2π·a)So, this vector in Cartesian coordinates is

{

*}*

**0; μ**_{0}·I_{1}/(2π·a); 0The second magnetic field has its force lines forming a cylindrical surface with an axis being the second wire. At point

*the direction of this magnetic field intensity vector is perpendicular to a radius from the second wire towards point*

**(0,0,0)***, that is along the X-axis towards its positive direction.*

**(0,0,0)**Considering the values of electric current

*and the distance to the source (the second wire) equaled to*

**I**_{2}*, the magnitude of this vector is*

**b**

**|B**_{2}| = μ_{0}·I_{2}/(2π·b)So, this vector in Cartesian coordinates is

{

*}*

**μ**_{0}·I_{2}/(2π·b); 0; 0Therefore, the combined vector of magnetic field intensity has its three coordinates

{

*}*

**μ**_{0}·I_{2}/(2π·b); μ_{0}·I_{1}/(2π·a); 0The magnitude of this vector is

[

*]*

**μ**_{0}/(2π)

**√(I**_{1}/a)²+(I_{2}/b)²
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