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Magnetic field due to current in infinite (long) straight wire

The expression for the magnitude of magnetic field due to infinite wire can be obtained by suitably putting appropriate values of angles in the expression of magnetic field due to finite wire. Here,

φ 1 = π 2 ; φ 2 = π 2

and

B = μ 0 I 4 π R sin φ 1 + sin φ 2 B = μ 0 I 4 π R sin π 2 + sin π 2 B = μ 0 I 2 π R

The important point to note here is that magnetic field is independent of the relative angular position of point of observation with respect to infinite wire. Magnetic field, however, depends on the perpendicular distance from the wire.

In reality, however, we always work with finite wire or at the most with long wire. A finite length wire is approximated as infinite or long wire for at least for close points around the wire.

Magnetic field at a point near the end of current carrying long wire

The wire here extends from an identified position to infinity in only one direction. In this case, the angles involved are :

φ 1 = 0 ; φ 2 = π 2

and

B = μ 0 I 4 π R sin 0 + sin π 2 B = μ 0 I 4 π R

Problem : Calculate magnetic field at point P due to current 5 A flowing through a long wire bent at right angle as shown in the figure. The point P lies at a linear distance 1 m from the corner.

Magnetic field due to current in wire

Magnetic field due to current in wire.

The point P lies on the extension of wire segment in x-direction. Here angle between current element and displacement vectors is zero i.e. θ =0 and sinθ. As such this segment does not produce any magnetic field at point P. On the other hand, the point P lies near one of the end of the segment of wire in y-direction. The wire being long, the magnetic field due to wire segment in y-direction is :

B = μ 0 I 4 π R

Putting values,

B = μ 0 I 4 π R = 10 - 7 X 5 1 = 5 X 10 - 7 T

Applying Right hand thumb rule, the magnetic field at P is perpendicular to xy plane and into the plane of drawing (i.e. negative z-direction).

B = - 5 X 10 - 7 k

Exercises

Two long straight wires at A and C, perpendicular to the plane of drawing, carry currents such that point D is a null point. The wires are placed at a linear distance of 10 m. If the current in the wire at A is 10 A and its direction is out of the plane of drawing, then find (i) the direction of current and (ii) magnitude of current in the second wire.

Two straight wires carrying current

Two straight wires carrying current

In order to nullify the magnetic field at D due to current in wire at A, the direction of magnetic field due to current in the wire at C should be equal and opposite. This means that the current in the wire at C should be opposite that of wire at A. Hence, the direction of current in the wire at C should be into the plane of drawing. Now, the magnitudes of magnetic fields due to currents are equal. Let the current in second wire be I, then:

Two straight wires carrying current

Two straight wires carrying current

μ 0 I 2 π X 5 = μ 0 X 10 2 π X 15 I = 50 15 = 3.34 A

Calculate magnetic field at the center due to current flowing in clockwise direction through a wire in the shape of regular hexagon. The arm of hexagon measures 0.2 m and current through the wire is 10 A.

Applying right hand rule for vector cross product, we realize that magnetic field due to each arm of the hexagon for given current direction (clockwise) is into the plane of hexagon. As such, we can algebraically add magnetic field due to each arm to obtain net magnetic field.

B = 6 B a

where B a is magnetic field due to current in one of the arms. Now, we consider one of the arms of hexagon as shown in the figure. Here,

Magnetic field at the center due current

Magnitude of magnetic field is six times the magnetic field due to one arm.

φ 1 = π 6 ; φ 2 = π 6 R = a 2 cot φ 1 = 0.2 2 cot φ 6 = 0.1 X 3 = 0.1732 m

and

B = 6 B a = 6 μ 0 I 4 π R sin π 6 + sin π 6

Putting values, we have :

B = 6 X 10 - 7 X 10 X 1 0.1732 = = 3.462 X 10 - 5 T

A current 10√2 ampere flows through the wire having configuration as shown in the figure. Determine magnetic field at P.

Magnetic field due to current in wire

Magnetic field due to current in wire

We shall determine magnetic field to different straight segments of wires. Let us consider the out of plane orientation as positive. Now, for wire segment AC, the point P is at the end of straight wire of length 4 m and is at a perpendicular linear distance of 4 m. The magnetic field at P due to segment AC is perpendicular and out of the plane of drawing. The magnetic field due to segment AC is :

B A C = 2 μ 0 I 8 π L = 2 μ 0 I 8 π X 4 = 2 μ 0 I 32 π

For the wire segment CD, the point P lies on the extended line passing through the wire. The magnetic field due to this segment, therefore, is zero.

B C D = 0

For the wire segment DE, the point P is at the end of straight wire of length 2 m and is at a perpendicular linear distance of 2 m. The magnetic field at P due to segment DE is perpendicular and into the plane of drawing. The magnetic field is :

B D E = - 2 μ 0 I 8 π L = - 2 μ 0 I 8 π X 2 = - 2 μ 0 I 16 π

For the wire segment EF, the point P is at the end of straight wire of length 2 m and is at a perpendicular linear distance of 2 m. The magnetic field at P due to segment DE is perpendicular and into the plane of drawing. The magnetic field is :

B E F = - 2 μ 0 I 16 π

For the wire segment FG, the point P lies on the extended line passing through the wire. The magnetic field due to this segment, therefore, is zero.

B F G = 0

For the wire segment GA, the point P is at the end of straight wire of length 4 m and is at a perpendicular linear distance of 4 m. The magnetic field at P due to segment GA is perpendicular and out of the plane of drawing. The magnetic field is :

B G A = 2 μ 0 I 32 π

The net magnetic field at P is :

B = B A C + B C D + B D E + B E F + B F G + B G A B = 2 μ 0 I 32 π + 0 - 2 μ 0 I 16 π - 2 μ 0 I 16 π + 0 + 2 μ 0 I 32 π B = - 2 μ 0 I 16 π B = - 2 X 4 π X 10 - 7 X 10 2 16 π = - 2 X 10 - 7 X 10 4 B = - 5 X 10 - 7 T

The net magnetic field is into the plane of drawing.

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Source:  OpenStax, Electricity and magnetism. OpenStax CNX. Oct 20, 2009 Download for free at http://cnx.org/content/col10909/1.13
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