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Magnetic field due to current in straight wire

Module by: Sunil Kumar Singh. E-mail the author

The Biot-Savart law allows us to calculate magnetic field due to steady current through a small element of wire. Since direction of magnetic field due to different current elements of an extended wire carrying current is not unique, we need to add individual magnetic vectors to obtain resultant or net magnetic field at a point. This method of determining the net magnetic field follows superposition principle, which says that magnetic fields due to individual small current element are independent of each other and that the net magnetic field at a point is obtained by vector sum of individual magnetic field vectors :

B = B i = B 1 + B 2 + B 3 + B = B i = B 1 + B 2 + B 3 +

We calculate magnetic field due to individual current element (I dl) using Biot-Savart law :

d B = μ 0 4 π I đ l X r r 3 d B = μ 0 4 π I đ l X r r 3

where "dl" is referred as “current length element” and "I dl" as “current element”.

In the case of a straight wire, the task of vector addition is simplified to a great extent because direction of magnetic field at a point due to all current elements comprising the straight wire is same.

Direction of magnetic field (Right hand thumb rule)

A straight line and a point constitute an unique plane. This is true for all points in three dimensional rectangular space (x,y,z). For convenience, let us consider that the point of observation (P) lies in xy plane as shown in the figure below. We can say that the straight wire along y-axis also lie in xy plane. Clearly, this plane is the plane of current length element dl and displacement vector r, which appear in the Biot-Savart expression. The direction of magnetic field is vector cross product dlX r, which is clearly perpendicular to the plane xy. This means that the magnetic field is along z-axis. This conclusion is independent of the relative positions of current length elements of the wire with respect to observation point P.

In a nutshell, we conclude that the directions of magnetic fields due to all current elements constituting straight wire at a point P are same. Though, magnitudes of magnetic fields are different as different current elements are located at different linear distance from the point i.e. displacement vectors (r) are different for different current length elements (dl).

Figure 1: Magnetic fields due to all current elements constituting straight wire at a point P are same.
Magnetic field due to current in straight wire
 Magnetic field due to current in straight wire  (ms1a.gif)

See in the figure how magnetic fields due to three current elements in positive y-direction are acting in negative z-direction. The magnetic fields due to different current elements are B 1 B 1 , B 2 B 2 and B 3 B 3 acting along PZ’ as shown in the figure. Note that magnitudes of magnetic fields are not equal as current elements are positioned at different linear distance.

The magnetic field is along z-axis either in positive or negative z direction depending on the direction of current and whether observation point is on right or left of the current carrying straight wire. By convention, magnetic field vector into the plane of drawing is denoted by a cross (X) and magnetic field vector out of the plane of drawing is denoted by a dot (.). Following this convention, magnetic field depicted on either side of a current carrying straight wire is as shown here :

Figure 2: Representation of magnetic field in terms of cross and dot.
Magnetic field due to current in straight wire
 Magnetic field due to current in straight wire  (ms2.gif)

Here B 1 B 1 , B 2 B 2 , B 3 B 3 and B 4 B 4 are the net magnetic fields at four different positions due to all current elements of the wire. If we draw a circular path around the straight wire such that its plane is perpendicular to the wire and its center lies on it, then each point on the perimeter is equidistant from the center. As such magnitudes of magnetic field on all points on the circle are equal. The direction of magnetic field as determined by right hand vector cross product rule is tangential to the circle.

Figure 3: Magnetic field on the perimeter of circle is tangential.
Magnetic field due to current in straight wire
 Magnetic field due to current in straight wire  (ms3.gif)

The observations as above are the basis of Right hand thumb rule for finding direction of magnetic field due to current in straight wire. If holding straight wire with right hand so that the extended thumb points in the direction of current, then curl of the fingers gives the direction of magnetic field around the straight wire.

Figure 4: If holding straight wire with right hand so that the extended thumb points in the direction of current, then curl of the fingers gives the direction of magnetic field around the straight wire.
Right hand thumb rule
Right hand thumb rule  (ms4.gif)

Magnetic field due to current in finite straight wire

Since directions of magnetic fields due to all current elements are same, we can integrate the expression of magnitude as given by Biot-Savart law for the small current element (we have replaced dl by dy in accordance with notation in the figure) :

B = đB = μ 0 4 π I đ y sin θ r 2 B = đB = μ 0 4 π I đ y sin θ r 2

Figure 5: Magnitude of magnetic field is obtained by integration of elemental magnetic field.
Magnetic field due current in finite straight wire
 Magnetic field due current in finite straight wire  (ms5a.gif)

In order to evaluate this integral in terms of angle φ, we determine đy, r and θ in terms of perpendicular distance “R” (which is a constant for a given point) and angle “φ”. Here,

y = R tan φ y = R tan φ d y = R sec 2 φ đ φ d y = R sec 2 φ đ φ r = R sec φ r = R sec φ θ = π 2 - φ θ = π 2 - φ

Substituting in the integral, we have :

B = μ 0 4 π I R sec 2 φ đ φ sin π 2 φ R 2 sec 2 φ = μ 0 4 π I cos φ đ φ R B = μ 0 4 π I R sec 2 φ đ φ sin π 2 φ R 2 sec 2 φ = μ 0 4 π I cos φ đ φ R

Taking out I and R out of the integral as they are constant :

B = μ 0 I 4 π R cos φ đ φ B = μ 0 I 4 π R cos φ đ φ

Integrating between angle “ φ 1 φ 1 ” and “ φ 2 φ 2 ”, we have :

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

We follow the convention whereby angle is measured from perpendicular line. The angle below perpendicular line is treated negative and angle above perpendicular line is positive. In case, we want to do away with the sign of angle, we put φ 1 = φ 1 φ 1 = φ 1 in above equation :

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

Note that angles being used with this expression are positive numbers only. Also note that the magnitude of magnetic field depends on where the point of observation P lies with respect to straight wire, which is reflected in the value of angle φ.

We can also express magnetic field due to current in a straight wire at a perpendicular distance “R” in terms of angles between straight wire and line joining point of observation and end points.

Figure 6: Magnetic field due current in finite straight wire
Magnetic field due to current in wire
 Magnetic field due to current in wire  (ms5d.gif)

B = μ 0 I 4 π R X [ sin π 2 θ 1 + sin π 2 θ 2 ] B = μ 0 I 4 π R X [ sin π 2 θ 1 + sin π 2 θ 2 ] B = μ 0 I 4 π R X [ cos θ 1 + cos θ 2 ] B = μ 0 I 4 π R X [ cos θ 1 + cos θ 2 ]

Magnetic field at a point on perpendicular bisector

In this case, angles on either side of the bisector are equal :

φ 1 = φ 2 = φ φ 1 = φ 2 = φ

Figure 7: Magnetic field at a point on perpendicular bisector
Magnetic field at a point on perpendicular bisector
 Magnetic field at a point on perpendicular bisector  (ms9.gif)

Magnetic field at a point on perpendicular bisector is :

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

Let “L” be the length of wire. Then,

sin φ = O C P C = L 2 { L 2 2 + R 2 } = L L 2 + 4 R 2 sin φ = O C P C = L 2 { L 2 2 + R 2 } = L L 2 + 4 R 2

Putting in the equation of magnetic field,

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

Example 1

Problem : A square loop of side “L” carries a current “I”. Determine the magnetic field at the center of loop.

Solution : The magnetic field due to each side of the square here is same as Magnetic field due to current in straight wire at a distance L/2 on the perpendicular bisector. The magnetic fields due to current in the four sides are in the same direction. Hence, magnitude of magnetic field due to current in loop is four times the magnetic field due to current in one side :

B = 4 X μ 0 I L 2 π R L 2 + 4 R 2 B = 4 X μ 0 I L 2 π R L 2 + 4 R 2

Here, R = L/2

Figure 8: Magnetic field at the center of square loop
Magnetic field at the center of square loop
 Magnetic field at the center of square loop  (ms10.gif)

B = 4 X μ 0 I L 2 π L 2 { L 2 + 4 L 2 2 } = 4 μ 0 I L π L 2 L B = 4 X μ 0 I L 2 π L 2 { L 2 + 4 L 2 2 } = 4 μ 0 I L π L 2 L B = 2 2 μ 0 I π L B = 2 2 μ 0 I π L

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

In this case, the angles involved are :

Figure 9: Magnitude of magnetic field is obtained by integration of elemental magnetic field.
Magnetic field due current in finite straight wire
Magnetic field due current in finite straight wire  (ms5c.gif)

φ 1 = 0 ; φ 2 = φ φ 1 = 0 ; φ 2 = φ

and

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

We can also get the expression for magnetic field in terms of length of wire. Here,

sin φ = O C P C = L L 2 + R 2 sin φ = O C P C = L L 2 + R 2

Putting in the expression of magnetic field, we have :

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

In case, R = L, then,

B = μ 0 I L 4 π L L 2 + L 2 = μ 0 I 4 π L 2 = 2 μ 0 I 8 π L B = μ 0 I L 4 π L L 2 + L 2 = μ 0 I 4 π L 2 = 2 μ 0 I 8 π L

Example 2

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

Figure 10: Magnetic field due current in the arrangment
Magnetic field due to current in the arrangment
 Magnetic field due to current in the arrangment  (ms12.gif)

Solution : 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 AC = 2 μ 0 I 8 π L = 2 μ 0 I 8 π X 4 = 2 μ 0 I 32 π B AC = 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 CD = 0 B CD = 0

For the wire segment DE, the angles between the line segment and line joining the point P with end points are known by geometry of the figure. Hence, Magnetic field due to this segment is :

B DE = - μ 0 I 4 π R cos θ 1 + cos θ 2 = - μ 0 I 4 π 2 X cos 45 0 + cos 45 0 B DE = - μ 0 I 4 π R cos θ 1 + cos θ 2 = - μ 0 I 4 π 2 X cos 45 0 + cos 45 0 B DE = - μ 0 I 4 π 2 X 2 2 = - μ 0 I 4 π B DE = - μ 0 I 4 π 2 X 2 2 = - μ 0 I 4 π

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

B EF = 0 B EF = 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 FA = 2 μ 0 I 32 π B FA = 2 μ 0 I 32 π

The net magnetic field at P is :

B = B AC + B CD + B DE + B EF + B FA B = B AC + B CD + B DE + B EF + B FA B = 2 μ 0 I 32 π + 0 μ 0 I 4 π + 0 + 2 μ 0 I 32 π B = 2 μ 0 I 32 π + 0 μ 0 I 4 π + 0 + 2 μ 0 I 32 π B = μ 0 I 2 4 16 π B = μ 0 I 2 4 16 π B = - 2.59 X 4 π X 10 - 7 X 10 16 π = - 2.59 X 10 - 7 X 10 4 B = - 2.59 X 4 π X 10 - 7 X 10 16 π = - 2.59 X 10 - 7 X 10 4 B = - 0.65 X 10 - 6 = - 6.5 X 10 - 5 T B = - 0.65 X 10 - 6 = - 6.5 X 10 - 5 T

The net magnetic field is into the plane of drawing.

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

Example 3

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.

Figure 11: Magnetic field due to current in wire.
Magnetic field due to current in wire
 Magnetic field due to current in wire  (ms8.gif)

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 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 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 B = - 5 X 10 - 7 k

Exercises

Exercise 1

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.

Figure 12: Two straight wires carrying current
Two straight wires carrying current
 Two straight wires carrying current  (ms13.gif)

Solution

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:

Figure 13: Two straight wires carrying current
Two straight wires carrying current
 Two straight wires carrying current  (ms14.gif)

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

Exercise 2

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.

Solution

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 B = 6 B a

where B a 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,

Figure 14: Magnitude of magnetic field is six times the magnetic field due to one arm.
Magnetic field at the center due current
 Magnetic field at the center due current  (ms6.gif)

φ 1 = π 6 ; φ 2 = π 6 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 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 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 B = 6 X 10 - 7 X 10 X 1 0.1732 = = 3.462 X 10 - 5 T

Exercise 3

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

Figure 15: Magnetic field due to current in wire
Magnetic field due to current in wire
 Magnetic field due to current in wire  (ms11.gif)

Solution

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 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 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 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 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 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 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 = 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 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 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 = - 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 B = - 5 X 10 - 7 T

The net magnetic field is into the plane of drawing.

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