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Magnetic field due to current in a circular wire

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

Magnetic field due to current in circular wire is largely axial. It means that we need to concentrate our investigation of magnetic field on axial positions. One of the important axial positions is center of the circular wire itself. We shall limit our discussion in this module to this case of magnetic field at the center of circular coil. The procedure for deriving expression for the magnetic field due to current in circular wire is same as that of current carrying straight wire. Here also, we make use of superposition principle whereby we combine the small magnetic fields due to each of the small current elements composing the circular coil.

Circular wire is considered to be composed of small linear current elements. We determine magnetic field due to each of the linear current elements applying Biot-Savart law. Finally, we determine net magnetic field using superposition principle (i.e. by determining vector sum of magnetic fields due to all current elements).

In general, the bending of current carrying wire in circular shape has the effect of strengthening or localizing magnetic field in narrower region about the axis.

Direction of magnetic field (Right hand thumb rule)

Let us consider two diametrically opposite small current elements on the circular wire. The magnetic field lines are compressed inside the circle as it accommodates all the circular closed lines drawn outside. This compression of magnetic field lines is maximum at the center. In the figure here, we consider the circular coil in horizontal plane. The magnetic field lines being perpendicular to current elements are in the plane of drawing.

Such is the case with any other pair of current elements as well. This means that magnetic field line passing though axis is reinforced by all such diametrically opposite pairs of current element. The magnetic field due to current in circular wire, therefore, is nearly axial.

The observations as above are the basis of Right hand thumb rule for current in circular wire. If we orient right hand such that curl of fingers follows the direction of current in the circular wire, then extended thumb points in the direction of magnetic field at its center.

Right hand thumb rules for straight wire and circular wire are opposite in the notations. The curl of hand represents magnetic field in the case of straight wire, whereas it represents current in the case of circular wire. Similarly, the extended thumb represents current in the case of straight wire, whereas it represents magnetic field in the case of circular wire.

There is yet another simple way to find the direction of axial magnetic field at the center. Just look at the circular loop facing it. If the current is clockwise, then magnetic field is away from you and if the current is anticlockwise, then magnetic field is towards you.

Current in circular wire and magnet

The directional attributes of the magnetic field due to current in circular wire have an important deduction. If the current in a circular loop is anticlockwise when we look from one end (face), then the same current is clockwise when we look from opposite end (face). What it means that if direction of magnetic field is towards you from one face, then the direction of magnetic field is away from you from the other end and vice versa.

The magnetic lines of force enters from the face in which current is clockwise and exits from the face in which current is anticlockwise. This is exactly the configuration with real magnet. The anticlockwise face of the circular wire is equivalent to north pole and clockwise face is equivalent to south pole of the physical magnet. For this reason, a current in a circular wire is approximately equivalent to a tiny bar magnet.

Magnitude of magnetic field due to current in circular wire

Evaluation of Biot-Savart expression at the center of circle for current in circular wire is greatly simplified. There are threefold reasons :

1: The directions of magnetic fields due to all current elements at the center are same just as in the case of straight wire.

2: The linear distance (r) between current length element (dl) and the point of observation (center of circular wire) is same for all current elements.

3: The angle between current length element vector (dl) and displacement vector (R) is right angle for all current elements. Recall that angle between tangent and radius of a circle is right angle at all positions on the perimeter of a circle.

The magnitude of magnetic field due to a current element according to Biot-Savart law is given by :

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

But, θ=90° and sin90°=1. Also, r = R = Radius of circular wire.

đB = μ 0 4 π I đ l R 2 đB = μ 0 4 π I đ l R 2

All parameters except "đl" in the right hand expression of the equation are constants and as such they can be taken out of the integral.

B = đB = μ 0 I 4 π R 2 đ l B = đB = μ 0 I 4 π R 2 đ l

The integration of dl over the complete circle is equal to its perimeter i.e. 2πR.

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

If the wire is a coil having N circular turns, then magnetic filed at the center of coil is reinforced N times :

B = μ 0 N I 2 R B = μ 0 N I 2 R

Example 1

Problem : A thin ring of radius “R” has uniform distribution of charge, q, on it. The ring is made to rotate at an angular velocity “ω” about an axis passing through its center and perpendicular to its plane. Determine the magnitude of magnetic field at the center.

Solution : A charged ring rotating at constant angular velocity is equivalent to a steady current in circular wire. We need to determine this current in order to calculate magnetic field. For this, let us concentrate at any cross section of the ring. All the charge passes through this cross section in one time period of revolution. Thus, equivalent current is :

I = q T = q ω 2 π I = q T = q ω 2 π

Now, magnetic field due to steady current in circular wire is :

B = μ 0 I 2 R B = μ 0 I 2 R

Substituting for current, we have :

B = μ 0 q ω 4 π R B = μ 0 q ω 4 π R

Example 2

Problem : Calculate magnetic field at the center O for the current flowing through wire segment as shown in the figure. Here, current through wire is 10 A and radius of the circular part is 0.1 m.

Solution : Magnetic field at O is contributed by long straight wire and circular wire. The direction of magnetic field at O due to straight part of the wire is into the plane of drawing as obtained by applying Right hand thumb rule for straight wire. The direction of current in the circular part is anticlockwise and hence magnetic field due to this part is out of the plane of drawing as obtained by applying Right hand thumb rule for circular wire.

The magnitude of magnetic field due to circular wire is :

B C = μ 0 I 2 R B C = μ 0 I 2 R

The magnitude of magnetic field due to straight wire is :

B S = μ 0 I 2 π R B S = μ 0 I 2 π R

Hence, magnitude of magnetic field at O is algebraic sum of two magnetic fields (we consider outward direction as positive) :

B = B C B S = μ 0 I 2 R μ 0 I 2 π R B = B C B S = μ 0 I 2 R μ 0 I 2 π R

Putting values :

B = 4 π 10 - 7 X 10 2 X 0.1 4 π 10 - 7 X 10 2 π X 0.1 B = 4 π 10 - 7 X 10 2 X 0.1 4 π 10 - 7 X 10 2 π X 0.1 B = 62.9 X 10 - 6 20 X 10 - 6 = 42.9 μ T B = 62.9 X 10 - 6 20 X 10 - 6 = 42.9 μ T

The net magnetic field is acting out of the plane of paper.

Magnitude of magnetic field due to current in circular arc

The magnitude of magnetic field due to current in arc shaped wire can be obtained by integrating Biot-Savart expression in an appropriate range. Now, the integral set up for current in circular wire is :

B = đ B = μ 0 I 4 π R 2 đ l B = đ B = μ 0 I 4 π R 2 đ l

Circular arc is generally referred in terms of the angle θ, it subtends at the center of the circle. From geometry, we know that :

đ l = R đ θ đ l = R đ θ Substituting in the integral and taking the constant R out of the integral, we have :

B = μ 0 I 4 π R đ θ B = μ 0 I 4 π R đ θ B = μ 0 I θ 4 π R B = μ 0 I θ 4 π R

This is the expression for the magnitude of magnetic field due to current in an arc which subtends an angle θ at the center. Note that the expression is true for the circle for which θ = 2π and magnetic field is :

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

Example 3

Problem : Find the magnetic field at the corner O due to current in the wire as shown in the figure. Here, radius of curvature is 0.1 m for the quarter circle arc and current is 10 A.

Solution :

Here the straight line wire segment AB and CD when extended meet at O. As such, there is no magnetic field due to current in these segments. The magnetic field at O is, therefore, solely due to magnetic field due to quarter arc AC. The arc subtends an angle π/2 at its center i.e O. Now,

B = μ 0 I θ 4 π R B = μ 0 I θ 4 π R

Putting values, we have :

B = 10 - 7 X 10 X π 0.1 X 2 = 0.157 X 10 - 6 = 0.157 μ T B = 10 - 7 X 10 X π 0.1 X 2 = 0.157 X 10 - 6 = 0.157 μ T

Since current in the arc is anticlockwise, magnetic field is perpendicular and out of the plane of drawing.

Current in straight wire .vs. current in circular wire

A length of wire, say L, is given and it is asked to maximize magnetic field in a region due to a current I in the wire. Which configuration would we consider – a straight wire or a circular wire? Let us examine the magnetic fields produced by these two configurations.

If we bend the wire in the circle, then the radius of the circle is :

R = L 2 π R = L 2 π

The magnetic field due to current I in the circular wire is :

B C = μ 0 I 2 R = 2 π μ 0 I 2 L = π μ 0 I L = 3.14 μ 0 I L B C = μ 0 I 2 R = 2 π μ 0 I 2 L = π μ 0 I L = 3.14 μ 0 I L

In the case of straight wire, let us consider that wire is long enough for a point around middle of the wire. For comparison purpose, we assume that perpendicular linear distance used for calculating magnetic field due to current in straight wire is equal to the radius of circle. The magnetic field at a perpendicular distance “R” due to current in long straight wire is given as :

B L = μ 0 I 4 π R = μ 0 I X 2 π 4 π L = μ 0 I 2 L = 0.5 μ 0 I L B L = μ 0 I 4 π R = μ 0 I X 2 π 4 π L = μ 0 I 2 L = 0.5 μ 0 I L

Clearly, the magnetic field due to current in circular wire is 6.28 times greater than that due to current in straight wire at comparable points of observations. Note that this is so even though we have given advantage to straight wire configuration by assuming it to be long wire. In a nutshell, a circular configuration tends to concentrate magnetic field along axial direction which is otherwise spread over the whole length of wire.

Example 4

Problem : A current 10 A flowing through a straight wire is split at point A in two semicircular wires of radius 0.1 m. The resistances of upper and lower semicircular wires are 10 Ω and 20 Ω respectively. The currents rejoin to flow in the straight wire again as shown in the figure. Determine the magnetic field at the center “O”.

Solution : The straight wire sections on extension pass through the center. Hence, magnetic field due to straight wires is zero. Here, the incoming current at A is distributed in the inverse proportion of resistances. Let the subscripts “1” and “2’ denote upper and lower semicircular sections respectively. The two sections are equivalent to two resistances in parallel combination as shown in the figure. Here, potential difference between “A” and “B” is :

V A B = I X R 1 X R 2 R 1 + R 2 = I 1 R 1 = I 2 R 2 V A B = I X R 1 X R 2 R 1 + R 2 = I 1 R 1 = I 2 R 2 I 1 = I X R 2 R 1 + R 2 = 10 X 20 30 = 20 3 A I 1 = I X R 2 R 1 + R 2 = 10 X 20 30 = 20 3 A I 2 = I X R 1 R 1 + R 2 = 10 X 10 30 = 10 3 A I 2 = I X R 1 R 1 + R 2 = 10 X 10 30 = 10 3 A

We see that current in the upper section is twice that in the lower section i.e. I 1 = 2 I 2 I 1 = 2 I 2 . Also, the magnetic field is perpendicular to the plane of semicircular section (plane of drawing). The current in the upper semicircular wire is clockwise. Thus, the magnetic field due to upper section is into the plane of drawing. However, the current in the lower semicircular is anticlockwise. Thus, the magnetic field due to lower section is out of the plane of drawing. Putting θ = π for each semicircular section, the net magnetic field due to semicircular sections at “O” is:

B = μ 0 I 1 π 4 π R μ 0 I 2 π 4 π R B = μ 0 I 1 π 4 π R μ 0 I 2 π 4 π R B = μ 0 I 1 π 4 π R μ 0 I 1 π 8 π R = μ 0 I 1 π 8 π R B = μ 0 I 1 π 4 π R μ 0 I 1 π 8 π R = μ 0 I 1 π 8 π R B = 10 - 7 X 20 3 X 8 X 0.1 = 8.3 X 10 - 7 T B = 10 - 7 X 20 3 X 8 X 0.1 = 8.3 X 10 - 7 T

The net magnetic field is into the plane of drawing.

Exercises

Exercise 1

An electron circles a single proton nucleus of radius 3.2 X 10 - 11 3.2 X 10 - 11 m with a frequency of 10 16 10 16 Hz. The charge on the electron is 1.6 X 10 - 19 1.6 X 10 - 19 Coulomb. What is the magnitude of magnetic field due to orbiting electron at the nucleus?

Solution

The equivalent current is given by :

I = q T = q ν I = q T = q ν

where ν and T are frequency and time period of revolutions respectively. The magnitude of magnetic field due to circular wire is given by :

B = μ 0 I 2 R B = μ 0 I 2 R

Substituting for I, we have :

B = μ 0 I 2 R = μ 0 q ν 2 R B = μ 0 I 2 R = μ 0 q ν 2 R

Putting values,

B = 4 π 10 - 7 X 1.6 X 10 - 19 X 10 16 2 X 3.2 X 10 - 11 B = 4 π 10 - 7 X 1.6 X 10 - 19 X 10 16 2 X 3.2 X 10 - 11 B = 31.4 T B = 31.4 T

Exercise 2

Calculate the magnetic field at O for the current loop shown in the figure.

Solution

The magnetic field due to linear part of the wire is zero as they pass through O when extended. The magnetic field due to inner arc is greater than outer arc. Further, magnetic field due to anticlockwise current in the inner arc is out of the plane of drawing and magnetic field due to clockwise current in the outer arc is into the plane of drawing. Net magnetic field due to the current in the wire is out of the plane of drawing, whose magnitude is :

B = μ 0 I θ 4 π r 1 μ 0 I θ 4 π r 2 B = μ 0 I θ 4 π r 1 μ 0 I θ 4 π r 2 B = μ 0 I π 4 π r 1 X 4 μ 0 I π 4 π r 2 X 4 B = μ 0 I π 4 π r 1 X 4 μ 0 I π 4 π r 2 X 4 B = μ 0 I 16 r 2 r 1 r 1 r 2 B = μ 0 I 16 r 2 r 1 r 1 r 2

Exercise 3

A current of 10 ampere flows in anticlockwise direction through the arrangement shown in the figure. Determine the magnetic field at the center “O”.

Solution

The magnetic field at “O” due to ¾ th of the circular arc is :

B C = μ 0 I X 3 π 4 π R X 2 = 3 μ 0 I 8 R = 3 X 4 π X 10 - 7 X 10 8 X 3 B C = μ 0 I X 3 π 4 π R X 2 = 3 μ 0 I 8 R = 3 X 4 π X 10 - 7 X 10 8 X 3 B C = 5 π X 10 - 7 = 15.7 X 10 - 7 T B C = 5 π X 10 - 7 = 15.7 X 10 - 7 T

Two linear part segments when extended pass through “O” and as such do not contribute to magnetic field. The magnetic field at “O” due to one 5 m segment is :

B L 1 = 2 μ 0 I 8 π R = 2 X 4 π X 10 - 7 X 10 8 π X 5 B L 1 = 2 μ 0 I 8 π R = 2 X 4 π X 10 - 7 X 10 8 π X 5 B L 1 = 2 X 10 - 7 T B L 1 = 2 X 10 - 7 T

The magnetic field at “O” due to two 5 m segments is :

B L = 2 X B L 1 = 2 X 2 X 10 - 7 = 2.83 X 10 - 7 T B L = 2 X B L 1 = 2 X 2 X 10 - 7 = 2.83 X 10 - 7 T

Magnetic fields due to both circular arc and linear segments are acting out of the plane of drawing, the net magnetic field at “O” is :

B = B C + B L = 15.7 X 10 - 7 + 2.83 X 10 - 7 = 18.53 X X 10 - 7 = 1.853 X 10 - 6 T B = B C + B L = 15.7 X 10 - 7 + 2.83 X 10 - 7 = 18.53 X X 10 - 7 = 1.853 X 10 - 6 T

Exercise 4

A current of 10 ampere flows in anticlockwise direction through the arrangement shown in the figure. The curved part is a semicircular arc. Determine the magnetic field at the center “O”.

Solution

The magnetic field at “O” due to semicircular arc acts upward and its magnitude is :

B C = μ 0 I X π 4 π R = μ 0 I 4 X 1 = 4 π X 10 - 7 X 10 4 = 31.4 X 10 - 7 T B C = μ 0 I X π 4 π R = μ 0 I 4 X 1 = 4 π X 10 - 7 X 10 4 = 31.4 X 10 - 7 T

The magnetic field due to lower straight conductor acts upward and its magnitude is :

B L 1 = 2 μ 0 I 8 π R = 2 X 4 π X 10 - 7 X 10 8 π X 1 B L 1 = 2 μ 0 I 8 π R = 2 X 4 π X 10 - 7 X 10 8 π X 1

B L 1 = 2 X 5 X 10 - 7 = 7.07 X 10 - 7 T B L 1 = 2 X 5 X 10 - 7 = 7.07 X 10 - 7 T

The magnetic field due to upper straight conductor acts upward and its magnitude is equal to that due to lower straight conductor :

B L 2 = 2 X 5 X 10 - 7 = 7.07 X 10 - 7 T B L 2 = 2 X 5 X 10 - 7 = 7.07 X 10 - 7 T

For the straight conductor at the far end, the center “O” lies on the bisector. The magnetic field acts upward and its magnitude is :

B L 3 = μ 0 I L 4 π R L 2 + 4 R 2 B L 3 = μ 0 I L 4 π R L 2 + 4 R 2

Here, R = 2 m, L = 2 m. Putting values in the equation, we have :

B L 3 = μ 0 I L 4 π R L 2 + 4 R 2 = 4 π X 10 - 7 X 10 X 2 4 π X 2 2 2 + 4 X 2 2 B L 3 = μ 0 I L 4 π R L 2 + 4 R 2 = 4 π X 10 - 7 X 10 X 2 4 π X 2 2 2 + 4 X 2 2 B L 3 = 5 X 10 - 7 = 2.24 X 10 - 7 T B L 3 = 5 X 10 - 7 = 2.24 X 10 - 7 T

The net magnetic field at “O” is :

B = B C + B L 1 + B L 2 + B L 3 B = B C + B L 1 + B L 2 + B L 3 B = 31.4 X 10 - 7 + 7.07 X 10 - 7 + 7.07 X 10 - 7 + 2.24 X 10 - 7 B = 31.4 X 10 - 7 + 7.07 X 10 - 7 + 7.07 X 10 - 7 + 2.24 X 10 - 7 B = 4.78 X 10 - 6 T B = 4.78 X 10 - 6 T

Exercise 5

A thin disc of radius “R” has uniform distribution of charge, q, on it. The ring is made to rotate at an angular velocity “ω” about an axis passing through its center and perpendicular to its plane. Determine the magnitude of magnetic field at the center of the disc.

Solution

We consider disc to be composed of infinite numbers of thin ring. We consider one such ring of thickness dr at a distance “r” from the center carrying charge dq. This ring carrying charge “dq” and rotating is equivalent to a current. The magnetic field at the center to this thin ring is (as obtained earlier in the example problem) :

đ B = μ 0 đ q ω 4 π r đ B = μ 0 đ q ω 4 π r

We need to determine “dq” in terms of given parameters. The current surface density, σ, is :

σ = q π R 2 σ = q π R 2

The area of the thin ring is :

đ A = 2 π r đ r đ A = 2 π r đ r

Hence, charge on the ring is :

đ q = σ đ A = 2 π r q đ r π R 2 = 2 r q đ r R 2 đ q = σ đ A = 2 π r q đ r π R 2 = 2 r q đ r R 2

Putting this espression for “dq”, the expression of magnetic field at the center due to rotating ring is :

đ B = μ 0 2 r q ω đ r 4 π r R 2 = μ 0 q ω đ r 2 π R 2 = μ 0 ω q đ r 2 π R 2 đ B = μ 0 2 r q ω đ r 4 π r R 2 = μ 0 q ω đ r 2 π R 2 = μ 0 ω q đ r 2 π R 2

In order to obtain magnetic field due to the rotating disc, we integrate the expression of magnetic field due to ring from r = 0 to r =R.

B = đB = 0 R μ 0 ω q đ r 2 π R 2 B = đB = 0 R μ 0 ω q đ r 2 π R 2

Taking out constants out of the integration sign, we have :

B = μ 0 w q 2 π R 2 0 R đ r B = μ 0 w q 2 π R 2 0 R đ r B = μ 0 w q 2 π R B = μ 0 w q 2 π R

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