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Physics Topics can be both theoretical and experimental, with scientists using a range of tools and techniques to understand the phenomena they investigate.
What are the Characteristics of Magnetic Field of a Circular Conductor?
i) Magnetic field at the centre of a circular conductor:
Let the current through a circular conductor of radius r be I.
Due to an element of length δl of the conductor, magnetic field at the centre of the circle [Fig.]
δB = \(\frac{\mu_0}{4 \pi} \cdot \frac{I \delta l \sin 90^{\circ}}{r^2}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{I \delta l}{r^2}\)
If the circumference of the circle be divided into a large number of such elements then, for each element, I and r remain the same and the angle between that element and the radius is 90°. Again sum of these elements = circumference of the circle = 2πr.
So, the magnetic field at the centre of the circular conductor,
B = \(\frac{\mu_0}{4 \pi} \frac{I}{r^2} \sum \delta l\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r^2} \cdot 2 \pi r\) = \(\frac{\mu_0 I}{2 r}\) …. (1)
If the circular conductor of single turn be replaced by a circular coil of N turns then,
B = \(\frac{\mu_0 N I}{2 r}\) …. (2)
The direction of magnetic field at the centre of the circular conductor can be determined by corkscrew rule. In the Fig., this direction is normally upward from the plane of the paper. If the conductor is in the form of a circular arc and if that arc makes an angle θ (rad) at the centre (Fig.), the magnetic field at the centre of the circle,
B = \(\frac{\mu_0 I}{2 r} \cdot \frac{\theta}{2 \pi}\) = \(\frac{\mu_0 I}{4 \pi r} \cdot \theta\)
ii) Magnetic field on the axis of a circular conductor: Let r = radius of a circular conductor, I = current through the conductor, O is the centre of the circular conductor [Fig.]. P is any point on the axis (OP = x). An element of length δl is considered at the topmost point C of the conductor. The line segment CP is perpendicular to this element. So, for the element of length δl at C, magnetic tìeld at the point P is,
From corkscrew rule we see that, the direction of δB is along PQ. The component of δB along the axis is δBsinθ and perpendicular to the axis is δBcosθ. If an element of length δl is now considered at the point D diametrically opposite to C on the circumference, the magnetic field at the point P will be δB and its direction will be along PR. Naturally its downward component δBcosθ neutralises the previous component δBcosθ, but two components
δBsinθ each will be added together along the axis. In this way, if the whole circular conductor is considered, the algebraic sum of the components δBsinθ along the axis will be the resultant magnetic field due to the circular conductor at the point P.
∴ B = \(\sum \delta B \sin \theta\) = \(\frac{\mu_0}{4 \pi} \sum \frac{I \delta l}{u^2} \cdot \frac{r}{u}\) = \(\frac{\mu_0}{4 \pi} \sum \frac{I \delta l}{u^3} \cdot r\)
Now, due to symmetry of the circular conductor, the magnitudes of I, r and u will be the same at every point on its circumference. Hence,
If a circular coil having N turns is taken instead of the circular conductor of single turn then,
B = \(\frac{\mu_0 N I}{2} \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\) …. (5)
Now, at the centre of the circle, i.e., at the point O, x = 0 and hence,
B = \(\frac{\mu_0 N I}{2 r}\), which is identical to the equation(2)
Expressions in CGS or Gaussian system: In equations (1) to (5) above, substituting B → H, I → i and µ0 → 4π we get, magnetic intensity at the centre nl a circular conductor,
H = \(\frac{2 \pi i}{r}\)
and magnetic intensity at the centre of a circular coil having N-turns,
H = \(\frac{2 \pi N i}{r}\)
In case of a conductor in the form of an arc of a circle, magneic intensity at the Centre of the conductor, H = \(\frac{i}{r} \theta\) where, θ is the angle made by the arc at the centre.
Magnetic intensity at any point on the axis of a circular conductor,
H = \(2 \pi i \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\)
In case of a circular coil having N turns, magnetic intensity at any point on its axis,
H = \(2 \pi N i \cdot \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\)
Angle (θ) subtended by a current carrying conductor of length 1 cm bent in the form of an arc of a circle of 1 cm radius is 1 rad. 1f the magnetic intensity at the centre of the arc is 1 Oe, then substituting r = 1, θ = 1, H = 1 in the relation H = \(\frac{i \theta}{r}\), we get, i = 1. The definition of emu of current is given in sec-tian 1.4 in the same fashion.
Characteristics of magnetic field of a circular conductor:
i) Direction of magnetic field: At any point on the axis of the conductor, the direction of the resultant magnetic field is always along the axis. For the direction of current shown in Fig., the magnetic field at the point P is along the axis and directed outward. But on the opposite side of the coil, at the point P’, magnetic field is along the axis and directed inward, i.e., along P’ O. If the direction of current in the coil is reversed, at the point P magnetic field will he inward while at the point P’, it will be outward.
ii) Magnitude of the magnetic field: From equation (5) we see that, the magnitude of the field becomes maximum at the centre of the circle, and decreases along the axis, on either B side [Fig.].
As a result, no uniform magnetic field is obtained at any point on the axis and thus a problem arises while constructing electrical instruments with a circular coil. This problem can be removed by using Helmholtz double coil.
Heimbotti double coil: Two circular coils having the same radius (= r) are placed coaxially at a distance equal to their radius [Fig.]. If a direct current is passed through them in the same direction, the magnetic fields generated between them will have the same direction. Thus the resultant magnetic field remains almost uniform in between the coils.
Numerical Examples
Example 1.
The radii of two concentric circular coils are 8 cm and 10 cm and the number of turns in them are 40 and 10, respectively. A 5 A current is passing through each of them in the same direction. Determine the magnetic field produced at the centre of the two coils.
Solution:
Since current flows in the same direction through the two coils, the directions of magnetic fields at the centre due to the coils will be the same. Hence, the resultant magnetic field will be obtained by adding these magnetic fields.
For the first coil, B1 = \(\frac{\mu_0 N_1 I}{2 r_1}\)
For the second coil, B2 = \(\frac{\mu_0 N_2 I}{2 r_2}\)
∴ The resultant magnetic field,
B = B1 + B2 = \(\frac{\mu_0 I}{2}\left(\frac{N_1}{r_1}+\frac{N_2}{r_2}\right)\)
= \(\frac{4 \pi \times 10^{-7} \times 5}{2}\left(\frac{40}{8}+\frac{10}{10}\right)\)
= 2π × 10-7 × 5 × 6 = 1.885 × 10-5 Wb ᐧ m-2
Example 2.
A current I is flowing through an infinitely long wire PQRS [Fig.]. The wire is bent at right angles so that the part QR becomes one-fourth of the circumference of a circle of radius r whose centre is at O. Determine the magnetic field at O.
Solution:
With respect to the point O, both the parts PQ and RS are semi-infinite wires and hence, for each part magnetic field at the point O will be, B1 = B2 = \(\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}\)
Again for a complete circular conductor, the magnetic field at its centre = \(\frac{\mu_0 I}{2 r}\)
So, for the one-fourth part QR of the circle, the magnetic field at O,
B3 = \(\frac{1}{4} \times \frac{\mu_0 I}{2 r}\) = \(\frac{\mu_0 I}{8 r}\)
Hence, the resultant magnetic field at O,
Example 3.
Determine the magnetic field at the point O due to the circuit shown in Fig.
Solution:
The point O lies on the same straight line with the two linear parts of the circuit. Hence, no magnetic field acts at O due to those two parts.
For a complete circular conductor, magnetic field at the centre of the circle = \(\frac{\mu_0 I}{2 r}\)
Hence, for a semicircular conductor, magnetic field at the centre = \(\frac{\mu_0 I}{4 r}\).
In the present figure, magnetic field at O due to the semicircular conductor of radius r is B1 = \(\frac{\mu_0 I}{4 r}\) and that due to the semicircular conductor of radius R is B2 = \(\frac{\mu_0 I}{4 R}\).
The corkscrew rule shows that, B1 and B2 are oppositely directed. Since B1 > B2, the resultant magnetic field at the point O (upward with respect to the page of the book), B = B1 – B2 = \(\frac{\mu_0 I}{4}\left(\frac{1}{r}-\frac{1}{R}\right)\) = \(\frac{\mu_0 I(R-r)}{4 r R}\)
Example 4.
What is the magnetic field produced at the centre of a hydrogen atom due to revolution of its electron in the first orbit (K-orbit)? Radius of the first orbit = 0.53 × 10-10 m, velocity of electron in that orbit = 2.19 × 106 m ᐧ s-1.
Solution:
Time period of revolution, T = \(\frac{2 \pi r}{v}\)
So, the circular ioop formed due to revolution of the electron carries an effective current,
I \(=\frac{\text { charge }}{\text { time period }}\) = \(\frac{e}{T}\) = \(\frac{e v}{2 \pi r}\)
So, magnetic field at the centre of the atom,
B = \(\frac{\mu_0 I}{2 r}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{e v}{r^2}\)
= 10-7 × \(\frac{\left(1.6 \times 10^{-19}\right) \times\left(2.19 \times 10^6\right)}{\left(0.53 \times 10^{-10}\right)^2}\) = 12.47 T
Example 5.
Calculate the magnetic induction at point O (centre of the partial circular conductor) shown in the figure.
Solution:
Magnetic induction at the centre due to whole conductor, B = Magnetic induction due to (straight part AB + curved part BCD + straight part DE)
∴ Btotal = BAB + BBCD + BDE
BAB = 0 (∵ the point O is along AB)
BBCD = \(\frac{3}{4}\) of the magnetic field due to the whole circle
Example 6.
H Two concentric but mutually perpendicular conducting coils are carrying current 3 A and 4 A, respectively. If the radius of each coil be 2π cm, what will be the magnetic induction at the centre of the coils? (μ0 = 4π × 10-7 Wb ᐧ A-1 ᐧ m-1) (AIEEE’05)
Solution:
r = 2π cm = \(\frac{\pi}{50}\) cm
Magnetic field at the centre of a circular coil, B = \(\frac{\mu_0 I}{2 r}\)
∴ For the first coil,
B1 = \(\frac{\left(4 \pi \times 10^{-7}\right) \times 3}{2 \times \frac{\pi}{50}}\) = 3 × 10-5 Wb ᐧ m-2
and for the second coil,
B2 = \(\frac{\left(4 \pi \times 10^{-7}\right) \times 4}{2 \times \frac{\pi}{50}}\) = 4 × 10-5 Wb ᐧ m-2
Since the coils are mutually perpendicular, B1 and B2 are also at right angles to each other. Hence, the resultant magnetic field,
B = \(\sqrt{B_1^2+B_2^2}\) = \(\sqrt{(3)^2+(4)^2} \times 10^{-5}\)
= 5 × 10-5 Wb ᐧ m-2
Example 7.
Each of two long straight wires, passing through the points A and B of Fig., carries a current I directed vertically upwards with respect to the plane of the paper. The separation between them is r. Find out the magnetic field at a point P on that plane, which is at a distance r from each of the wires.
Solution:
Here PA = PB = r
Due to the current carrying conductor, which passes through the point A, the magnetic field intensity at P is,
B1 = \(\frac{\mu_0}{4 \pi} \frac{2 I}{P A}\) = \(\frac{\mu_0}{4 \pi} \frac{2 I}{r}\)
(along \(\overrightarrow{P B_1}\) from, Fig.)
Similarly, due to the current carrying conductor which passes through the point B, the magnetic field intensity at point P is,
B2 = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{P B}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 I}{r}\) (along \(\overrightarrow{P B_2}\))
The resultant magnetic field intensity at point P is given by,
Example 8.
Two small identical circular loops marked (1) and (2) carrying equal currents, are placed with their geometrical axes perpendicular to each other as shown in figure. Find the magnitude and direction of the net magnetic field produced at O. Also determine the field when radius of the loop Is very large as compared to the distance of the point from centre.
Solution:
Let R be the radius of each loop. Magnetic field at O due to loop 1,
∴ Net Magnetic Field at O due to both the loops,
Example 9.
Two circular coils of radii a and 2a having a common centre, carry identical current i, but in opposite directions. Number of turns of the second conductor is 8. Show that magnetic field intensity at the centre is 3 times that due to the smaller one. Also find out the change in the previous result when current flows in the same direction through both the coils.
Solution:
Magnetic field intensity at the centre O due to the smaller loop is,
B1 = \(\frac{\mu_0 i}{2 a}\)(upwards)
Similarly magnetic field intensity at the centre due to the bigger loop.
Hence resultant field is 3 times that due to the smaller loop.
Now if the direction of the Current is same for both the loops, the resultant field will be, B = B1 + B2 = \(\frac{\mu_0 i}{2 a}+\frac{2 \mu_0 i}{a}\) = \(5 \frac{\mu_0 i}{2 a}\) = 5B1
Hence the resultant field wifi be 5 times that due to the smaller loop if current flows in the same direction.
Example 10.
A wire loop is formed by joining two semicircular wires of radii r1 and r2 as shown in the figure. If the loop carries a current I, find the magnetic field at the centre O.
Solution:
Magnetic field at the point O due to the semicircular part MNP is,
BMNP = \(\frac{\mu_0}{4} \cdot \frac{I}{r_1}\) (upwards)
Similarly magnetic field at O due to the semicircular part GFE is,
BGFE = \(\frac{\mu_0}{4} \cdot \frac{I}{r_2}\) (upwards)
As the point O lies along the straight parts ME and PG of the loop, the magnetic field due to them at O is zero.
So net magnetic field at O due to the whole loop,
B = B1 + B2
= \(\frac{\mu_0}{4} \cdot \frac{I}{r_1}\) + \(\frac{\mu_0}{4} \cdot \frac{I}{r_2}\)
∴ B = \(\frac{\mu_0}{4} I\left[\frac{1}{r_1}+\frac{1}{r_2}\right]\) (upwards)
Example 11.
Radius and number of turns of a circular coil are 10 cm and 25 respectively. What should be the current through the coil that will produce a magnetic field of 6.28 × 10-5Wb ᐧ m-2 at its centre?
Soluton:
Radius of the coil, r = \(\frac{10}{2}\) = 5 cm = 0.05 m; number of turns, N = 25 and magnetic field at the centre of the coil B = 6.28 × 10-5 Wb ᐧ m-2.
Now, B = \(\frac{\mu_0 N I}{2 r}\)
∴ I = \(\frac{2 r B}{\mu_0 r}\) = \(\frac{2 \times 0.05 \times\left(6.28 \times 10^{-5}\right)}{\left(4 \times \pi \times 10^{-7}\right) \times 25}\) = 0.2A
Example 12.
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54µT. What will be its value at the centre of the loop? [AIEEE ‘04]
Solution:
Magnetic field on the axis of a circular conductor; B = \(\frac{\mu_0 I}{2} \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}}\) and magnetic field at the centre (x = 0) is,
B’ = \(\frac{\mu_0 I}{2 r}\)
∴ \(\frac{B^{\prime}}{B}\) = \(\frac{\left(r^2+x^2\right)^{3 / 2}}{r^3}\)
or B’ = \(\frac{\left(r^2+x^2\right)^{3 / 2}}{r^3} B\)
= \(\frac{\left(3^2+4^2\right)^{3 / 2}}{3^3} \times 54\)
∴ B’ = 250µT