Magnetic Field on the Axis of a Circular Current Loop: Let’s understand how a magnetic field on the axis of a circular current loop works. There are two methods of calculating magnetic fields in magnetics at some point. One is Biot-Savart law, and the other is Ampere’s law. To find the magnetic field formula due to an infinitesimally small current-carrying wire at some point, we use the Biot-Savart law to calculate the magnetic field of a highly symmetric configuration carrying a steady current Ampere’s Circuital Law. 

Check here to use the Biot-Savart law to calculate a small current element’s magnetic field produced at some point in space. Using this formalism and the principle of superposition, we will define a magnetic field and calculate the total magnetic field due to the circular current loop.

Biot-Savart Law

Here, we will study the relationship between current and the magnetic field it produces. It is given by Biot-Savart’s law. Let us assume a finite conductor \(XY\) which is carrying a current \(I\), as shown in figure (a). Then, the magnetic field \(d\overrightarrow B \) due to an infinitesimal element \(\left( {d\overrightarrow l } \right)\) of the conductor is to be determined at a point \(P\) which is at a distance of \(\overrightarrow r \) from it. Let’s say that the angle between \({d\overrightarrow l }\) and the position vector \(\left( {\overrightarrow r } \right)\) is \(\theta\). Now, according to Biot-Savart’s law, the magnitude of the magnetic field \(d\overrightarrow B \) is proportional to the current \((I)\), the elemental length \(\left| {d\overrightarrow l } \right|\), and it is inversely proportional to the square of the distance \((r)\). And the direction is perpendicular to the plane, which contains \(d\overrightarrow l \) and \(\overrightarrow r\). Thus, in vector notation, we can write this as:

Fig. (a)

Now,
\({\text{d}}\overrightarrow B \propto \frac{{{\text{d}}\overrightarrow l \times \overrightarrow r}}{{{r^3}}}\)
\( {\text{d}}\overrightarrow B = \frac{{{\mu _0}}}{{4\pi }}\frac{{I\;{\text{d}}\overrightarrow l \times \overrightarrow r}}{{{r^3}}}\) ……(i)
Where, \(μ_0/4π\) is a constant of proportionality. The above expression holds when the medium is the vacuum. And by using the property of cross-product, we can also write
\(\left| {{\text{d}}\overrightarrow B} \right| = \;\frac{{{\mu _0}}}{{4\pi }}\frac{{Idl\sin \theta }}{{{r^2}}}\) ……(ii)
And,
\(\frac{{{\mu _0}}}{{4\pi }} = {10^{ – 7}}\;{\text{T}\;\text{m}}{{\text{A}}^{ – 1}}\)
Where, \(μ_0\) is the permeability of space.

Magnetic Field on the Axis of a Circular Current Loop

To determine the magnetic field due to a circular coil along its axis. Let us assume that the current \((I)\) is steady and that the evaluation is carried out in space (i.e., vacuum) by summing up the effect of infinitesimal current elements \(\left( {I{\text{d}}\overrightarrow l } \right)\). Let consider a circular loop carrying a steady current \((I)\), as shown in Fig. (b). The loop is placed in the \(y-z\) plane with its centre at the origin \(O\) and has a radius \((R)\). Now we will calculate the magnetic field at the point P on the axis of the loop, i.e. \(x-\)axis. Then, from the centre \(O\) of the loop, the distance of \(P\) be \(x\). Now, consider a small conducting element \(\left( {{\text{d}}\overrightarrow l } \right)\) of the loop. Now, according to Biot-Savart law, the magnitude of the magnetic field \(\left( {{\text{d}}\overrightarrow B } \right)\) due to element \(\left( {{\text{d}}\overrightarrow l } \right)\) will be
\({\text{d}}B\; = \frac{{{\mu _0}}}{{4\pi }}\frac{{I\left| {{\text{d}}\overrightarrow l \times \overrightarrow r} \right|}}{{{r^3}}}\) ……..(1)

Fig. (b)

Now, \(r^2 = x^2 + R^2\). Further, from the element to the axial point, any element of the loop will be perpendicular to the displacement vector. For example, the element \({\text{d}}\overrightarrow l \) in Fig. (b) is in the \(y-z\) plane, whereas the displacement vector \(\left( {\overrightarrow r } \right)\) is in the \(x-y\) plane. Hence, \(\left| {{\text{d}}\overrightarrow l \times \overrightarrow r } \right| = rdl\)
Thus,
\({\text{d}}B = \frac{{{\mu _0}}}{{4\pi }}\frac{{Idl}}{{\left( {{x^2} + {R^2}} \right)}}\) ……(2)
The direction of \({\text{d}}\overrightarrow B\) is shown in Fig.(c). It is perpendicular to the plane formed by \({\text{d}}\overrightarrow l \) and \(\overrightarrow r\). It has an \(x-\)component \({\text{d}}\overrightarrow B_x \) and a component perpendicular to the \(x-\)axis, \({\text{d}}\overrightarrow B_⊥\). And when the components perpendicular to the \(x-\)axis are summed over, they cancel out each, and then we obtain a null result. For example, the \({\text{d}}\overrightarrow B_⊥\) component due to \({\text{d}}\overrightarrow l \) is cancelled by the contribution due to the diametrically opposite to \({\text{d}}\overrightarrow l \) element, shown in Fig.(b). Thus, only the \(x-\)component survives. On integrating \({\text{d}}{\overrightarrow B _x} = {\text{d}}\overrightarrow B \,\cos \,\theta \) over the loop the net contribution along \(x-\)direction can be obtained as
\(\cos \theta = \frac{R}{{{{\left( {{x^2} + {R^2}} \right)}^{1/2}}}}\) ……(3)
From Equation (2) and (3), we have
\(d{\overrightarrow B_{\text{x}}} = \frac{{{\mu _0}Idl}}{{4\pi }}\frac{R}{{{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}} \hat i\)
The summation of elements \(({\text{d}}\overrightarrow l )\) over the loop is the circumference of the loop, i.e. \(2πR\). Thus, we can write the magnetic field at P due to the entire circular loop is
\(\overrightarrow B = {B_{\text{x}}}\hat i = \frac{{{\mu _0}I{R^2}}}{{2{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}\widehat i\) ……..(4)
If we put \(x = 0\) in this above equation we can get the magnetic field at the centre of the loop that is:
\(B_0 = \frac{μ_0 I}{2R} \widehat i\) ……..(5)
The magnetic field lines due to a current-carrying circular wire form a closed loop which is shown in Fig.(c). We can get the direction of the magnetic field by the right-hand thumb rule, which states that curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current, then the right-hand thumb will give the direction of the magnetic field.

Fig. (c)

Summary

According to the Biot-Savart law, we can get that the magnetic field \({\text{d}}\overrightarrow B \) at a point \(P\), due to an element \(({\text{d}}\overrightarrow l )\) which is carrying a steady current \((I)\) at a distance \((\overrightarrow r )\) from the current element is given as
\({\text{d}}\overrightarrow B = \frac{{{\mu _0}}}{{4\pi }}I\frac{{{\text{d}}\overrightarrow l \times \overrightarrow r}}{{{r^3}}} \)
And, we will integrate this vector expression over the entire length of the conductor to obtain the total field at \(P\).
The magnitude of the magnetic field due to a circular coil of radius \((R)\) carrying a current \((I)\) at an axial distance \((x)\) from the centre is given as
\(B = \frac{{{\mu _0}I{R^2}}}{{2{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}\), and
At the centre \((x = 0)\), this equation reduces to
\(B = \frac{{{\mu _0}I}}{{2R}}\)

Solved Problems

Q 1. The magnetic field \((\overrightarrow B )\) due to a current-carrying circular loop of radius \(12\;\rm{cm}\) at its centre is \(0.5 × 10^{-4}\;\rm{T}\). Find the magnetic field at a point on the axis at a distance of \(5\;\rm{cm}\) from the centre due to this loop?
Ans: Magnitude of the magnetic field at the centre of a circular loop is
\({B_1} = \frac{{{\mu _0}I}}{{2R}}\)
And, at the axial point, we have
\({B_2} = \frac{{{\mu _0}I{R^2}}}{{2{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}\)
Thus,
\(\frac{{{B_2}}}{{{B_1}}} = \frac{{{R^3}}}{{{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}\)
\( \Rightarrow {B_2} = {B_1}\left[ {\frac{{{R^3}}}{{{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}} \right]\)
\(\Rightarrow {B_2} = \left( {0.5 \times {{10}^{ – 4}}} \right)\left[ {\frac{{{{\left( {12} \right)}^3}}}{{{{\left( {{5^2} + {{12}^2}} \right)}^{3/2}}}}} \right]\)
\( \Rightarrow {B_2} = 3.9 \times {10^{ – 5}}\;{\text{T}}.\)

Q 2. Find is the magnitude of the net magnetic field at point \(P\) as shown in the figure? If two loops of wire carry the same current of \(10\,\rm{mA}\), but the flow of current is in opposite directions. One loop has a radius of \(R_1 = 50\;\rm{cm}\) and the other loop has a radius of \(R_2 = 100\;\rm{cm}\). From the first loop to the point \((P)\), the distance is \(0.25\;\rm{m}\), and from the second loop to the point, the distance is \(0.75\,\rm{m}\) from the point\((P)\).

Ans: The net magnetic field is the difference between the two fields generated by the coils because the currents are flowing in opposite directions. Using the given quantities in the problem, the net magnetic field at point \(P\) can be calculated by the equation given below:
Solving for the magnitude of net magnetic field using the equation and the quantities in the problem yields:
\({B_x} = \frac{{{\mu _0}I{R_x}^2}}{{2{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}\)
\(B = B_1 – B_2\)
\( \Rightarrow B = \frac{{{\mu _0}I{R_1}^2}}{{2{{\left( {{x_1}^2 + {R_1}^2} \right)}^{3/2}}}} – \frac{{{\mu _0}I{R_2}^2}}{{2{{\left( {{x_1}^2 + {R_2}^2} \right)}^{3/2}}}}\)
\( \Rightarrow B = \frac{{\left( {4{{\pi }} \times {{10}^{ – 7}}} \right)\left( {0.010} \right){{\left( {0.5} \right)}^2}}}{{2{{\left( {{{0.25}^2} + {{0.5}^2}} \right)}^{3/2}}}} – \frac{{\left( {4{{\pi }} \times {{10}^{ – 7}}} \right)\left( {0.010} \right){{\left( 1 \right)}^2}}}{{2{{\left( {{{0.75}^2} + {1^2}} \right)}^{3/2}}}}\)
\(⇒ B = 5.77 × 10^{-9}\;\rm{T}\) to the right.

FAQs

Q.1: How do you find the direction of the magnetic field on the axis of the circular loop?
Ans: The direction of the magnetic field is given by the right-hand thumb rule. That is Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current.

Q.2: What is the magnetic field due to a current-carrying loop?
Ans: The magnetic field on the axis of a current-carrying loop is given by:
\({B_x} = \frac{{{\mu _0}I{R_x}^2}}{{2{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}\widehat i\)
where \(x\) is the distance of that point on the axis from the centre of the loop, \(R\) is the radius of the loop, and \(I\) is the current flowing in the loop. For the magnetic field at the centre of the loop, put \((x = 0)\) in the above expression.

Q.3: Where is the magnetic field due to current through a circular loop uniform?
Ans: The magnetic field due to current through the circular loop is uniform at the centre of the current loop and non-uniform near the circular coil.

Q.4: How is the variation of the magnetic field on the axis of a circular loop?
Ans: The magnetic field is maximum at the centre, and it goes on decreasing as we move away from the centre of the circular loop on the axis of the loop.

Q.5: What is a circular loop?
Ans: A circular loop is made up of a large number of very small straight wires. When an electric current, flows through a circular coil of wire, then a magnetic field is produced. The field lines become straight and perpendicular to the plane of the coil at the centre of the circular wire.

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