A Chain Is Looped Around Two Gears of Radius 40 Mm That
Let's sympathize howMagnetic Field on the Centrality of a Circular Current Loopworks.In that location are ii methods of calculating magnetic fields in magnetics at some point. One isBritish indian ocean territory-Savart law, and the other isAmpere's constabulary.To discover the magnetic field due to an infinitesimally modest current-carrying wire at some point, we use Biot-Savart constabulary to calculate the magnetic field of a highly symmetric configuration carrying a steady current Ampere's Circuital Law.
Cheque hither to apply the Biot-Savart law to calculate a small current element'south magnetic field produced at some betoken in space. Using this ceremonial and the principle of superposition, we will calculate the full magnetic field due to the circular electric current loop.
Biot-Savart Police force
Here, nosotros will report the relationship between current and the magnetic field information technology produces. Information technology is given by Biot-Savart's police. Let us assume a finite usher \(XY\) which is conveying a current \(I\), as shown in figure (a). Then, the magnetic field \(d\overrightarrow B \) due to an infinitesimal element \(\left( {d\overrightarrow l } \correct)\) of the conductor is to be determined at a indicate \(P\) 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 police force, the magnitude of the magnetic field \(d\overrightarrow B \) is proportional to the current \((I)\), the elemental length \(\left| {d\overrightarrow 50 } \right|\), and it is inversely proportional to the square of the distance \((r)\). And the direction is perpendicular to the plane, which contains \(d\overrightarrow fifty \) and \(\overrightarrow r\). Thus, in vector note, nosotros can write this as:
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Fig. (a)
At present,
\({\text{d}}\overrightarrow B \propto \frac{{{\text{d}}\overrightarrow fifty \times \overrightarrow r}}{{{r^3}}}\)
\( {\text{d}}\overrightarrow B = \frac{{{\mu _0}}}{{iv\pi }}\frac{{I\;{\text{d}}\overrightarrow fifty \times \overrightarrow r}}{{{r^3}}}\) ……(i)
Where, \(μ_0/4π\) is a constant of proportionality. The higher up expression holds when the medium is the vacuum. And by using the holding of cantankerous-product, we can also write
\(\left| {{\text{d}}\overrightarrow B} \correct| = \;\frac{{{\mu _0}}}{{4\pi }}\frac{{Idl\sin \theta }}{{{r^2}}}\) ……(2)
And,
\(\frac{{{\mu _0}}}{{iv\pi }} = {ten^{ – 7}}\;{\text{T}\;\text{thou}}{{\text{A}}^{ – 1}}\)
Where, \(μ_0\) is the permeability of complimentary space.
Magnetic Field on the Axis of a Circular Electric current Loop
To determine the magnetic field due to a circular curl forth its axis. Let usa assume that the current \((I)\) is steady and that the evaluation is carried out in free space (i.e., vacuum) by summing up the effect of minute current elements \(\left( {I{\text{d}}\overrightarrow l } \right)\). Permit consider a circular loop carrying a steady electric 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 volition summate the magnetic field at the point P on the axis of the loop, i.east. \(x-\)axis. Then, from the middle \(O\) of the loop, the distance of \(P\) be \(ten\). Now, consider a minor conducting chemical element \(\left( {{\text{d}}\overrightarrow fifty } \right)\) of the loop. Now, co-ordinate to Biot-Savart law, the magnitude of the magnetic field \(\left( {{\text{d}}\overrightarrow B } \right)\) due to element \(\left( {{\text{d}}\overrightarrow l } \correct)\) will exist
\({\text{d}}B\; = \frac{{{\mu _0}}}{{4\pi }}\frac{{I\left| {{\text{d}}\overrightarrow l \times \overrightarrow r} \correct|}}{{{r^3}}}\) ……..(1)
Fig. (b)
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Now, \(r^2 = x^ii + R^2\). Further, from the element to the axial bespeak, any element of the loop volition be perpendicular to the deportation vector. For case, the chemical 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 \(x-y\) airplane. Hence, \(\left| {{\text{d}}\overrightarrow 50 \times \overrightarrow r } \right| = rdl\)
Thus,
\({\text{d}}B = \frac{{{\mu _0}}}{{iv\pi }}\frac{{Idl}}{{\left( {{ten^two} + {R^two}} \correct)}}\) ……(2)
The direction of \({\text{d}}\overrightarrow B\) is shown in Fig.(c). It is perpendicular to the aeroplane 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 \(10-\)centrality, \({\text{d}}\overrightarrow B_⊥\). And when the components perpendicular to the \(x-\)axis are summed over, they cancel out each, and so we obtain a nil 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 \({\text{d}}\overrightarrow 50 \) chemical element, shown in Fig.(b). Thus, just 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
\(\cos \theta = \frac{R}{{{{\left( {{x^2} + {R^2}} \right)}^{1/2}}}}\) ……(iii)
From Equation (2) and (3), we have
\(d{\overrightarrow B_{\text{10}}} = \frac{{{\mu _0}Idl}}{{iv\pi }}\frac{R}{{{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}} \hat i\)
The summation of elements \(({\text{d}}\overrightarrow fifty )\) over the loop is the circumference of the loop, i.due east. \(2πR\). Thus, we can write the magnetic field at P due to the unabridged round loop is
\(\overrightarrow B = {B_{\text{10}}}\chapeau i = \frac{{{\mu _0}I{R^two}}}{{two{{\left( {{x^ii} + {R^2}} \right)}^{3/2}}}}\widehat i\) ……..(4)
If we put \(x = 0\) in this above equation we can get the magnetic field at the heart 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 grade a closed loop which is shown in Fig.(c). We tin get the direction of the magnetic field by the right-paw thumb dominion, which states that curl the palm of your right hand around the circular wire with the fingers pointing in the management of the current, and then the right-hand thumb will give the management of the magnetic field.
Fig. (c)
Summary
According to the Biot-Savart law, we tin can get that the magnetic field \({\text{d}}\overrightarrow B \) at a betoken \(P\), due to an element \(({\text{d}}\overrightarrow 50 )\) which is conveying a steady current \((I)\) at a altitude \((\overrightarrow r )\) from the electric current element is given every bit
\({\text{d}}\overrightarrow B = \frac{{{\mu _0}}}{{four\pi }}I\frac{{{\text{d}}\overrightarrow fifty \times \overrightarrow r}}{{{r^3}}} \)
And, we volition integrate this vector expression over the unabridged length of the conductor to obtain the total field at \(P\).
The magnitude of the magnetic field due to a circular gyre of radius \((R)\) carrying a current \((I)\) at an axial distance \((x)\) from the centre is given as
\(B = \frac{{{\mu _0}I{R^2}}}{{2{{\left( {{x^ii} + {R^two}} \right)}^{3/ii}}}}\), and
At the heart \((ten = 0)\), this equation reduces to
\(B = \frac{{{\mu _0}I}}{{2R}}\)
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Solved Bug
Q 1. The magnetic field \((\overrightarrow B )\) due to a current-carrying circular loop of radius \(12\;\rm{cm}\) at its center is \(0.five × 10^{-iv}\;\rm{T}\). Discover the magnetic field at a indicate on the axis at a altitude of \(5\;\rm{cm}\) from the centre due to this loop?
Ans: Magnitude of the magnetic field at the middle of a round loop is
\({B_1} = \frac{{{\mu _0}I}}{{2R}}\)
And, at the axial point, nosotros have
\({B_2} = \frac{{{\mu _0}I{R^two}}}{{2{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}\)
Thus,
\(\frac{{{B_2}}}{{{B_1}}} = \frac{{{R^3}}}{{{{\left( {{x^ii} + {R^ii}} \correct)}^{3/2}}}}\)
\( \Rightarrow {B_2} = {B_1}\left[ {\frac{{{R^3}}}{{{{\left( {{x^2} + {R^ii}} \right)}^{3/two}}}}} \right]\)
\(\Rightarrow {B_2} = \left( {0.v \times {{x}^{ – 4}}} \right)\left[ {\frac{{{{\left( {12} \correct)}^3}}}{{{{\left( {{5^2} + {{12}^2}} \right)}^{3/2}}}}} \right]\)
\( \Rightarrow {B_2} = 3.9 \times {x^{ – v}}\;{\text{T}}.\)
Q 2. Find is the magnitude of the net magnetic field at point \(P\) every bit shown in the effigy? If two loops of wire carry the same current of \(10\,\rm{mA}\), only the catamenia of current is in contrary directions. 1 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 starting time loop to the point \((P)\), the distance is \(0.25\;\rm{m}\), and from the second loop to the betoken, the distance is \(0.75\,\rm{g}\) 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 contrary directions. Using the given quantities in the problem, the cyberspace magnetic field at point \(P\) can exist calculated past the equation given below:
Solving for the magnitude of net magnetic field using the equation and the quantities in the problem yields:
\({B_x} = \frac{{{\mu _0}I{R_x}^ii}}{{two{{\left( {{x^2} + {R^2}} \correct)}^{three/2}}}}\)
\(B = B_1 – B_2\)
\( \Rightarrow B = \frac{{{\mu _0}I{R_1}^2}}{{two{{\left( {{x_1}^two + {R_1}^two} \right)}^{three/2}}}} – \frac{{{\mu _0}I{R_2}^two}}{{2{{\left( {{x_1}^2 + {R_2}^2} \correct)}^{3/ii}}}}\)
\( \Rightarrow B = \frac{{\left( {4{{\pi }} \times {{ten}^{ – 7}}} \correct)\left( {0.010} \correct){{\left( {0.5} \correct)}^ii}}}{{two{{\left( {{{0.25}^2} + {{0.5}^2}} \correct)}^{3/2}}}} – \frac{{\left( {4{{\pi }} \times {{x}^{ – 7}}} \right)\left( {0.010} \right){{\left( 1 \correct)}^2}}}{{2{{\left( {{{0.75}^2} + {1^ii}} \correct)}^{3/2}}}}\)
\(⇒ B = 5.77 × 10^{-nine}\;\rm{T}\) to the right.
FAQs
Q.one. How do you notice the management of the magnetic field on the centrality of the circular loop?
Ans: The direction of the magnetic field is given by the right-hand thumb rule. That is Scroll 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 electric current-carrying loop?
Ans: The magnetic field on the axis of a current-carrying loop is given by:
\({B_x} = \frac{{{\mu _0}I{R_x}^2}}{{ii{{\left( {{10^2} + {R^2}} \right)}^{3/2}}}}\widehat i\)
where \(x\) is the distance of that point on the centrality 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 middle of the loop, put \((x = 0)\) in the above expression.
Q.3. Where is the magnetic field due to electric current through a circular loop compatible?
Ans: The magnetic field due to current through the round loop is uniform at the centre of the current loop and not-uniform near the round roll.
Q.four. How is the variation of the magnetic field on the axis of a circular loop?
Ans: The magnetic field is maximum at the center, and it goes on decreasing as we motion abroad from the centre of the circular loop on the axis of the loop.
Q.five. What is a circular loop?
Ans: A circular loop is made up of a big number of very modest straight wires. When an electric electric current, flowing through a circular coil of wire, and so 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.
We hope this detailed article on Magnetic Field on the Axis of a Circular Electric current Loop helps you in your preparation. If you become stuck do allow us know in the comments section below and we volition get back to you at the earliest.
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