Magnetic Field at the axis of a current-carrying Circular Loop

Derivation→

Let us consider, A circular loop which have radius $a$, carrying a current $i$. Let take a point $P$ on the axis of the loop at a distance $x$ from the center $O$ of the loop.

Now to find the magnetic field at point $P$, take a small current element of length $dl$ at the top of the loop. Let $r$ is the distance between the current element and point $P$. Now apply the Biot-Savart's Law to find the magnitude of the magnetic field due to small current-element $dl$ at point $P$ i.e.

$dB=\frac{\mu_{\circ}}{4 \pi} \frac{i \: dl\: sin\theta}{r^{2 }}$

Where $\theta$ is the angle between the length of the element $dl$ and the line joining the element to the point $P$. Here this angle is $90^{\circ}$

$dB=\frac{\mu_{\circ}}{4\pi} \frac{i \: dl}{r^{2}}\qquad(1)$

The direction of the magnetic field $d\overrightarrow{B}$ is in the plane of the paper and at right angles to the line $r$ as shown in the figure below (By applying the right-hand screw rule). It can be resolved into two components, one is vertical component of magnitude $dB\: sin\phi$ along the axis of the loop and the other is horizontal component of magnitude $dB\: cos\phi$ at right angles to the axis.
Magnetic Field at the Axis of a Current Carrying Circular Loop
Magnetic Field at the Axis of a Current-Carrying Circular Loop
Let us consider another current element of the same length at the bottom of the loop, directly opposite to the first element, which is at right angles to the plane of the page but directed inwards. the field due to $d\overrightarrow{B}$ but directed as shown in the figure above. It is clear that components of both the fields at right angles to the axis are equal and opposite. Hence they cancel each other. The component along the axis are in the same direction and so they are added up.

If we imagine the whole of the loop to be divided into such current elements, the resultant field $\overrightarrow{B}$ at $P$ is directed along the axis and its magnitude is given by

$B=\int dB \: \sin\phi$

Now substitute the value of $dB$ from equation $(1)$

$B=\int \frac{\mu_{\circ}}{4\pi} \frac{i \: dl}{r^{2}} \: \sin\phi$

From figure, $sin\phi=\frac{a}{r}$, we get

$dB=\int \frac{\mu_{\circ}}{4 \pi} \frac{i}{r^{2}} \frac{a}{r}dl$

$dB=\int \frac{\mu_{\circ}}{4 \pi} \frac{ia}{r^{3}} dl$

$dB= \frac{\mu_{\circ}}{4 \pi} \frac{ia}{r^{3}} \int dl$

$dB= \frac{\mu_{\circ}}{4 \pi} \frac{ia}{r^{3}} 2 \pi \: a \qquad \left( \int dl = 2 \pi \: a \right)$

$dB= \frac{\mu_{\circ}ia^{2}}{2r^{3}}$

Agian from figure $r^{2}=\left( a^{2} + x^{2}\right)^{3}$ or $r= \left( a^{2} + x^{2}\right)^{3/2}$. Now substitute the value of $r$ in above equation.

$B=\frac{\mu_{\circ}\: i\: a^{2}}{2(a^{2}+x^{2})^{3/2}}$

If it is a coil of $N$ turns, then each turn will contribute equally to $B$. Thus

$B=\frac{\mu_{\circ}\: N \: i \: a^{2}}{2(a^{2}+x^{2})^{3/2}}$

The direction of the magnetic field $\overrightarrow{B}$ is along the axis of the loop and that of the coil.

Magnetic field at the Centre of the Loop→

At the centre of the coil, we have $x=0$

$B=\frac{\mu_{\circ}\: N \: i}{2a}$

Again the direction of the magnetic field $\overrightarrow{B}$ is perpendicular to the plane of the coil i.e. Along the axis of the coil.

Popular Posts

Study-Material













  • Classical world and Quantum world

  • Inadequacy of classical mechanics

  • Drawbacks of Old Quantum Theory

  • Bohr's Quantization Condition

  • Energy distribution spectrum of black body radiation

  • Energy distribution laws of black body radiation

  • The Compton Effect | Experiment Setup | Theory | Theoretical Expression | Limitation | Recoil Electron

  • Davisson and Germer's Experiment and Verification of the de-Broglie Relation

  • Significance of Compton's Effect

  • Assumptions of Planck’s Radiation Law

  • Derivation of Planck's Radiation Law

  • de-Broglie Concept of Matter wave

  • Definition and derivation of the phase velocity and group velocity of wave

  • Relation between group velocity and phase velocity ($V_{g}=V_{p}-\lambda \frac{dV_{p}}{d\lambda }$)

  • Group velocity is equal to particle velocity($V_{g}=v$)

  • Product of phase velocity and group velocity is equal to square of speed of light ($V_{p}.V_{g}=c^{2}$)

  • Heisenberg uncertainty principle

  • Generation of wave function for a free particle

  • Physical interpretation of the wave function

  • Derivation of time dependent Schrodinger wave equation

  • Derivation of time independent Schrodinger wave equation

  • Eigen Function, Eigen Values and Eigen Vectors

  • Postulate of wave mechanics or Quantum Mechanics

  • Quantum Mechanical Operators

  • Normalized and Orthogonal wave function

  • Particle in one dimensional box (Infinite Potential Well)

  • Minimum Energy Or Zero Point Energy of a Particle in an one dimensional potential box or Infinite Well

  • Normalization of the wave function of a particle in one dimension box or infinite potential well

  • Orthogonality of the wave functions of a particle in one dimension box or infinite potential well

  • Eigen value of the momentum of a particle in one dimension box or infinite potential well

  • Schrodinger's equation for the complex conjugate waves function

  • Probability Current Density for a free particle in Quantum Mechanics

  • Ehrenfest's Theorem and Derivation

  • Momentum wave function for a free particle

  • Wave function of a particle in free state

  • One dimensional Step Potential Barrier for a Particle

























  • Blog Archive