Name: Magnetic Flux
Uploaded: 2019-07-19
Duration: 6 min 44 s
Description: Magnetic flux is the total number of magnetic lines of force that are passing through a given surface. To learn more about this intriguing concept, watch the...

Question

Show that a current-carrying loop is equivalent to a magnet. Obtain an expression for the magnetic moment of the loop in terms of the loop parameters

Accepted Answer

The magnetic field at a distance x from the centre of a current carrying loop of $N$ number of turns and the radius $R$ , when $x > > R$ , (i.e at a long distance from O along the x-axis) is written as:

$B = \frac{μ_{\circ} N i R^{2}}{2 x^{3}} B = \frac{μ_{\circ} N i R^{2} π}{2 x^{3} π} B = \frac{2 μ_{\circ} N i A}{4 x^{3}}$

$\Rightarrow B = \frac{μ_{\circ} 2 m}{4 x^{3}}$

Where $N i A = m$ is called the Magnetic dipole moment of the loop and $A = π R^{2}$ is the area of the circular loop.

Similarly, in electrostatics, for an electric dipole, electric field due to the dipole along its axis,

$E = \frac{2 p_{e}}{4 {πε}_{\circ} x^{3}}$

Here, $p_{e}$ is the electric dipole moment. This shows the current carrying circular loop is equivalent to a magnetic dipole.

For a revolving electron, for 1 revolution number of turns $= 1$ , the equivalent current:

$i = \frac{e}{t} = \frac{e V }{2 πR} The magnetic moment of the loop : m = i N A m = \frac{e v}{2 πR} \times 1 \times {πR}^{2} m = \frac{e v R}{2}$

This is expression for the magnetic moment of the loop in terms of loop parameters.

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