$m=iA$ |

$\overrightarrow{m} = i\overrightarrow{A}$ |

$m=NiA$ |

$\overrightarrow{m} =N i\overrightarrow{A}$ |

$B=\frac{\mu_{\circ}}{4\pi} \frac{2m}{a^{3}}$ |

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Magnetic Dipole Moment of Current carrying loop

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Current carrying Loop or Coil or Solenoid:

The current carrying loop (or Coil or solenoid) behaves like a bar magnet. A bar magnet with the north and south poles at its ends is a magnetic dipole, so a current -loop is also a magnetic dipole.

Equation of Magnetic Dipole Moment of Current carrying Loop:

When a current loop is suspended in a magnetic field, it experiences the torque which tends to rotate the current loop to a position in which the axis of the loop is parallel to the field. So the magnitude of the torque acting on the current loop in the uniform magnetic field $\overrightarrow{B}$ is given by:

$\tau=iAB sin\theta \qquad(1)$

Where $A$ - Area of the current loop

We also know that when the electric dipole is placed in the electric field, it also experiences the torque which tends to rotate the electric dipole in the electric field. So the magnitude of the torque on the electric dipole in the uniform electric field $\overrightarrow{E}$ is given by:

$\tau=pE sin\theta \qquad(2)$

Where $p$ - The magnitude of the electric dipole moment

Now compare the equation $(1)$ and equation $(2)$ and we can conclude that the current loop also has a magnetic dipole moment just like an electric dipole have an electric dipole moment. The magnetic dipole moment is associated with the current in the loop and the area of the current loop. It is represented by $\overrightarrow {m}$. So the magnitude of the magnetic dipole moment of current carrying loop is:

The vector form of the magnetic dipole moment current carrying loop is

The magnetic dipole moment of current carrying coil:
If the current-carrying loop has $N$ number of turns (i.e current carrying coil) then the magnetic dipole moment of current carrying coil:

The vector form of the magnetic dipole moment of the current carrying coil is

The magnetic dipole moment of Circular Loop:
Let us consider the circular loop of radius $a$ in which current $i$ is flowing the magnitude of the magnetic dipole moment of the circular loop:

$m=i A$

Here the area $A$ of the circular loop is $\pi a^{2}$ then the magnitude of the magnetic dipole moment of the circular loop is:

$m=i \pi a^{2} \qquad(3)$

The magnetic field at the center of the current carrying a circular loop in terms of current is:

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

Now substitute the value of $i$ from equation $(3)$ in the above equation then the magnetic field at the center of the current carrying circular loop in terms of magnetic dipole moment is:

$B=\frac{\mu_{\circ}m}{2\pi a^{3}}$

$m=iA$ |

$\overrightarrow{m} = i\overrightarrow{A}$ |

$m=NiA$ |

$\overrightarrow{m} =N i\overrightarrow{A}$ |

$B=\frac{\mu_{\circ}}{4\pi} \frac{2m}{a^{3}}$ |

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