Name: Earth's Magnetic Poles
Uploaded: 2019-07-19
Duration: 3 min 53 s
Description: Let us explore the concept of true north and the magnetic north pole of the Earth and try to locate the poles using the concepts behind them in this video.

Question

Compare current carrying loop and a bar-magnet as a magnetic dipole. Obtain an expression for magnetic field due to a magnetic dipole at equatorial line.

Accepted Answer

A current-carrying solenoid or a coil, or a current-loop, behaves like a bar magnet. When a current loop is suspended in a magnetic field, it experiences a torque that tends to rotate the loop to a position in which the axis of the loop is parallel to the field. The magnitude of the torque acting on the current-loop in a uniform magnetic field $\vec{B}$ with its axis (which is perpendicular to the plane of the loop) at an angle 0 with the direction of $\vec{B}$ is given by

$𝜏 = IABsinθ . . . . . . . . . . . (i)$

Where A is the area of the current loop.
We have read in electrostatics that an electric dipole placed in an electric field also experiences a torque that tends to align the dipole along the field. The magnitude of the torque acting on an electric dipole in a uniform electric field $\vec{E}$ with its axis at an angle 0 with the direction of $\vec{E}$ is given by

$𝜏 = PEsinθ . . . . . . . . (ii)$

Where p is the magnitude of the electric dipole moment.
Comparing Eqs. (i) and (ii), we can associate a magnetic dipole moment $\vec{m}$ with the current-loop of area A and carrying a current /. The magnitude of $\vec{m}$ is

$m = IA$

The SI unit of m is ${ampere-metre}^{2} ({Am}^{2})$ and its dimensions are $[{AL}^{2}]$ .

The magnetic dipole moment equation is written in vector form as $\vec{m} = I \vec{A}$

where $\vec{A}$ is the area vector perpendicular to the plane of the loop. Thus, the direction of the magnetic dipole moment is perpendicular to the plane of the loop. This direction is given by a curled-straight right-hand rule : “If we grasp the loop in our right hand such that the fingers curl around the loop in the
Direction of the current, then our extended thumb will point in the direction of the dipole moment $\vec{m}$ .” The above result is true for any planar current-loop.

Hence, a current-loop is equivalent to a magnetic dipole.
If, instead of a current-loop, there is a current-carrying coil having N turns, then its magnetic moment would be given by

$\vec{m} = NI \vec{A}$

An expression for the magnetic field at a point on the equatorial due to a dipole:

A line that is the perpendicular bisector of the bar magnet is known as an equatorial line. Let's consider $Q$ be the point on the broad-side-on position of the bar magnet SN, at a distance $d$ from its mid point $' Q'$ . Let the pole strength of the magnet be $m$ and its magnetic length $2 l$ .

The magnetic field at the point $Q$ due to the north pole of the magnet is

$\vec{B_{N}} = \frac{μ_{0}}{4 π} \frac{m \vec{N Q}}{{(N Q)}^{3}}$

The magnetic field at point $Q$ due to the south pole of the magnet is

$\vec{B_{S}} = \frac{μ_{0}}{4 π} \frac{m \vec{Q S}}{{(Q S)}^{3}}$

But, $N Q = Q S = \sqrt{d^{2} + l^{2}}$

The resultant magnetic field at the equatorial point is

$\vec{B_{Q}} = \vec{B_{N}} + \vec{B_{S}} = \frac{μ_{0}}{4 π} \frac{m}{{(d^{2} + l^{2})}^{\frac{3}{2}}} (\vec{N Q} + \vec{S Q}) According to the traingle rule, \vec{NQ} + \vec{SQ} = \vec{NS} = 2 l \vec{B_{Q}} = \frac{μ_{0}}{4 π} \frac{m \times 2 l}{{(d^{2} + l^{2})}^{\frac{3}{2}}} \vec{B_{Q}} = \frac{μ_{0}}{4 π} \frac{- M}{{(d^{2} + l^{2})}^{\frac{3}{2}}}$

The negative sign in the above expression implies the magnetic field is antiparallel to the magnetic moment.

Compare current carrying loop and a bar-magnet as a magnetic dipole. Obtain an expression for magnetic field due to a magnetic dipole at equatorial line.

Important Questions on Magnetic Field and Earth's Magnetism

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