Magnetic Dipole Assignment Help

Electromagnetism - Magnetic Dipole

A magnetic dipole consists of two unlike poles of equal strength and separated by a small distance.

For example a bar magnet a compass needle etc. are magnetic dipoles. We shall show that a current loop behaves as a magnetic dipole an atom of a magnetic material behaves as a dipol3e due to electrons revolving around the nucleus.

The two poles of a magnetic dipole (or a magnet), called North Pole and South Pole are always of equal strength and of opposite nature.

Further such tow magnetic poles always exist in pair and cannot be separated from each other.

The distance between the two poles of a bar magnet is called the magnetic length of the magnet. It is a vector directed form S pole of magnet to ties N-pole and is represented by 2I‾

Magnetic dipole moment is the product of strength of either pole (m) or the magnetic length (2I‾) of the magnet it is represented by  M‾ .

Magnetic dipole moment = strength of either pole X magnetic length

M‾  = m (2I‾)

Magnetic dipole moment is a vector quantity directed form south to North Pole of the magnet

We shall show that the S.I units of M are joule/tesla or ampere meter.

∴ SI unit of pole strength is Am.

Magnetic field strength at a point due to bar magnet
When point lies on axial line of bar magnet let 2l be the magnetic length of bar magnet with centre O. M‾ is the magnetic dipole moment of the magnet OP = d is the distance of the point P on the axial line from the centre of the magnet if m is the strength of each pole then magnetic field strength at p due to N pole of magnet is 

B = μ0/4π = m x 1/Np2 = μ0/4π m/(d - I)2

Along NP produced.

Magnetic field strength at p due to S pole of magnet

B = μ0/4π m x 1/Sp2 = μ0/4π m/(d + I)2 along PS

∴ Magnetic field strength at p due to the bar magnet

B1 = B - B = μ0 /4π m/(d - I)2 - μ0/4π m/(d + I)2 = μ0m/4π [ 1/(d - I)2 - 1/(d + i)2

Bi = μ0m/4π (d + I)2 - (d - I)2/(d2 - I2)2 = μ0m/4π 4Id/(d2 - I2)2 = μ0/4π (m X 2I) 2d/(d2 - I2)

B1 μ0/4π 2Md/(d2 - I2)2

When the magnet is short

I2 <<d2

∴ B1  μ0 /4π2Md/d4 =  μ0/4π2M/ d

The direction of B1 is along SN produced.
When point lies on equatorial line of bar magnet

In the point p is shown on equatorial line of the same bar magnet where OP = d magnetic field strength at P due to N pole of magnet

B = μ0/4π m X 1/NP2 = μ0/4π m/(d2 + I2)

Along NP produced. Magnetic field strength at p due to s pole of magnet B = μ0/4π m x 1/Si2 = μ0/4π m/(d2 + I2) along Ps.

AS B = B in magnitude their components B sin θ along OP produced and B sin θ along PO cancel. However components along PX parallel to NS add. Therefore,

Magnetic field strength at p due to the bar magnet B2 = B cos θ+B cos θ = 2B cosθ,

Along PX = 2 μ0/4π m/(d2 + I2) X 1 d2 + I2 = μ0/4π m x 2I (d2 + I2)3/2

B= μ0/4π M/(d2 + I23/2

If the magnet is short I2 < < d2

∴ B2 =  μ0/4π M/(d2) 3/2 =  μ0/4π M/d

The direction of B2 is along PX a line parallel to NS as shown in dividing the two equations we get B1 /B2 = 2 or B1 = 2 B2 - Magnetic Dipole Assignment Help, Magnetic Dipole Homework Help, Magnetic Dipole Assignment Tutors, Magnetic Dipole Solutions, Magnetic Dipole Answers, Electromagnetism Assignment Tutors

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