Advertisements
Advertisements
Question
Calculate the magnetic field at a point on the axial line of a bar magnet.
Solution
Consider a bar magnet NS. Let N be the North Pole and S be the south pole of the bar magnet, each of pole strength qm and separated by a distance of 2l. The magnetic field at a point C (lies along the axis of the magnet)
Magnetic field at a point along the axial line due to magnetic dipole
at a distance from the geometrical center O of the bar magnet can be computed by keeping unit north pole (qMC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength can be computed using Coulomb’s law of magnetism as follows:
The force of repulsion between north pole of the bar magnet and unit north pole at point C (in free space) is
`vec "F"_"N" = mu_0/(4pi) "q"_"m"/("r" - l)^2 hatl` ...(1)
where r – l is the distance between north pole of the bar magnet and unit north pole at C. The force of attraction between South Pole of the bar magnet and unit North Pole at point C (in free space) is
`vec "F"_"S" = - mu_0/(4pi) "q"_"m"/("r" - l)^2 hatl` ...(2)
where r + 1 is the distance between south pole of the bar magnet and unit north pole at C.
From equation (1) and (2), the net force at point C is `vec"F" = vec"F"_"N" + vec"F"_"S"`.
From definition, this net force is the magnetic field due to magnetic dipole at a point `"C"(vec"F" = vec"B")`
`vec"B" = mu_0/(4pi (r - l)^2) hat"i" + ((mu_0)/(4pi) "q"_"m"/(4 + "i")^2 hat"i")`
`= (mu_0 "q"_"m")/(4pi) (1/(r - l)^2 - 1/(r + l)^2)hat"i"`
`vec"B" = (mu_0 "2r")/(4pi) (("q"_"m" * (2l))/(r^2 - l^2)^2) hat"i"` ...(3)
Since, magnitude of magnetic dipole moment is `|vec"P"_"m"| "p"_"m" = "q"_"m"`.
2l the magnetic field point C equation (3) can be written as
`vec"B"_"axial" = mu_0/(4pi) ((2"rp"_"m")/("r"^2 - l^2)^2)hat"i"` ....(4)
If the distance between two poles in a bar magnet are small (looks like short magnet) compared to the distance between geometrical centre O of bar magnet and the location of point C i.e.,
r >>1 then, (r2 – l2)2 ≈ r4 ….. (5)
Therefore, using equation (5) in equation (4), we get
`vec"B"_"axial" = mu_0/(4pi) ((2"rp"_"m")/"r"^3) hat"i"`
`= mu_0/(4pi) 2/"r"^3 vec"p"_"m"`
Where `vec"p"_"m" = "p"_"m" hat"i"` ...(6)
APPEARS IN
RELATED QUESTIONS
A circular coil of radius 5 cm and 50 turns carries a current of 3 ampere. The magnetic dipole moment of the coil is nearly
A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is
Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and same electric current is passed through them. Which of the following loop configuration will experience greater torque?
Define magnetic dipole moment.
State Coulomb’s inverse law.
Obtain the magnetic induction at a point on the equatorial line of a bar magnet. Magnetic field at a point along the equatorial line due to a magnetic dipole (bar magnet).
