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Question: The mid points of two small magnetic dipoles of length \( d \) in end-on positions, are separated by...

The mid points of two small magnetic dipoles of length dd in end-on positions, are separated by a distance xx, (x>>d)\left( {x > > d} \right). The force between them is proportional to xn{x^{ - n}} where nn is:

(A) 11
(B) 22
(C) 33
(D) 44

Explanation

Solution

Hint To solve this question, we have to use the formula for the magnetic field at the axial position due to a magnet. The parameters of the magnets are given in the question, which can be used to get the final answer.

Formula Used: The formulae used to solve this question are given by
BA=μ0(2m)4πr3\Rightarrow {B_A} = \dfrac{{{\mu _0}\left( {2m} \right)}}{{4\pi {r^3}}}
Here mm is the magnetic dipole moment, rr is the distance from the centre of the magnet, and μ0{\mu _0} is the magnetic permeability in vacuum.
F=dUdr\Rightarrow F = - \dfrac{{dU}}{{dr}}
Here UU is the potential energy corresponding to the conservative force FF, and rr is the distance.

Complete step by step answer

As both the dipoles are placed axially, they will exert force on each other due to the axial magnetic field.
We know that the axial magnetic field is given by
BA=μ0(2m)4πr3\Rightarrow {B_A} = \dfrac{{{\mu _0}\left( {2m} \right)}}{{4\pi {r^3}}}
Here, according to the question, the dipoles are of length dd and their mid points are separated by a distance xx. But it is also given that (x>>d)\left( {x > > d} \right). So the length of the dipoles can be neglected and approximately we take r=xr = x. So, we have the magnetic field
BA=μ0(2m)4πx3\Rightarrow {B_A} = \dfrac{{{\mu _0}\left( {2m} \right)}}{{4\pi {x^3}}}
BA=μ0m2πx3\Rightarrow {B_A} = \dfrac{{{\mu _0}m}}{{2\pi {x^3}}} ……...(i)
Now, the magnetic potential energy is given by
U=mB\Rightarrow U = - mB
Substituting (i), we get
U=μ0m22πx3\Rightarrow U = - \dfrac{{{\mu _0}{m^2}}}{{2\pi {x^3}}} ……...(ii)
Also, as the magnetic force is conservative in nature, so it is related to the potential energy by F=dUdrF = - \dfrac{{dU}}{{dr}}
Here r=xr = x
F=dUdx\therefore F = - \dfrac{{dU}}{{dx}}
Substituting UU from (ii), we have
F=d[μ0m22πx3]dx\Rightarrow F = - \dfrac{{d\left[ { - \dfrac{{{\mu _0}{m^2}}}{{2\pi {x^3}}}} \right]}}{{dx}}
F=3μ0m22πx4\Rightarrow F = - \dfrac{{3{\mu _0}{m^2}}}{{2\pi {x^4}}}
As we can see from the final expression of the force that it is proportional to 1x4\dfrac{1}{{{x^4}}} or x4{x^{ - 4}} .
Comparing xn{x^{ - n}}, we get n=4n = 4 .
Hence, the correct answer is option D.

Note
If we do not remember the formula for the equatorial magnetic field, then we can take the help of the electrostatic analogy. We know that the electric field due to a dipole at its equatorial position is given by E=14πε02pr3E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2p}}{{{r^3}}}. Replacing the electric field EE with the magnetic field BB, the electric dipole moment pp with the magnetic dipole moment mm and the constant 14πε0\dfrac{1}{{4\pi {\varepsilon _0}}} with the constant μ04π\dfrac{{{\mu _0}}}{{4\pi }} we can get the corresponding expression for the axial magnetic field.