Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The mid points of two small magnetic dipoles of length $d$ in end-on positions, are separated by a distance $x$, $\left( {x > > d} \right)$. The force between them is proportional to ${x^{ - n}}$ where $n$ is:(A) $1$ (B) $2$ (C) $3$ (D) $4$

Last updated date: 20th Jun 2024
Total views: 394.5k
Views today: 8.94k
Verified
394.5k+ views
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
$\Rightarrow {B_A} = \dfrac{{{\mu _0}\left( {2m} \right)}}{{4\pi {r^3}}}$
Here $m$ is the magnetic dipole moment, $r$ is the distance from the centre of the magnet, and ${\mu _0}$ is the magnetic permeability in vacuum.
$\Rightarrow F = - \dfrac{{dU}}{{dr}}$
Here $U$ is the potential energy corresponding to the conservative force $F$, and $r$ is the distance.

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
$\Rightarrow {B_A} = \dfrac{{{\mu _0}\left( {2m} \right)}}{{4\pi {r^3}}}$
Here, according to the question, the dipoles are of length $d$ and their mid points are separated by a distance $x$. But it is also given that $\left( {x > > d} \right)$. So the length of the dipoles can be neglected and approximately we take $r = x$. So, we have the magnetic field
$\Rightarrow {B_A} = \dfrac{{{\mu _0}\left( {2m} \right)}}{{4\pi {x^3}}}$
$\Rightarrow {B_A} = \dfrac{{{\mu _0}m}}{{2\pi {x^3}}}$ ……...(i)
Now, the magnetic potential energy is given by
$\Rightarrow U = - mB$
Substituting (i), we get
$\Rightarrow U = - \dfrac{{{\mu _0}{m^2}}}{{2\pi {x^3}}}$ ……...(ii)
Also, as the magnetic force is conservative in nature, so it is related to the potential energy by $F = - \dfrac{{dU}}{{dr}}$
Here $r = x$
$\therefore F = - \dfrac{{dU}}{{dx}}$
Substituting $U$ from (ii), we have
$\Rightarrow F = - \dfrac{{d\left[ { - \dfrac{{{\mu _0}{m^2}}}{{2\pi {x^3}}}} \right]}}{{dx}}$
$\Rightarrow F = - \dfrac{{3{\mu _0}{m^2}}}{{2\pi {x^4}}}$
As we can see from the final expression of the force that it is proportional to $\dfrac{1}{{{x^4}}}$ or ${x^{ - 4}}$ .
Comparing ${x^{ - n}}$, we get $n = 4$ .
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 $E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2p}}{{{r^3}}}$. Replacing the electric field $E$ with the magnetic field $B$, the electric dipole moment $p$ with the magnetic dipole moment $m$ and the constant $\dfrac{1}{{4\pi {\varepsilon _0}}}$ with the constant $\dfrac{{{\mu _0}}}{{4\pi }}$ we can get the corresponding expression for the axial magnetic field.