Question

What is the magnitude of the equatorial and axial field due to the bar magnet of length 5cm at a distance of 50 cm from its midpoint? The magnetic moment of a bar magnet is $0.4Am^2$

Answer 1

Hint: Field at equatorial point is given as:
$B = \dfrac{{{\mu _0}M}}{{4\pi {{({r^2} + {l^2})}^{\dfrac{3}{2}}}}}$ ($M$ is the magnetic moment, $l$ is the distance of the bar magnet from the centre)
Field at axial point is given as:
$B = \dfrac{{{\mu _0}2Mr}}{{4\pi {{({r^2} - {l^2})}^2}}}$ ($M$, is the magnetic dipole moment, $r$ is the distance where field is calculated)
Magnetic dipole is given as $2lm$, where $m$ is the magnetic moment; $l$ is the length of the magnet from the centre point.
Using the above relations we will proceed for the problem.

Complete step by step answer:
Let’s define magnetic dipole moment first.
Magnetic moment: The product of the strength of either pole and the magnetic length of the magnet is called magnetic dipole moment. It is a vector quantity and is represented by $M=2lm$. The SI unit magnetic dipole moment is $Am^2$.
Now come to the calculation part:
First we will calculate the magnetic field on the equatorial point.
$
   \Rightarrow B = \dfrac{{{\mu _0}M}}{{4\pi {{({r^2} + {l^2})}^{\dfrac{3}{2}}}}} \\
   \Rightarrow B = \dfrac{{{{10}^{ - 7}} \times 0.4}}{{{{({{0.5}^2} + {{(0.05)}^2})}^{\dfrac{3}{2}}}}} \\
 $($\dfrac{{{\mu _0}}}{{4\pi }} = 10^{-7}$, we substituted the values of $M, r$ and $l$ ,also $1m=100cm $)
$ \Rightarrow B = \dfrac{{{{10}^{ - 7}} \times 0.4}}{{{{({{0.5}^2})}^{\dfrac{3}{2}}}}}$ (length of the bar magnet is too small as compared to $r$, so we neglected it)
$
   \Rightarrow B = \dfrac{{{{10}^{^{ - 7}}} \times 0.4}}{{0.125}} \\
   \Rightarrow B = 3.2 \times {10^{^{ - 7}}}T \\
 $ (when 2 and 2 are cancelled power of 5 becomes 3)
Magnetic field at the equatorial point is 3.2$ \times 10^{-7} T$.
Now, we will calculate the field at the axial point.
$
   \Rightarrow B = \dfrac{{{\mu _0}2Mr}}{{4\pi {{({r^2} - {l^2})}^2}}} \\$
On substituting the corresponding values,
$ \Rightarrow B = \dfrac{{{{10}^{ - 7}} \times 2 \times 0.4 \times 0.5}}{{{{({{.5}^2})}^2}}} $
On simplification,
$
   \Rightarrow B = \dfrac{{0.4 \times {{10}^{ - 7}}}}{{0.0625}} \\
   \Rightarrow B = 6.4 \times {10^{ - 7}}T \\
 $

Magnetic field at the axial point is $6.4 \times 10^{-7} T$.

Note:
Magnetic dipole also exists for bar magnetic, which has two equal and opposite poles. The bar magnet will always have an even number of magnetic poles because one single pole can never exist. Poles always exist in pairs even if we break the bar magnetic in any number of pieces.