Question

Magnetic intensity for an axial point due to a short bar magnet of magnetic moment M is given by
( a ) ${\displaystyle \frac{u_0}{4\pi }}\times {\displaystyle \frac{M}{d^3}}$
( b ) ${\displaystyle \frac{u_0}{4\pi }}\times {\displaystyle \frac{M}{d^2}}$
( c ) ${\displaystyle \frac{u_0}{2\pi }}\times {\displaystyle \frac{M}{d^2}}$
( d ) ${\displaystyle \frac{u_0}{2\pi }}\times {\displaystyle \frac{M}{d^3}}$

Answer 1

Hint:
In this question, we have to find out the magnetic intensity for an axial point. First we suppose a magnet and it is at the same axis of the point. Then we will find the magnetic intensity for a short bar magnet by neglecting the length of the magnet.



Complete step by step solution:

We have to find out the magnetic field intensity for an axial point due to a short bar magnet.
We know the magnetic lines flow from the north pole towards the south pole forming circular loops. This induces a magnetic moment and the strength or the intensity of this magnetic field must be calculated.
Let us consider the magnetic field due to some magnet. As we can see that the magnet is placed axially, that is in the same axis of the point.
 Then the equation for the magnetic field intensity at a point on the axis of the bar magnet is as follow:-
${\overrightarrow{B}}_{axial}={\displaystyle \frac{u_0}{4\pi }}{\displaystyle \frac{2Md}{{(d^2-l^2)}^2}}$
Where r is the distance between the point and the centre of the magnet
L is the length of the magnet.
As the magnet used in this question is of short bar magnet then the length of the bar magnet is neglected as it is very small in comparison to the d value.
Then the equation be
${\overrightarrow{B}}_{axial}={\displaystyle \frac{u_0}{4\pi }}{\displaystyle \frac{2Md}{{\left(d^2\right)}^2}}$
${\overrightarrow{B}}_{axial}={\displaystyle \frac{u_0}{2\pi }}{\displaystyle \frac{M}{d^3}}$


Therefore, the correct option is D.




Note:
 In this question, we also remember that at an axial point, the magnetic field has the same direction as that of the magnetic dipole moment vector due to a bar magnet. Since, the north pole is nearer to the point, the magnetic field at that point is higher than the south pole.