Question

Four small identical bar magnets each of magnetic dipole moment 'M' are placed on vertices of a square of side 'a' such that diagonals of the square coincide with perpendicular bisector of respective magnets. The net magnetic field at the centre of the square is

• A. zero
• B. $\frac{{\mu }_0}{\sqrt{2\pi }}\frac{M}{a^3}$
• C. $\frac{2\sqrt{2}{\mu }_0}{\pi }\frac{M}{a^3}$
• D. $\frac{2{\mu }_0}{\pi }\frac{M}{a^3}$

Answer 1

The correct option is A zero

Solution:

In the figure I marked the vertices of square as $A,B,C,D.$

FOR HORIZONTAL MAGNETS:

now both the horizontal magnets i.e. those magnets which are

placed at corners A and C having

magnetic moment in same direction i.e. in $-Xdi{r}^n.$

and also $OA=OC$

At centre O: magnetic fields due to both horizontal magnets

are equal in magnitude but in opposite direction.

so due to horizontal magnets net field is, $B_1=0$ $.....\left(1\right)$

FOR VERTICAL MAGNETS:

now both the vertical magnets i.e. those magnets which are

placed at corners B and D having

magnetic moment in same direction i.e. in $+Ydi{r}^n.$

and also $OB=OD$

At centre O: magnetic fields due to both vertical magnets

are equal in magnitude but in opposite direction.

so due to vertical magnets net field is, $B_2=0$ $.....\left(2\right)$

For net magnetic field at O:

$B=B_1+B_2$

using $e{q}^n\left(1\right)$ and $e{q}^n\left(2\right),$ we get

$B=0$

hence,

The correct opt:A