Q. Two short bar magnets of length $1\, cm$ each have magnetic moments $1.20 \,Am^2 $ and $1.00 \,Am^2 $ respectively. They are placed on a horizontal table parallel to each other with their $N$ poles pointing towards the south. They have a common magnetic equator and are separated by a distance of $20.0\, cm$. The value of the resultant horizontal magnetic induction at the mid-point $O$ of the line joining their centres is close to (Horizontal component of the earth's magnetic induction is $ 3.6 \times {10}^{-5} W b/m^2$
Solution:
$B_{net} =B_1 +B_2 +B_H$
$B_{net} = \frac{ {\mu}_0}{4 \pi } \frac{(M_1+M_2)}{r^3} +B_H$
$ \, \, \, \, \, = \frac{ {10}^{-7} (1.2+1)}{(0.1)^3}+ 3.6 \times {10}^{-5}$
$ \, \, \, \, \, \, \, \, \, \, \, \, =2.56 \times {10}^{-4} Wb/m^2$
