Question

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 $3.6 \times {10}^{-5} W b/m^2$

• A. $3.6 \times {10}^{-5} W b/m^2$
• B. $2.56 \times {10}^{-4} W b/m^2$
• C. $3.50 \times {10}^{-4} W b/m^2$
• D. $5.80 \times {10}^{-4} W b/m^2$

Answer 1

Answer: (B)

$2.56 \times {10}^{-4} W b/m^2$

Answer 2

$B_{net} =B_1 +B_2 +B_H$
$B_{net} = \frac{ {\mu}_0}{4 \pi } \frac{(M_1+M_2)}{r^3} +B_H$
$\, \, \, \, \, = \frac{ {10}^{-7} (1.2+1)}{(0.1)^3}+ 3.6 \times {10}^{-5}$
$\, \, \, \, \, \, \, \, \, \, \, \, =2.56 \times {10}^{-4} Wb/m^2$