Question

Two short bar magnets of length $1cm$ each have magnetic moments $1.20A{m}^2$ and $1.00A{m}^2$ 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 $20.0cm$. 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 earth's magnetic induction is $3.6\times {10}^{-5}Wb/m^2$)

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

Answer 1

The correct option is A $2.56\times {10}^{-4}Wb/m^2$

$B_{net}=B_1+B_2+B_H$

$B_{net}=\frac{{\mu }_0}{4\pi }\frac{\left(M_1+M_2\right)}{r^3}+B_H$

$=\frac{{10}^{-7}\left(1.2+1\right)}{{\left(0.1\right)}^3}+3.6\times {10}^{-5}=2.56\times {10}^{-4}Wb/m^3$

Hence,

option $\left(A\right)$ is correct answer.