Two short bar magnets of length 1 cm each have magnetic moments $1.20 \mathrm{~A} \mathrm{~m}^{2}$ a

Question

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

Professor PC Thomas Classes · Answer

(c) : The situation is as shown in the figure.\ As the point $O$ lies on broad-side position with respect to both the magnets. Therefore, the net magnetic field at point $O$ is,\ } $$ \begin{aligned} & B_{	ext {net }}=B_{1}+B_{2}+B_{H} \ & \begin{aligned} B_{	ext {net }}=\frac{\mu_{0}}{4 \pi} \frac{M_{1}}{r^{3}}+\frac{\mu_{0}}{4 \pi} & \frac{M_{2}}{r^{3}}+B_{H} \ & =\frac{\mu_{0}}{4 \pi r^{3}}\left(M_{1}+M_{2}ight)+B_{H} \end{aligned} \end{aligned} $$ Substituting the given values, we get\ $B_{	ext {net }}=\frac{4 \pi 	imes 10^{-7}}{4 \pi 	imes\left(10 	imes 10^{-2}ight)^{3}}$\ $=\frac{10^{-7}}{10^{-3}} 	imes 2.2+3.6 	imes 10^{-5}$\ $=2.56 	imes 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}$\