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Two short bar magnets oflength $1\ cm$ 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 resultand 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 $3.6 \times 10^{-5}$ $Wbm^{-2}$ )
$3.6 \times 10^{-5} $ $Wbm^{-2}$
$2.56 \times 10^{-4}$ $ Wbm^{-2}$
$3.50 \times 10^{-4} $ $Wbm^{-2}$
$5.80 \times 10^{-4}$ $Wbm^{-2}$
Solution
Given $: M_{1}=1.20 \,A m^{2}$ and $M_{2}=1.00 \,A m^{2}$
$r=\frac{20}{2}\, c m=0.1\, \mathrm{m}$
$\mathrm{B}_{\mathrm{net}}=\mathrm{B}_{1}+\mathrm{B}_{2}+\mathrm{B}_{\mathrm{H}}$
$B_{n e t}=\frac{\mu_{0}}{4 \pi} \frac{\left(M_{1}+M_{2}\right)}{r^{3}}+B_{H}$
$=\frac{10^{-7}(1.2+1)}{(0.1)^{3}}+3.6 \times 10^{-5}$
$=2.56 \times 10^{-4}\, \mathrm{wb} / \mathrm{m}^{2}$
