Two short bar magnets have their magnetic moments $1.2 \mathrm{Am}^2$ and $1.0 \mathrm{Am}^2$. They are…
Two short bar magnets have their magnetic moments $1.2 \mathrm{Am}^2$ and $1.0 \mathrm{Am}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \mathrm{~cm}$ between their centres, such that their north poles pointing towards geographic south. They have common magnetic equatorial line. Horizontal component of earth's field is $3.6 \times 10^{-5} \mathrm{~T}$. Then, the resultant horizontal magnetic induction at mid point of the line joining their centers is $\left(\frac{\mu_0}{4 \pi}=10^{-7} \mathrm{~N} / \mathrm{m}\right)$
$3.6 \times 10^{-5} \mathrm{~T}$
$1.84 \times 10^{-4} \mathrm{~T}$
$2.56 \times 10^{-4} \mathrm{~T}$
$5.8 \times 10^{-5} \mathrm{~T}$
Solution
We knows, $B=\frac{\mu_0}{4 \pi} \frac{M}{r^3}$
Hence the resultant horizontal magnetic induction point of the line joining their conters is
$\begin{aligned}
B & =B_1+B_2+B_H \\
& =\frac{10^{-7} \times 1.2}{\left(10 \times 10^{-2}\right)^3}+\frac{10^{-7} \times 1}{\left(10 \times 10^{-2}\right)^3}+3.6 \times 10^{-5} \\
& =1.2 \times 10^{-4}+1 \times 10^{-4}+0.36 \times 10^{-4} \\
& =2.56 \times 10^{-4} \mathrm{~T}
\end{aligned}$