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Two short magnets $A B$ and $C D$ are in the $X-Y$ plane and are parallel to $X$-axis and co-ordinates of their centres respectively are $(0,2)$ and $(2,0)$. Line joining the north-south poles of $C D$ is opposite to that of $A B$ and lies along the positive $X$-axis. The resultant field induction due to $A B$ and $C D$ at a point $P(2,2)$ is $100 \times 10^{-7} \mathrm{~T}$. When the poles of the magnet $C D$ are reversed, the resultant field induction is $50 \times 10^{-7} \mathrm{~T}$. The value of magnetic moments of $A B$ and $C D$ (in $\mathrm{Am}^2$ ) are :
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Verified Answer
The correct answer is:
$300 ; 200$
As shown in figure,
$B=B_1+B_2=\frac{\mu_0}{4 \pi}\left(\frac{2 M_1}{r_1^3}+\frac{M_2}{r_2^3}\right)$
$100 \times 10^{-7}=10^{-7}\left(\frac{2 M_1}{8}+\frac{M_2}{8}\right)$
$\Rightarrow \quad 2 M_1+M_2=800$ $\ldots$ (i)
If the poles of the magnet $C D$ are reversed, then
$50 \times 10^{-7}=10^{-7}\left(\frac{2 M_1}{8}-\frac{M_2}{8}\right)$
$\Rightarrow \quad 2 M_1-M_2=400$ $\ldots$ (ii)
Solving Eqs. (i) and (ii), we obtain
$M_1=300 \mathrm{Am}^2, M_2=200 \mathrm{Am}^2$
$B=B_1+B_2=\frac{\mu_0}{4 \pi}\left(\frac{2 M_1}{r_1^3}+\frac{M_2}{r_2^3}\right)$
$100 \times 10^{-7}=10^{-7}\left(\frac{2 M_1}{8}+\frac{M_2}{8}\right)$
$\Rightarrow \quad 2 M_1+M_2=800$ $\ldots$ (i)
If the poles of the magnet $C D$ are reversed, then
$50 \times 10^{-7}=10^{-7}\left(\frac{2 M_1}{8}-\frac{M_2}{8}\right)$
$\Rightarrow \quad 2 M_1-M_2=400$ $\ldots$ (ii)
Solving Eqs. (i) and (ii), we obtain
$M_1=300 \mathrm{Am}^2, M_2=200 \mathrm{Am}^2$
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