Two short magnets AB and CD are in the X-Y plane and are parallel to X-axis and co-ordinates of their centers respectively are (0, 2) and (2, 0). Line joining the north-south poles of CD is opposite to that of AB and lies along the positive X-axis. The resultant field induction due to AB and CD at a point P (2, 2) is $100 imes 10^{- 7}\ T$. When the poles of the magnet CD are reversed, the resultant field induction is $50 imes 10^{- 7}\ T.$ The value of magnetic moments of AB and CD (in ${Am}^{2}$) are

Question

Two short magnets AB and CD are in the X-Y plane and are parallel to X-axis and co-ordinates of their centers respectively are (0, 2) and (2, 0). Line joining the north-south poles of CD is opposite to that of AB and lies along the positive X-axis. The resultant field induction due to AB and CD at a point P (2, 2) is $100 	imes 10^{- 7}\ T$. When the poles of the magnet CD are reversed, the resultant field induction is $50 	imes 10^{- 7}\ T.$ The value of magnetic moments of AB and CD (in ${Am}^{2}$) are

Dr. Abubacker · Accepted Answer

(a) As shown in figure,
[image:5_image49.emf]
$B = B_{1} + B_{2} = \frac{\mu_{0}}{4\pi}\left( \frac{2M_{1}}{r_{1}^{3}} + \frac{M_{2}}{r_{2}^{3}} ight)$
$100 	imes 10^{- 7} = 10^{- 7}\left( \frac{2M_{1}}{8} + \frac{M^{2}}{8} ight)$
$\Longrightarrow \ 2M_{1} + M_{2} = 800\ \ \ \ \ \ \ \ \ \ \ \ \ ..(i)$
If the poles of the magnet CD are reversed, then
$50 	imes 10^{- 7} = 10^{- 7}\left( \frac{2M_{1}}{8} - \frac{M_{2}}{8} ight)$
$\Longrightarrow \ 2M_{1} - M_{2} = 400\ \ \ \ \ \ \ \ldots(ii)$
Solving Eqs. (i) and (ii), we obtain
$M_{1} = 300\ Am^{2},\ M_{2} = 200\ Am^{2}$

Explanation

Reviewed by Dr. Abubacker

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