Two short magnets AB and CD are in the X-Y pl | vedclass

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(A) Let $M_1$ be the magnetic moment of magnet $AB$ and $M_2$ be the magnetic moment of magnet $CD$. Point $P(2,2)$ lies on the axial line of magnet $

Vedclass · Accepted Answer

$300; 200$

Vedclass · Answer

(A) Let $M_1$ be the magnetic moment of magnet $AB$ and $M_2$ be the magnetic moment of magnet $CD$. Point $P(2,2)$ lies on the axial line of magnet $AB$ at a distance $r_1 = 2$ from its center $(0,2)$,and on the equatorial line of magnet $CD$ at a distance $r_2 = 2$ from its center $(2,0)$.The magnetic field at $P$ due to $AB$ (axial) is $B_1 = \frac{\mu_0}{4\pi} \frac{2M_1}{r_1^3} = 10^{-7} 	imes \frac{2M_1}{2^3} = 10^{-7} 	imes \frac{M_1}{4}$.The magnetic field at $P$ due to $CD$ (equatorial) is $B_2 = \frac{\mu_0}{4\pi} \frac{M_2}{r_2^3} = 10^{-7} 	imes \frac{M_2}{2^3} = 10^{-7} 	imes \frac{M_2}{8}$.Given the resultant field $B = B_1 + B_2 = 100 	imes 10^{-7} \ T$:$10^{-7} (\frac{M_1}{4} + \frac{M_2}{8}) = 100 	imes 10^{-7} \Rightarrow 2M_1 + M_2 = 800$ (Eq. $i$).When the poles of $CD$ are reversed,the direction of $B_2$ is reversed,so $B' = B_1 - B_2 = 50 	imes 10^{-7} \ T$:$10^{-7} (\frac{M_1}{4} - \frac{M_2}{8}) = 50 	imes 10^{-7} \Rightarrow 2M_1 - M_2 = 400$ (Eq. $ii$).Adding (Eq. $i$) and (Eq. $ii$): $4M_1 = 1200 \Rightarrow M_1 = 300 \ Am^2$.Substituting $M_1$ in (Eq. $i$): $2(300) + M_2 = 800 \Rightarrow M_2 = 200 \ Am^2$.

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