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(D) Let the two magnets be $M_1$ and $M_2$ with magnetic moment $M = 10 \, Am^2$. The distance between their centres is $0.2 \, m$,so the distance fro

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$\sqrt{5} 	imes 10^{-3} \, T$

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(D) Let the two magnets be $M_1$ and $M_2$ with magnetic moment $M = 10 \, Am^2$. The distance between their centres is $0.2 \, m$,so the distance from the centre of each magnet to the midpoint $P$ is $d = 0.1 \, m$.For magnet $1$,point $P$ lies on its axial line. The magnetic field $B_a$ is given by $B_a = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$.For magnet $2$,point $P$ lies on its equatorial line. The magnetic field $B_e$ is given by $B_e = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3}$.Since these fields are perpendicular,the resultant magnetic field is $B_{net} = \sqrt{B_a^2 + B_e^2}$.$B_{net} = \sqrt{\left( \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3} ight)^2 + \left( \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} ight)^2} = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} \sqrt{2^2 + 1^2} = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} \sqrt{5}$.Substituting the values: $B_{net} = 10^{-7} \cdot \frac{10}{(0.1)^3} \cdot \sqrt{5} = 10^{-7} \cdot \frac{10}{0.001} \cdot \sqrt{5} = 10^{-7} \cdot 10^4 \cdot \sqrt{5} = \sqrt{5} 	imes 10^{-3} \, T$.

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