Two identical short bar magnets each of magnetic moment $12.5\, Am^2$ are placed with 10 cm separation between their centers as shown in the figure. The magnitude of magnetic field at O is

$\sqrt{5} × 10^{-2} T$

The correct answer is Option (1) → $\sqrt{5} × 10^{-2} T$

Given magnetic moment of each short bar magnet: $M=12.5\ \mathrm{A\,m^2}$. Separation between centers = $10\ \mathrm{cm}$, so distance of O from each center = $r=5\ \mathrm{cm}=0.05\ \mathrm{m}$. Use $\frac{\mu_0}{4\pi}=10^{-7}\ \mathrm{T\,m/A}$.

Field on the axial line of a dipole at distance $r$:

$B_{\text{axial}}=\frac{\mu_0}{4\pi}\frac{2M}{r^3}$

Field on the equatorial line of a dipole at distance $r$:

$B_{\text{eq}}=\frac{\mu_0}{4\pi}\frac{M}{r^3}$

For the left magnet (axial at O):

$B_1=B_{\text{axial}}=10^{-7}\frac{2M}{r^3}$

For the right magnet (equatorial at O):

$B_2=B_{\text{eq}}=10^{-7}\frac{M}{r^3}$

Numerical values:

$B_2=10^{-7}\frac{12.5}{(0.05)^3}=0.01\ \mathrm{T}$

$B_1=2B_2=0.02\ \mathrm{T}$

Since $B_1$ and $B_2$ are perpendicular, resultant magnitude at O is:

$B=\sqrt{B_1^2+B_2^2}=\sqrt{(2B_2)^2+B_2^2}=\sqrt{5}\,B_2=0.01\sqrt{5}\ \mathrm{T}$

Final value:

$B=0.01\sqrt{5}\ \mathrm{T}\approx 2.236\times10^{-2}\ \mathrm{T}$