Bar Magnetic Field

Let the magnetic moment of the bar magnet be $\mu$ .

The magnetic field at a point that is at distance $r$ from the center of the magnet and makes an angle of $\theta$ with the axis of the magnet, is given by

$B = \frac{\left( \frac{\mu _{0} }{4 \pi}\right) \mu}{r^{3}} \sqrt{1 + 3 \cos ^{2} \theta}$

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For point A, $r = 2$ , $\theta = \frac{\pi}{2}$ . Hence magnetic field at A is $B_{\text{A}} = \frac{\left( \frac{\mu _{0} }{4 \pi}\right) \mu}{2^{3}} \sqrt{1 + 3 \cos ^{2} \frac{\pi}{2}} = \frac{1}{8} \times \left( \frac{\mu _{0} }{4 \pi}\right) \mu$

For point B, $r = 3$ , $\theta = 0$ . Hence magnetic field at B is

$B_{\text{B}} = \frac{\left( \frac{\mu _{0} }{4 \pi}\right) \mu}{3^{3}} \sqrt{1 + 3 \cos ^{2} 0} = \frac{2}{27} \times \left( \frac{\mu _{0} }{4 \pi}\right) \mu$

For point C, $r = \sqrt{5}$ , $\theta = \cos^{-1} \frac{2} {\sqrt{5}}$ . Hence magnetic field at C is

$B_{\text{C}} = \frac{\left( \frac{\mu _{0} }{4 \pi}\right) \mu}{\sqrt{5}^{3}} \sqrt{1 + 3 \times \frac{4}{5}} = \frac{\sqrt{17}}{25} \times \left( \frac{\mu _{0} }{4 \pi}\right) \mu$

Since $\left( \frac{\mu _{0} }{4 \pi}\right) \mu$ is a constant, we can compare its coefficients.

$\begin{aligned} B_{\text{A}} : B_{\text{B}} : B_{\text{C}} & = \frac{1}{8} : \frac{2}{27} : \frac{\sqrt{17}}{25}\\ & \approx 0.125 : 0.074 : 0.164 \end{aligned}$

We see that the magnetic field is the strongest at point $\boxed{\text{C}}$ $_\square$