The ratio of magnetic fields due to a bar mag | vedclass

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Question

(B) The magnetic field on the axis of a bar magnet is given by $B = \frac{\mu_0}{4\pi} \frac{2Md}{(d^2-l^2)^2}$,where $d$ is the distance from the cen

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$10$

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(B) The magnetic field on the axis of a bar magnet is given by $B = \frac{\mu_0}{4\pi} \frac{2Md}{(d^2-l^2)^2}$,where $d$ is the distance from the center and $2l$ is the magnetic length.Given $d_1 = 10 \ cm$ and $d_2 = 10 + 10 = 20 \ cm$.The ratio is $\frac{B_1}{B_2} = \frac{d_1}{d_2} \left( \frac{d_2^2 - l^2}{d_1^2 - l^2} ight)^2 = \frac{25}{2}$.Substituting the values: $\frac{10}{20} \left( \frac{20^2 - l^2}{10^2 - l^2} ight)^2 = \frac{25}{2}$.$\frac{1}{2} \left( \frac{400 - l^2}{100 - l^2} ight)^2 = \frac{25}{2} \Rightarrow \left( \frac{400 - l^2}{100 - l^2} ight)^2 = 25$.Taking the square root: $\frac{400 - l^2}{100 - l^2} = 5$.$400 - l^2 = 500 - 5l^2 \Rightarrow 4l^2 = 100 \Rightarrow l^2 = 25$.So,$l = 5 \ cm$.The magnetic length is $2l = 2 	imes 5 \ cm = 10 \ cm$.

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