Question

The field intensity of two points on the axial line of a bar magnet at distances $40cm$ and $20cm$ is $10:125$. The length of the magnet is

• A. $10cm$
• B. $5cm$
• C. $15cm$
• D. $20cm$

Answer 1

The correct option is D $20cm$

$\frac{{\theta }_1}{2}=ta{n}^{-1}\left(\frac{l}{20}\right),\frac{{\theta }_2}{2}=ta{n}^{-1}\left(\frac{l}{40}\right)$

$d_1=\sqrt{l^2+{20}^2}$

$d_2=\sqrt{l^2+{40}^2}$

$B_3=\frac{{\mu }_{0}p}{4\pi d_2^2}$

$B_4=\frac{{\mu }_{0}p}{4\pi d_2^2}$

$B^{\prime \prime }=\sqrt{B_3^2+B_4^2+2B_3{B}_4(cos\pi -{\theta }_2)}$

$cos\theta =\frac{1-ta{n}^2\theta /2}{1+ta{n}^2\theta /2}$

$B_1=\frac{{\mu }_{0}p}{4\pi d_1^2}$

$B_1=\frac{{\mu }_{0}p}{4\pi d_1^2}$

$B_2=\frac{{\mu }_{0}p}{4\pi d_1^2}$

$B^{\prime }=\sqrt{B_1^2+B_2^2+2B_1{B}_{2}COS\left(\pi {\theta }_1\right)}$

$\frac{B^{\prime }}{B^{\prime \prime }}=\frac{10}{125}=\frac{\frac{{\mu }_{0}p}{4\pi d_1^2}\sqrt{2(1+cos{\theta }_1)}}{\frac{{\mu }_{0}p}{4\pi d_2^2}\sqrt{2(1-cos{\theta }_2)}}$

$\frac{2}{25}=\frac{(l^2+{40}^2)\sqrt{1-\left(\frac{l}{20}p\right)/(l/20{)}^2}}{(l^2+{20}^2)\sqrt{1-(1-(\frac{l}{40}{)}^2}/(1+4d^2)}$

$l=20cm.$