Question

The ratio of magnetic fields due to a bar magnet at the two axial points $P_1$ and $P_2$ which are separated from each other by $10$cm is $25:2$. Point $P_1$ is situated at $10$cm from the centre of the magnet.Points $P_1$and $P_2$ are on the same side of magnet and distance of $P_2$ from the centre is greater than distance of $P_1$ from the centre of magnet,then magnetic length is

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

Answer 1

Answer: (B)

$10$ cm

We know that the magnetic field at a distance $d$ from centre on the axis of a magnet of length $2l$ and magnetic moment $M$ is given by:-

$B=\frac{{\mu }_o}{4\pi }\frac{2Md}{(d^2-l^2{)}^2}$

For $P_1$, $d_1=10cm$

for $P_2$, $d_2=d_1+10=20cm$

$B_1=\frac{{\mu }_o}{4\pi }\frac{2M{d}_1}{(d_1^2-l^2{)}^2}$

$B_2=\frac{{\mu }_o}{4\pi }\frac{2M{d}_2}{(d_2^2-l^2{)}^2}$

Given that :$\frac{B_1}{B_2}=\frac{25}{2}$

$⟹\frac{d_1}{d_2}\frac{(d_2^2-l^2{)}^2}{(d_1^2-l^2{)}^2}=\frac{25}{2}$

$⟹\frac{10}{20}\frac{({20}^2-l^2{)}^2}{({10}^2-l^2{)}^2}=\frac{25}{2}$

$⟹\frac{({20}^2-l^2{)}^2}{({10}^2-l^2{)}^2}=25$

$⟹\frac{{20}^2-l^2}{{10}^2-l^2}=5$

$⟹400-l^2=500-5l^2$

$⟹4l^2=100$

$⟹l^2=25$

$⟹l=5cm$

Hence,magnetic length = $2l=10cm$

Hence, answer is option-(B).