Question

If the ratio of magnetic fields on the axial line of a long magnet at distances of $10\text{cm}$ and $20\text{cm}$ is $12.5:1$, then the length of the magnet will be

• A. $\text{20 cm}$
• B. $\text{40 cm}$
• C. $\text{30 cm}$
• D. $\text{10 cm}$

Answer 1

The correct option is D $\text{10 cm}$

Let $2l$ be the length of the magnet.

The magnetic field due to a long magnet at an axial point is given by

$B_{ax}=\left(\frac{{\mu }_0}{4\pi }\right)\frac{2Mx}{(x^2-l^2{)}^2}$

$\therefore B_{10cm}=\frac{{\mu }_0}{4\pi }\frac{2M\left(10\right)}{[{10}^2-l^2{]}^2}.....\left(1\right)$

Similarly,

$B_{20cm}=\frac{{\mu }_0}{4\pi }\frac{2M\left(20\right)}{{\left[\right(20{)}^2-l^2]}^2}......\left(2\right)$

After equations $\left(1\right)÷\left(2\right)$, we get

$\frac{B_{10cm}}{B_{20cm}}=\left(\frac{10}{20}\right)\frac{{\left[\right(20{)}^2-l^2]}^2}{{\left[\right(10{)}^2-l^2]}^2}$

$\because \frac{B_{10cm}}{B_{20cm}}=\frac{12.5}{1}$

$\frac{12.5}{1}=\frac{1}{2}\times {\left(\frac{400-l^2}{100-l^2}\right)}^2$

$\frac{25}{1}={\left(\frac{400-l^2}{100-l^2}\right)}^2$

$\pm 5=\frac{400-l^2}{100-l^2}$

$\pm 500\mp 5l^2=400-l^2$

$l=\pm 5or\pm 5\sqrt{6}$

Since, the length of magnet can't be negative.

$\therefore l=5\text{cm}or5\sqrt{6}\text{cm}$

Therefore, the length of the magnet will be $2l=10\text{cm}or10\sqrt{6}\text{cm}$.

Hence, option $\left(D\right)$ is correct.