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. Magnetic length of the bar magnet is (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)

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

Answer 1

The Correct Option is B

Let magnetic field at $P_{1}$ is $B_{1}$ and at $P_{2}$ is $B_{2}$

$\therefore \,\,\,\, \frac{B_{1}}{B_{2}}=\frac{\frac{x_{1}}{\left(x_{1}^{2}-l^{2}\right)^{2}}}{\frac{x_{2}}{\left(x_{2}^{2}-l^{2}\right)^{2}}}=\frac{\left(x_{2}^{2}-1^{2}\right)^{2}}{\left(x_{1}^{2}-1^{2}\right)^{2}} \times \frac{x_{1}}{x_{2}}$
$\Rightarrow \,\,\,\,\ \frac{25}{2}=\frac{10}{20}\left[\frac{x_{2}^{2}-l^{2}}{x_{1}^{2}-l^{2}}\right]^{2}$
or $\,\,\,\,\,\left[\frac{x_{2}^{2}-l^{2}}{x_{1}^{2}-l^{2}}\right]=5$
Or $ \,\,\,\,\,x_{2}^{2}-l^{2}=5 x_{1}^{2}-5 l^{2}$
or $4 l^{2}=5 x_{1}^{2}-x_{2}^{2}$
$= 5 \times(10)^{2}-(20)^{2} $
$=500-400=100 $
$\Rightarrow\,\,\,\, l^{2} =25 $
$\Rightarrow \,\,\,\,l=5 \,cm$
$\therefore\,\,\,\,2 l=10 \,cm$