Question

Two short bar magnets $\text{A}$ and $\text{B}$ are identical and these are arranged as shown in figure. Their length is negligible as compared to the separation between them. A magnetic needle is placed between the magnets at point $\text{P}$ which gets deflected through an angle $\theta$ under the influence of the magnets. The ratio of $x_1$ and $x_2$ will be

• A. $(2\tan \theta {)}^{1/3}$
• B. $(2\cot \theta {)}^{1/3}$
• C. $(2\tan \theta {)}^{-1/3}$
• D. $(2\sec \theta {)}^{1/3}$

Answer 1

The correct option is B $(2\cot \theta {)}^{1/3}$


At point $\text{P}$, field due to magnets $\text{A}$ and $\text{B}$ will be mutually perpendicular at the equilibrium condition.

$\therefore \tan \theta =\frac{B_A}{B_B}$

$B_A=B_B\tan \theta ......\left(1\right)$

And we know that, magnetic field due to bar magnet at axial and equatorial point for $x>>l$ is given by

$B_A=\frac{{\mu }_0}{4\pi }\left(\frac{2M}{x_1^3}\right)$

[Point $\text{P}$ is an axial point for magnet $\text{A}$]

$B_B=\frac{{\mu }_0}{4\pi }\left(\frac{M}{x_2^3}\right)$

[Point $\text{P}$ is an equatorial point for magnet $\text{B}$]

putting the values of $B_A\&B_A$ in equations $\left(1\right)$, we get

$\frac{{\mu }_0}{4\pi }\frac{2M}{x_1^3}=\frac{{\mu }_0}{4\pi }\frac{M}{x_2^3}\left(\tan \theta \right)$

$\frac{2}{\tan \theta }={\left(\frac{x_1}{x_2}\right)}^3$

$\therefore \frac{x_1}{x_2}=(2\cot \theta {)}^{\frac{1}{3}}$

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