Question

A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic N - S direction. Null points are found on the axis of magnet at r distance from the center of magnet. If the bar magnet is turned by ${180}^o$, Where will the new null points be located?(From center).

• A. $\frac{1}{2^{3}r}ataxialposition$
• B. $\frac{1}{2^{\frac{1}{3}}}ataxialposition$
• C. $\frac{1}{2^{3}r}ataxialequationalposition$
• D. $\frac{r}{2^{\frac{1}{3}}}ataxialequationalposition$

Answer 1

The correct option is D $\frac{r}{2^{\frac{1}{3}}}ataxialequationalposition$

Initially, the null points are on the axis of magnet. So,

$B=\frac{{\mu }_0}{4\pi }\frac{2m}{r^3}.............\left(1\right)$

When the bar magnet is turned by 180⁰, then the null points are obtained

$B^{{}^{\prime }}=\frac{{\mu }_0}{4\pi }\frac{m}{d^3}.........\left(2\right)$

From equation (1) and (2)

$\frac{{\mu }_0}{4\pi }\frac{2m}{r^3}=\frac{{\mu }_0}{4\pi }\frac{m}{d^3}$

$\frac{2}{r^3}=\frac{1}{d^3}$

$d^3=\frac{r^3}{2}$

$d=\frac{r}{{\left(2\right)}^{\frac{1}{3}}}$

New null pont is obtained at equatorial position.