Question

A short bar magnet of moment $4{\text{Am}}^2$ is placed in the magnetic meridian with its south-pole pointing geographic north. The distance between two null points is found to be $20\text{cm}$.

$\left(iii\right)$If the same magnet is placed with its north pole pointing geographic north, then the distance between the null points $(\text{use}{4}^{\frac{1}{3}}=1.59)$.

• A. $1.59\text{m}$
• B. $0.159\text{m}$
• C. $15.59\text{m}$
• D. $0.0159\text{m}$

Answer 1

The correct option is B $0.159\text{m}$



At null points, $B_H=B_{eq}.......\left(1\right)$

Where,

$B_H=\text{Horizontal component of earth's field}{B}_{eq}=\text{Equatorial field due to a bar magnet}$

$B_{eq}=\frac{{\mu }_0}{4\pi }\frac{M}{d^3}.......\left(2\right)[\because B_H=8\times {10}^{-4}\text{T}]$

From (1) and (2) we get,

$\Rightarrow 8\times {10}^{-4}=\frac{4\pi \times {10}^{-7}}{4\pi }\times \frac{4}{d^3}$

$\Rightarrow d^3=\frac{1}{2}\times {10}^{-3}$

$\Rightarrow d={\left(\frac{1}{2}\right)}^{\frac{1}{3}}\times {10}^{-1}\text{m}=\frac{1}{10\times (2{)}^{\frac{1}{3}}}\text{m}$

The separation between the two null points $D$ is given by,

$D=2d=\frac{2}{10\times (2{)}^{\frac{1}{3}}}=\frac{4^{\frac{1}{3}}}{10}$

$\therefore D=\frac{1.59}{10}=0.159\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.