Question

A bar magnet $30cm$ long is placed in the magnetic meridian with its north pole pointing south. The neutral point is observed at a distance of $30cm$ from its one end. Calculate the pole strength of the magnet. Given horizontal component of earth's field $=0.34G.$

• A. 4.3 Am
• B. 5.2 Am
• C. 6.9 Am
• D. 8.6 Am

Answer 1

Answer: (D)

8.6 Am

Here, $2l=30cm$
or $l=15cm=0.15m,$
$t=30cm=0.30cm$,
$B_H=0.34G=0.34\times {10}^{-4}T$
When magnet is placed with its north pole pointing south, neutral point is obtained on its axial line. Therefore, at the neutral point.
$B_{axial}=B_H$
or $\frac{{\mu }_0}{4\pi }\times \frac{2Mr}{(r^2-l^2{)}^2}=B_H$
or $M=\frac{4\pi }{{\mu }_0}\times \frac{B_H(r^2-l^2{)}^2}{2r}$
$=\frac{1}{{10}^{-7}}\times \frac{0.34\times {10}^{-4}\times ({0.30}^2-{0.15}^2{)}^2}{2\times 0.30}$
$=\frac{0.34\times {10}^{-4}\times (0.0675{)}^2}{{10}^{-7}\times 2\times 0.30}$
$=2.582A{m}^2$
The pole strength of the magnet,
$m=\frac{M}{2l}=\frac{2.582}{0.30}=8.606Am$