Question

Two magnetic poles of equal pole strengths each of $1.732Am$ should be placed at ___ $cms$ apart so that the force of attraction between them is $0.5dynes$ (nearly) :

• A. $6$
• B. $2.45$
• C. $24.5$
• D. $\sqrt{3}$

Answer 1

The correct option is C $24.5$

$F=0.5dynes$
$=0.5\times {10}^{-5}N$
$m_1=m_2=m=1.732Am$
$F=\frac{\mu m_1{m}_2}{\pi d^2}$
$0.5\times {10}^{-5}=\frac{{10}^{-7}\times 1.732\times 1.732}{\pi \times d^2}$
$d^2=6\times {10}^{-2}$
$=0.245m$
$d=24.5cm$