Question

A magnet of magnetic moment $1.525\times {10}^{-3}A.m^2$. is placed in a magnetic field of induction field strength $0.1gauss$, perpendicular to the field direction. Then each pole experiences a force of $0.305dynes$. The distance between the two poles of the magnet is

• A. $0.5\times {10}^{-2}cm$
• B. $0.5\times {10}^{-1}cm$
• C. $0.5cm$
• D. $5cm$

Answer 1

The correct option is C $0.5cm$

$1gauss={10}^{-4}T$
$M=1.525\times {10}^{-3}A{m}^2$
$M=0.1gauss$
$=0.1\times {10}^{-4}T$
$F=0.305dynes$
$=0.305\times {10}^{-5}N$
$F=MP$
$0.305\times {10}^{-5}\times P$
$N=Pl$
$1.525\times {10}^{-3}=0.305\times l$
$l=5\times {10}^{-3}m$
$l=0.5cm$