Question

Two magnetic poles of strengths 40 Am, 10 Am are separated by a distance of 20 cm in air. Find the force between them. The distance is reduced to 10 cm, find the force

• A. ${10}^{-3}N,4\times {10}^{-3}N$
• B. $2\times {10}^{-3}N,2\times {10}^{-3}N$
• C. $4\times {10}^{-3}N,{10}^{-3}4N$
• D. ${10}^{-3}N,{10}^{-3}N$

Answer 1

The correct option is A ${10}^{-3}N,4\times {10}^{-3}N$

Given : $m_1=40Am$

$m_2=10Am$

initial distance, $x_1=20cm=2\ast {10}^{-1}m$

second distance,$x_2=10cm={10}^{-1}m$

To Find: force for distance $x_{1}is{F}_1=?$

force for distance $x_{2}is{F}_2=?$

Solution: A we know that,

force between two magnetic poles is, $F=\frac{{\mu }_0}{4\pi }\frac{m_1{m}_2}{r^2}$

$r$ is the distance between them,

$For$ $F_1:-$

$F_1=\frac{{\mu }_0}{4\pi }\frac{40\ast 10}{(2\ast (10{)}^{-1}{)}^2}$

$F_1={10}^{-7}\ast \frac{4\ast {10}^2}{4\ast (10{)}^{-2}}$

$F_1={10}^{-3}N$

$For$ $F_2:-$

$F_2=\frac{{\mu }_0}{4\pi }\frac{40\ast 10}{({10}^{-1}{)}^2}$

$F_2={10}^{-7}\ast \frac{4\ast {10}^2}{(10{)}^{-2}}$

$F_2=4\ast {10}^{-3}N$

hense,

The correct opt:A