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5.Magnetism and Matter
easy
The small magnets each of magnetic moment $10 \,A-m^2$ are placed end-on position $0.1\,m$ apart from their centres. The force acting between them is....$N$
A
$0.6 \times {10^7}$
B
$0.06 \times {10^7}$
C
$0.6$
D
$0.06$
Solution
(c)$F = \frac{{{\mu _0}}}{{4\pi }}\left( {\frac{{6MM'}}{{{d^4}}}} \right)$ in end-on position between two small magnets.
$\therefore \;F = {10^{ – 7}}\left( {\frac{{6 \times 10 \times 10}}{{{{\left( {0.1} \right)}^4}}}} \right) = 0.6\,N$
Standard 12
Physics
