Question

Two small magnets each of magnetic moment $10A{m^2}$ are placed in end on position $0.1m$ part from their centres. The force acting between them is
(A) $0.6 \times {10^7}N$
(B) $0.06 \times {10^7}N$
(C) $0.6$
(D) $0.06$

Answer 1

Hint:
In order to solve this question, we should know that when two magnets are bring close to each other at some distance, they start to attract or repel each other with a definite magnitude of force and here using given data we will find the force between two given magnets.



Formula used:
If m and m’ be the magnetic moment of two small magnets and the distance between their centres is r then force F between them is calculated using relation $F = \dfrac{{{\mu _o}}}{{4\pi }}\left( {\dfrac{{6mm'}}{{{r^4}}}} \right)$ where, $\dfrac{{{\mu _o}}}{{4\pi }} = {10^{ - 7}}N{A^{ - 2}}$



Complete step by step solution:
According to the question, we have given that two small magnets have the magnetic moment $m = m' = 10A{m^2}$ and distance between their centres is $r = 0.1m$ then using the relation $F = \dfrac{{{\mu _o}}}{{4\pi }}\left( {\dfrac{{6mm'}}{{{r^4}}}} \right)$ the force between the two given magnets can be calculated by substituting the values of the parameters we get,
$F = {10^{ - 7}}\left( {\dfrac{{6(10)(10)}}{{{{(0.1)}^4}}}} \right)$
on solving for F we get,
$F = 0.6N$
So, the force between the two short magnets which have the same value of magnetic moments of $m = m' = 10A{m^2}$ and distance between their centres is $r = 0.1m$ has the magnitude of $0.6N$
Hence, the correct answer is option (C) $0.6$




Therefore, the correct option is C.




Note:
It should be remembered that, all the physical quantities used in this problem are in their respective SI units, always make sure that all physical quantities are in same system of units before proceeding any numerical question.