Question

Two magnets, each of magnetic moment $'M'$ are placed so as to form a cross at right angles to each other. The magnetic moment of the system will be
A. $2M$
B. $M\sqrt 2 $
C. $\dfrac{M}{2}$
D. $\dfrac{M}{{\sqrt 2 }}$

Answer 1

Hint:
This problem is based on the Magnetic Field, we know that the magnetic moment is a vector quantity. Also, we know that the resultant of two vectors (inclined at a certain angle from each other) can be calculated by using the formula $R = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } $ hence, use this relation to get the solution to the given problem.




Complete step by step solution:
Let us consider two magnets A and B placed perpendicular to each other as shown in the figure below: -

The magnetic moment of both magnets A and B is the same i.e., ${M_A} = {M_B} = M$ (given)
Since, Magnetic moment is a vector quantity and we know that the resultant of two vectors which are inclined at a certain angle from each other is defined as: -
$R = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } $
Therefore, the net magnetic moment of two magnets given can be calculated as: -
${M_{net}} = \sqrt {{M_A}^2 + {M_B}^2 + 2{M_A}{M_B}\cos \theta } $
Substituting ${M_A} = {M_B} = M$ in the above expression, we get
$ \Rightarrow {M_{net}} = \sqrt {{M^2} + {M^2} + 2MM\left( 0 \right)} $ $(\therefore \cos {90^ \circ } = 0)$
$ \Rightarrow {M_{net}} = \sqrt {2{M^2}} = M\sqrt 2 $
Thus, the resultant magnetic moment of the system is$M\sqrt 2 $.
Hence, the correct option is (B) $M\sqrt 2 $ .



Therefore, the correct option is B.




Note:
Since this is a problem related to Magnetic Moment and Resultant of vectors hence, quantities that are required to calculate the resultant Magnetic Moment of both the magnets must be identified on a prior basis as it gives a better understanding of the problem and helps to further solve the question.