Two square loops, P and Q , carry currents in the ratio 3 : 2 . If...

The magnetic dipole moment of a current-carrying loop is given by $M = I A$, where $I$ is the current and $A$ is the area of the loop.

For a square loop of side length $a$, the area is $A = a^2$.

The ratio of the magnetic dipole moments of loops $P$ and $Q$ is

$\dfrac{M_P}{M_Q} = \dfrac{I_P A_P}{I_Q A_Q} = \left(\dfrac{I_P}{I_Q}\right) \left(\dfrac{a_P}{a_Q}\right)^2$

Given $\dfrac{I_P}{I_Q} = \dfrac{3}{2}$ and $\dfrac{a_P}{a_Q} = \dfrac{1}{3}$, substituting these values gives

$\dfrac{M_P}{M_Q} = \left(\dfrac{3}{2}\right) \left(\dfrac{1}{3}\right)^2 = \dfrac{3}{2} \times \dfrac{1}{9} = \dfrac{1}{6}$

The ratio of the magnetic dipole moment of loop $P$ to that of loop $Q$ is $1 : 6$.

Answer: $1 : 6$