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(B) The magnetic energy $U$ stored in a solenoid is given by $U = \frac{B^2 V}{2 \mu_0}$, where $B$ is the magnetic field, $V$ is the volume, and $\mu

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$2$ : $1$

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(B) The magnetic energy $U$ stored in a solenoid is given by $U = \frac{B^2 V}{2 \mu_0}$, where $B$ is the magnetic field, $V$ is the volume, and $\mu_0$ is the permeability of free space.Given that the magnetic energy stored in both solenoids is the same, $U_X = U_Y$.Therefore, $\frac{B_X^2 V_X}{2 \mu_0} = \frac{B_Y^2 V_Y}{2 \mu_0}$, which simplifies to $B_X^2 V_X = B_Y^2 V_Y$.Since volume $V = A 	imes L$, we have $B_X^2 A_X L_X = B_Y^2 A_Y L_Y$.Given $A_Y = 2 A_X$ and $L_Y = 2 L_X$, we substitute these into the equation:$B_X^2 A_X L_X = B_Y^2 (2 A_X) (2 L_X)$.$B_X^2 A_X L_X = 4 B_Y^2 A_X L_X$.Dividing both sides by $A_X L_X$, we get $B_X^2 = 4 B_Y^2$.Thus, $\frac{B_X^2}{B_Y^2} = 4$, which implies $\frac{|B_X|}{|B_Y|} = \sqrt{4} = 2$.Therefore, the ratio is $2 : 1$.

Consider two solenoids $X$ and $Y$ such that the area and length of $Y$ are twice that of $X$ respectively and the magnetic energy stored in both the solenoids is same, then the ratio of magnitude of magnetic fields of the two solenoids $\frac{|B_X|}{|B_Y|}$ is

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