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Step-by-Step Solution
Step 1: Express the Dipole Moment of the Circular Loop
The magnetic dipole moment $m$ (sometimes denoted as $M$) of a current-carrying circular loop is given by:
$m = I \times A$
where:
$I$ is the current flowing in the loop.
$A$ is the area of the loop.
For a circular loop of radius $R$:
$A = \pi R^2$
Therefore, we have:
$m = I \times \pi R^2$
Step 2: Relate the Increase in Dipole Moment to the Change in Radius
We are told that the dipole moment is doubled (from $m$ to $2m$) while keeping the current constant. If the new radius of the loop is $R_1$, we can write:
$2m = I \times \pi R_1^2$
But originally,
$m = I \times \pi R^2$
Combining these two equations:
$2 \bigl(I \pi R^2 \bigr) = I \pi R_1^2$
$I \pi$ cancels out from both sides, giving:
$2 R^2 = R_1^2$
Hence,
$R_1 = \sqrt{2}\, R
Step 3: Write the Magnetic Field at the Center for Each Case
The magnetic field at the center of a circular current loop of radius $R$ is given by:
$B = \dfrac{\mu_0 \, I}{2 R}$
Therefore:
When the loop radius is $R$, the magnetic field at the center is
$B_1 = \dfrac{\mu_0 I}{2 R}$.
When the loop radius is $R_1 = \sqrt{2}\,R$, the magnetic field at the center becomes
$B_2 = \dfrac{\mu_0 I}{2 ( \sqrt{2}\, R )} = \dfrac{\mu_0 I}{2 \sqrt{2} \, R}.$
Step 4: Find the Ratio $B_1 / B_2$
Taking the ratio of $B_1$ to $B_2$:
$\displaystyle \dfrac{B_1}{B_2}
= \dfrac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{2 \sqrt{2}\,R}}
= \dfrac{\mu_0 I}{2R} \times \dfrac{2 \sqrt{2}\,R}{\mu_0 I}
= \sqrt{2}$
Final Answer
$\displaystyle \sqrt{2}$
