Question

A circular loop carrying a current I has a dipole moment of $\mu$ and a magnetic field of $B_1$ at its center. The magnetic field at the loop's center is $B_2$ when the dipole moment is doubled while maintaining the current constant. The $\frac{B_1}{B_2}$ ratio is

• A. $\surd 2$
• B. $\frac{1}{\sqrt{2}}$
• C. $2$
• D. $\sqrt{3}$

Answer 1

The correct option is A

$\surd 2$

Step 1: Given Data:

Current in the loop = $I$

Magnetic field at center = $B_1$

Magnetic field at center when dipole moment is doubled = $B_2$

Step 2: Formula used:

DIpole moment for $n$ turns of the wire loop $\left(\mathrm{\mu }\right)$$=niA$

Where $n$is number of turns, $i$ is the current, $A$is surface area

The magnetic field at the center of the loop $\left(\mathrm{B}\right)$ $B=\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{i}}{2R}$.

Step 3: Calculating the new radius

The dipole moment at center can be calculated as-

${\mu }_1=i\times \left(\pi R^2\right)$

When the dipole moment is doubled suppose new dipole moment is-

${\mu }_2=2{\mu }_1$

keeping current constant.

Let the new radius is $R\text{'}$then-

$$\begin{gathered} i\times \left(\pi R{\text{'}}^2\right)=2\times i\times \left(\pi R^2\right) \\ R\text{'}=\sqrt{2}R \end{gathered}$$

Thus, If dipole moment is doubled and keeping current constant ,$\mathrm{R}\text{'}$ becomes-

$\mathrm{R}\text{'}=\sqrt{2}R$

Step 4: Calculating the ratio

The magnetic field at the center of the loop -

Using formula,

$B_1=\frac{\mu i}{2R}$…………(1)

The magnetic field $B_2$ will be--

$$\begin{gathered} B_2=\frac{\mu I}{2\left(\sqrt{2}\right)R} \\ {B}_2=\frac{B_1}{\sqrt{2}} \end{gathered}$$From equation (1)

or we can calculate the ratio as-

$\frac{{\mathrm{B}}_1}{{\mathrm{B}}_2}=\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}=\frac{R\text{'}}{R}=\sqrt{2}$.

Hence option A is the correct answer.