Question

When the radius of a circular current carrying coil is doubled and current in it is halved, the magnetic dipole moment of coil originally $4\text{unit}$, now it becomes

• A. $6\text{unit}$
• B. $10\text{unit}$
• C. $8\text{unit}$
• D. $12\text{unit}$

Answer 1

The correct option is C $8\text{unit}$

Magnetic moment is, $M=iA$

Given, $r_i=r;r_f=2r;i_i=i{i}_f=\frac{i}{2}$

$M_i=4\text{unit}$

The initial magnetic moment of the loop is,

$M_i=i_i\times A_i=i\pi r^2.....\left(1\right)$

The final magnetic moment of the loop is,

$M_f=i_f\times A_f=\frac{i}{2}\times \pi (2r{)}^2=2i\pi r^2.....(2)$

On dividing $\left(2\right)÷\left(1\right)$ we get,

$=\frac{M_f}{M_i}=\frac{2i\pi r^2}{i\pi r^2}$

$=\frac{M_f}{4}=2\Rightarrow M_f=8\text{unit}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.