Question

Consider a small cube of length $1\text{cm}$ at the centre of a circular loop of radius $10\text{cm}$ carrying a current of $1\text{A}$. The energy stored inside the cube is $n\pi \times {10}^{-12}\text{J}$. The value of $n$ is
(Give integer value only)

Answer 1

Given:
$a=1\text{cm};r=10\text{cm};i=1\text{A}$

Volume of cube, $V=a^3=(0.01{)}^3{\text{m}}^3$

Magnetic field at the center of a circular loop is,

$B=\frac{{\mu }_{0}i}{2r}$

$\therefore$ Energy stored in the cube is,

$E=\frac{B^2}{2{\mu }_0}\times \text{Volume of cube}$

$={\left(\frac{{\mu }_{0}i}{2r}\right)}^2\times \frac{1}{2{\mu }_0}\times V=\frac{{\mu }_0{i}^{2}V}{8r^2}$

Substituting the values,

$E=\frac{4\pi \times {10}^{-7}\times 1^2\times (0.01{)}^3}{8\times {0.1}^2}$

$\therefore E=5\pi \times {10}^{-12}\text{J}$

Correct answer: $5$