Question 5: Magnetism and Matter

Question

A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A . The magnetic energy stored inside the cube is $\alpha 	imes 10^{-14} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ .
$$ \left(\mu_{\mathrm{o}}=4 \pi 	imes 10^{-7} \mathrm{Tm} / \mathrm{A}, \pi=3.14ight) $$

Accepted Answer

The magnetic field produced at the center of a circular wire of radius $r$ carrying current $I$ is:
$$ B=\frac{\mu_0 I}{2 r} $$
The electric current in the circular loop is $\mathrm{I}=2 \mathrm{~A}$, the radius of the loop is $\mathrm{r}=10 \mathrm{~cm}=0.1 \mathrm{~m}$.
$$ \mathrm{B}=\frac{4 \pi 	imes 10^{-7} 	imes 2}{2 	imes 0.1}=4 \pi 	imes 10^{-6} \mathrm{~T} $$
Magnetic energy is stored in the space where a magnetic field exists. The energy per unit volume is:
$$ u_B=\frac{B^2}{2 \mu_0} $$
For a very small volume V where the field B is approximately uniform:
$$ \mathrm{U}=\mathrm{u}_{\mathrm{B}} 	imes \mathrm{V}=\frac{\mathrm{B}^2}{2 \mu_0} 	imes \mathrm{V} $$
The volume of the cube is,
$$ V=(1 \mathrm{~mm})^3=\left(10^{-3} \mathrm{~m}ight)^3=10^{-9} \mathrm{~m}^3 $$
So, the total energy stored in the cube is,
$$ U=\frac{\left(4 \pi 	imes 10^{-6}ight)^2}{2 	imes 4 \pi 	imes 10^{-7}} 	imes 10^{-9} $$
$$\Rightarrow $$ $U=\frac{16 \pi^2 	imes 10^{-12}}{8 \pi 	imes 10^{-7}} 	imes 10^{-9}=2 \pi 	imes 10^{-5} 	imes 10^{-9}$
$$\Rightarrow $$ $\mathrm{U}=2 	imes 3.14 	imes 10^{-14}=6.28 	imes 10^{-14} \mathrm{~J}=\alpha 	imes 10^{-14} \mathrm{~J}$
$$ \Rightarrow \alpha=6.28 $$
Therefore, the correct option is (A).

Answer

6.28

Answer

$$6. 28 	imes 10^{-6} $$

Answer

628

Answer

$$ 6 .28 	imes 10^{-4} $$

Magnetism and Matter

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