The magnetic field at the centre of a current-carrying circular coil of radius $R$ is $B_c$ and the magnetic field at a point on its axis at a distance $R$ from its centre is $B_a$. The value of $\frac{B_c}{B_a}$ is

$A$ small cube of side $1 \text{ mm}$ is placed at the centre of a circular loop of radius $10 \text{ cm}$ carrying a current of $2 \text{ A}$. The magnetic energy stored inside the cube is $\alpha \times 10^{-14} \text{ J}$. The value of $\alpha$ is . . . . . . . ($\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}$, $\pi = 3.14$)

Explain the experiment which demonstrates the magnetic field produced due to a straight long current-carrying wire.

Due to $10 \ A$ of current flowing in a circular coil of $10 \ cm$ radius,the magnetic field produced at its centre is $3.14 \times 10^{-3} \ Wb/m^2$. The number of turns in the coil will be:

$A$ long solenoid has $100 \, \text{turns/m}$ and carries current $i$. An electron moves within the solenoid in a circle of radius $2.30 \, \text{cm}$ perpendicular to the solenoid axis. The speed of the electron is $0.046 \, c$ ($c = 3 \times 10^8 \, \text{m/s}$ is the speed of light). Find the current $i$ in the solenoid (approximate). (in $ \text{A}$)

$A$ current $I$ flows in an anticlockwise direction in a circular arc of a wire having $\left(\frac{3}{4}\right)^{\text{th}}$ of the circumference of a circle of radius $R$. The magnetic field $B$ at the centre of the circle is $(\mu_0 = \text{permeability of free space})$