$A$ uniform circular wire loop is connected to the terminals of a battery. The magnetic field induction at the centre due to $ABC$ portion of the wire will be (length of $ABC = l_1$, length of $ADC = l_2$):

$A$ current carrying circular coil of radius $R$ has a point $P$ situated on its axis at a distance $x$ from its centre $O$ of the coil. The magnetic induction at point $P$ is $\left(\frac{1}{8}\right)^{\text{th}}$ of the magnetic field at its centre $O$. The value of $x$ is

$A$ current-carrying loop $ABCD$ has two circular arcs $AD$ and $BC$ with radii $1 \text{ cm}$ and $2 \text{ cm}$ respectively, as shown in the figure. The two arcs $AD$ and $BC$ subtend a common angle of $30^{\circ}$ at the centre $O$. If the current flowing in the loop is $\frac{1.2}{\pi} \text{ A}$, then the magnitude of the net magnetic field at $O$ is (Given $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$): (in $\mu \text{T}$)

$PQRS$ is a square loop made of a uniform conducting wire. The current enters the loop at $P$ and leaves at $S$. Then the magnetic field will be

In a hydrogen atom, an electron is revolving in an orbit of radius $0.53 \, \text{\AA}$ with $6.6 \times 10^{15} \, \text{rotations/second}$. The magnetic field produced at the centre of the orbit is ....... $Wb/m^2$.

$A$ coil having $9$ turns carrying current $I$ produces a magnetic field $B_1$ at the centre. If the same wire is rewound into a coil of $3$ turns carrying the same current $I$,then the magnetic field at the centre $B_2$ is: