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

$A$ circular coil of wire consisting of $100$ turns,each of radius $8.0 \; cm$,carries a current of $0.40 \; A$. What is the magnitude of the magnetic field $B$ at the centre of the coil?

$A$ 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}$)

$A$ particle is moving with velocity $\overrightarrow{v} = \hat{i} + 3\hat{j}$ and it produces an electric field at a point given by $\overrightarrow{E} = 2\hat{k}$. It will produce a magnetic field at that point equal to (all quantities are in $SI$ units):

$A$ straight conductor of length $32 \,cm$ carries a current of $30 \,A$. Magnetic induction at a point in air at a perpendicular distance of $12 \,cm$ from the mid-point of the conductor is (in $G$)

There are two infinitely long straight current-carrying conductors held at right angles to each other such that their common ends meet at the origin, as shown in the figure. The ratio of current in both conductors is $1:1$. The magnetic field at point $P(x, y)$ is: