$A$ straight wire of diameter $0.4 \,mm$ carrying a current of $2 \,A$ is replaced by another wire of $0.8 \,mm$ diameter carrying the same current. The magnetic field at distance $R$ from both the wires is $B_1$ and $B_2$ respectively. The relation between $B_1$ and $B_2$ is

The maximum magnetic field produced by a current of $12 \ A$ passing through a copper wire of diameter $1.2 \ mm$ is (in $mT$)

$A$ long solenoid carrying current $I_1$ produces a magnetic field $B_1$ along its axis. If the current is reduced to $20 \%$ and the number of turns per $cm$ is increased five times,then the new magnetic field $B_2$ is equal to:

$A$ toroid has a number of turns per unit length $n$ and current $i$. What is the magnetic field inside the toroid?

$A$ long solenoid has $70 \text{ turns } cm^{-1}$ and carries current $I$. An electron moves within the solenoid in a circle of radius $2.5 \text{ cm}$ perpendicular to the solenoid axis. If the speed of the electron is $4.4 \times 10^6 \text{ m s}^{-1}$, then the current $I$ in the solenoid is (Take $\mu_0 = 4 \pi \times 10^{-7} \text{ SI unit}$, mass of electron $= 9 \times 10^{-31} \text{ kg}$, charge of electron $= 1.6 \times 10^{-19} \text{ C}$) (in $\text{ mA}$)

$A$ long straight wire of circular cross-section (radius $a$) is carrying steady current $I$. The current $I$ is uniformly distributed across this cross-section. The magnetic field is