$A$ 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

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

$A$ rod with a circular cross-section area of $2\,cm^2$ and a length of $40\,cm$ is wound uniformly with $400$ turns of an insulated wire. If a current of $0.4\,A$ flows in the wire windings,the total magnetic flux produced inside the windings is $4\pi \times 10^{-6}\,Wb$. The relative permeability of the rod is (Given: Permeability of vacuum $\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A$)

$A$ solenoid of length $2 \ m$ carries a current of $20 \ A$. The diameter of the solenoid is $3 \ cm$. If the magnetic field inside the solenoid is $20 \ mT$, then the length of wire forming the solenoid is (assume $\mu_0 = 4 \pi \times 10^{-7} \ H/m$) (in $m$)

$A$ particle of mass $1 \times 10^{-27} \ kg$ and charge $1 \times 10^{-16} \ C$ enters a uniform magnetic field within a solenoid at a speed of $1000 \ m/s$. The velocity vector makes an angle of $60^{\circ}$ with the axis of the solenoid. The solenoid has $5000$ turns along its length $L$ and carries a current of $5 \ A$. The number of revolutions the particle makes along the helical path within the solenoid by the time it emerges from the solenoid's opposite end is:

The figure shows a long straight wire of circular cross-section (radius $a$) carrying a steady current $I$. The current $I$ is uniformly distributed across this cross-section. Calculate the magnetic field in the region $r < a$ and $r > a$.