$A$ long straight wire of radius $a$ carries a steady current $I$. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields $B$ and $B'$,at radial distances $\frac{a}{2}$ and $2a$ respectively,from the axis of the wire is

$PQ$ and $RS$ are long parallel conductors separated by a certain distance. $M$ is the midpoint between them (see the figure). The net magnetic field at $M$ is $B$. Now,the current $2 \text{ A}$ is switched off. The field at $M$ now becomes

$A$ closely packed coil having $1000$ turns has an average radius of $62.8\,cm$. If the current carried by the wire of the coil is $1\,A$,the value of the magnetic field produced at the centre of the coil will be (permeability of free space $\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A$) nearly:

$A$ coil of $12$ turns is made from a wire of constant length carrying a current $I$. If the number of turns is changed to $3$,what is the percentage change in the magnetic field produced at its centre?

$A$ long current-carrying wire produces a magnetic field of $1 \ T$ at a distance of $r$. What will be the magnetic field at distances of $(a)$ $\frac{r}{2}$,$(b)$ $2r$,and $(c)$ $3r$?

An electron moves in a circular orbit with uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is (where $\mu_{0} =$ permeability of free space,$e =$ electronic charge):