A coil having 9 turns carrying a current produces magnetic field $B_1$ at the centre. Now the coil is rewounded into 3 turns carrying same current. Then, the magnetic field at the centre $B_2=$

19. A particle of specific charge $q / m=\pi \mathrm{C} \mathrm{kg}^{-1}$ is projected the origin towards positive $X$-axis with the velocity $10 \mathrm{~ms}^{-1}$ in a uniform magnetic field $\mathbf{B}=-2 \hat{\mathbf{k} T}$. The velocity $\mathbf{v}$ of particle after time $t=\frac{1}{12} \mathrm{~s}$ will be (in $\mathrm{ms}^{-1}$)

The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is 64 times the magnetic field at a distance $x$ on its axis from the centre of the coil. Then, the value of $x$ is

A uniform magnetic field of strength $B=2 \mathrm{mT}$ exists vertically downwards. These magnetic field lines pass through a closed surface as shown in the figure. The closed surface consists of a hemisphere $S_1$, a right circular cone $S_2$ and a circular surface $S_3$. The magnetic flux through $S_1$ and $S_2$ are respectively.