$B_{x}$ and $B_{y}$ are the magnetic fields at the center of two coils $x$ and $y$ respectively,each carrying equal current. If coil $x$ has $200$ turns and $20 \ cm$ radius,and coil $y$ has $400$ turns and $20 \ cm$ radius,the ratio of $B_{x}$ and $B_{y}$ is $:-$

$A$ current $I$ enters a circular coil of radius $R$,branches into two parts and then recombines as shown in the circuit diagram. The resultant magnetic field at the centre of the coil is

$A$ small cube of side $1 \text{ mm}$ is placed at the centre of a circular loop of radius $10 \text{ cm}$ carrying a current of $2 \text{ A}$. The magnetic energy stored inside the cube is $\alpha \times 10^{-14} \text{ J}$. The value of $\alpha$ is . . . . . . . ($\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}$,$\pi = 3.14$)

$A$ straight wire carrying a current of $12 \,A$ is bent into a semi-circular arc of radius $2 \,cm$ as shown in the figure. Then the magnetic field due to the straight segments at the centre of the arc is

The fractional change in the magnetic field intensity at a distance $r$ from the centre on the axis of a current-carrying coil of radius $a$ to the magnetic field intensity at the centre of the same coil is: (Take $r << a$)

For the adjoining figure,the magnetic field at point $P$ will be: