The net magnetic field at the centre $O$ of the circle due to the current-carrying loop as shown in the figure is $(\theta < 180^\circ)$.

Consider a current-carrying wire as shown in the figure. If the radius of the curved part of the wire is $R$ and the linear parts are assumed to be very long, then the magnetic induction of the field at the point $O$ is

Two very long straight parallel wires,parallel to the $y-$axis,carry currents $4I$ and $I$ along the $+y$ direction and $-y$ direction,respectively. The wires pass through the $x-$axis at the points $(d, 0, 0)$ and $(-d, 0, 0)$ respectively. The graph of the magnetic field $z-$component as one moves along the $x-$axis from $x=-d$ to $x=+d$ is best given by:

Choose the correct statement.

$A$ very long wire $ABDMNDC$ is shown in the figure carrying current $I$. $AB$ and $BC$ parts are straight,long,and at a right angle. At $D$,the wire forms a circular turn $DMND$ of radius $R$. $AB$ and $BC$ parts are tangential to the circular turn at $N$ and $D$. The magnetic field at the centre of the circle is

An electron (mass $9 \times 10^{-31} \ kg$ and charge $1.6 \times 10^{-19} \ C$) moving with speed $v = c/100$ $(c = 3 \times 10^8 \ ms^{-1})$ is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4} \ T$ perpendicular to its direction of motion. We wish to apply a uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then: