Correct Answer is Option (B)

NOTE : When a charged particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant).

Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by $\frac{1}{2}m{v}^2$ and $v^2$ is the square of the magnitude of velocity which does not change.

Correct Answer is Option (B)

The charged particle will move along the lines of electric field (and magnetic field). Magnetic field will exert no force. The force by electric field will be along the lines of uniform electric field. Hence the particle will move in a straight line.

Correct Answer is Option (A)

$\frac{B_2}{B_1}=\frac{{\mu }_0{n}_2{i}_2}{{\mu }_0{n}_1{i}_1}$

$\Rightarrow \frac{B_2}{6.28\times {10}^{-2}}=\frac{100\times \frac{i}{3}}{200\times i}$

$\Rightarrow B_2=\frac{6.28\times {10}^{-2}}{6}$

$=1.05\times {10}^{-2}Wb/m^2$

Correct Answer is Option (C)

Equating magnetic force to centripetal force.

$\frac{m{V}^2}{r}=qvB\sin {90}^{\circ }$

Time to complete one revolution.

$T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}$

Correct Answer is Option (B)

it will turn towards right of direction of motion