Inside a region of non uniform magnetic field given by the formula $\vec{B} = - B_{0} x k^{^}$ , a particle of mass $M$ and charge $q$ is projected horizontally as shown. It is given that $q B_{0} / M = 1$ . At any instant the velocity vector of the particle is found to $(\frac{v}{2}) i^{^} + a j^{^} + b k^{^}$ If the initial value of the velocity is $v \frac{m}{s}$ , then

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$x$ coordinate of the particle at the instant when its velocity in the $x -$ direction becomes equal $x$ to zero, is ${(2 v)}^{\frac{1}{2}}$

$a = \frac{v}{2} a n d b = 0$

$x$ coordinate of the particle at the instant when its velocity in the $x -$ direction becomes equal to zero, is ${(3 v)}^{\frac{1}{2}}$

$a = \frac{\sqrt{3} v}{2} a n d b = 0$

answer is B, C.

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Detailed Solution

As the magnetic field is acting along $z - a x i s$ , and particle initial velocity is along $x - a x i s$ , so particle will remain in $x y - p l a n e$ . Also its speed must be constant. Thus

$b = 0 a n d v = \sqrt{{v^{2}}_{x} + {v^{2}}_{y}} o r v^{2} = {(\frac{v}{2})}^{2} + a^{2} ∴ a = \frac{\sqrt{3}}{2} v N o w, q v B = M (v \frac{d v}{d x}) o r q v \times (- B_{0} x) = M v \frac{d v}{d x} o r - \int_{v}^{0} d v = \frac{q B_{0}}{M} \int_{0}^{x} x d x o r v = (\frac{q B_{0}}{M}) \frac{x^{2}}{2} o r v = 1 \times \frac{x^{2}}{2} ∴ x = \sqrt{2 v}$