Two infinitely long straight current carrying wires are placed parallel to y-axis in xy plane. Each wire carries current I in opposite directions as shown. A charged particle of charge +q and mass m is projected from position P (0, 0, a) with initial velocity $\vec{v} = v \vec{j}$ .Then

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The magnitude of initial acceleration of the particle is $\frac{μ_{0} Iqv}{2 π a m}$

The direction of initial acceleration of the particle is parallel to positive x-axis

The direction of initial acceleration of the particle is parallel to positive y-axis

The radius of curvature of the path of the particle just after it is projected is $\frac{2 π a m v}{μ_{0} Iq}$

answer is A, B, D.

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Detailed Solution

${\vec{B}}_{n e t} = 2 B \cos θ \hat{k}$

$\begin{array}{l} = 2 \frac{μ_{0} I}{2 π \sqrt{2} a} \frac{1}{\sqrt{2}} \hat{k} \\ = \frac{μ_{0} I}{2 π a} \hat{k} \\ \vec{F} = q (v \hat{j}) \times (\frac{μ_{0} I}{2 π a} \hat{k}) = \frac{μ_{0} I q v}{2 π a} \hat{i} \\ \vec{a} = \frac{\vec{F}}{m} = \frac{μ_{0} I q v}{2 π a m} \hat{i} \\ Radius of curvature R = \frac{v^{2}}{a} = \frac{2 π a m v^{2}}{μ_{0} I q v} = \frac{2 π a m v}{μ_{0} I q} \end{array}$