Consider a long cylindrical wire of radius R having uniform current density $\vec{J} = J_{0} \hat{k}$ . Axis of this cylindrical wire is along z-axis. A cylindrical wire of radius $(\frac{R}{4})$ is removed from the cylindrical wire which lead to a cavity. (Axis of this cavity is parallel to z-axis which is passing through the point $O^{'} (4, 3)$ . Cross sectional view of the wire is shown in diagram. On the basis of above paragraph answer the following questions

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Detailed Solution

Magnetic field inside the cavity is uniform
$\begin{array}{l} \vec{B} = (\frac{μ_{0} J_{0}}{2}) (OO) (\frac{- 3 \hat{i} + 4 \hat{j}}{5}) \\ \vec{B} = (\frac{μ_{0} J_{0}}{2}) (5) (\frac{- 3 \hat{i} + 4 \hat{j}}{5}) \\ \vec{B} = \frac{μ_{0} J_{0}}{2} (- 3 \hat{i} + 4 \hat{j}) \end{array}$
For No collision
$\begin{array}{l} R_{0} \leq \frac{R}{4} \\ \Rightarrow \frac{2 {mv}_{0}}{μ_{0} J_{0} q} \leq \frac{R}{4} \\ \Rightarrow v_{0} \leq \frac{{Rμ}_{0} J_{0} q}{8 m} \\ v_{0} \leq \frac{{Rμ}_{0} J_{0}}{8} \end{array}$