A ring of radius R carries a current i as shown. It is situated in the yz plane with its center at the origin. The current is clockwise when viewed from the side of positive x-axis (right to left). An electron is projected with a velocity $3 \hat{i} + 4 \hat{j} - \hat{k}$ from the point ‘p’ having coordinates $(\sqrt{3} R, 0, 0)$ . What is the force experienced by the electron at this moment, if ‘e’ is the magnitude of charge on electron

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$\frac{μ_{o} e i}{16 R} (- \hat{j} - 4 \hat{k})$

$\frac{μ_{o} e i}{16 R} (\hat{j} + 4 \hat{k})$

$\frac{μ_{o} e i}{4 R} (- \hat{j} - 4 \hat{k})$

$\frac{μ_{o} e i}{32 R} (\hat{j} + 4 \hat{k})$

answer is A.

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Detailed Solution

$\begin{array}{l} B due to ring at x = \sqrt{3} R is \\ B = \frac{μ_{0} i R^{2}}{2 {(R^{2} + x^{2})}^{3 / 2}} \\ B = \frac{μ_{0} i R^{2}}{2 {((R)^{2} + (\sqrt{3} R)^{2})}^{3 / 2}} \\ \Rightarrow B = \frac{μ_{0} i R^{2}}{2 \times 8 R^{3}} = \frac{μ_{0} i}{16 R}, along the negative x - direction \\ ∴ \vec{B} = - \frac{μ_{0} i}{16 R} \hat{i} \end{array}$

Velocity of the particle is
$\vec{v} = 3 \hat{i} + 4 \hat{j} - \hat{k}$
Force on the particle is
$\begin{array}{rcl} \vec{F} & = & q (\vec{v} \times \vec{B}) \\ \Rightarrow & \vec{F} = (- e) [(3 \hat{i} + 4 \hat{j} - \hat{k}) \times (- \frac{μ_{o} i}{16 R} \hat{i})] \\ \Rightarrow & \vec{F} = \frac{μ_{o} i e}{16 R} [3 (\hat{i} \times \hat{i}) + 4 (\hat{j} \times \hat{i}) + (- \hat{k} \times \hat{i})] \\ \Rightarrow & \vec{F} = \frac{μ_{o} i e}{16 R} [0 - 4 \hat{k} - \hat{j}] \\ ∴ \vec{F} & = & \frac{μ_{o} i e}{16 R} [- \hat{j} - 4 \hat{k}] \end{array}$