A charged particle is projected in magnetic field $\vec{B} = 10 \hat{k}$ from origin in x-y plane. The particle moves in a circle and just touches a line $y = 5 m at x = 5 \sqrt{3} m$ . Then (mass of particle = 5 × 10^–5 kg, charge = 1μc) -

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workdone by magnetic force on the particle is zero.

The radius of circle is 10 m

speed of particle is 2m/s

The particle is projected at an angle 60° with x-axis

answer is A, B, C, D.

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Detailed Solution

$R^{2} - (R - 5)^{2} = (5 \sqrt{3})^{2}$

$\begin{array}{l} R^{2} - R^{2} (R - 5)^{2} = (5 \sqrt{3})^{2} \\ R^{2} - R^{2} - 25 + 10 R = 75 \\ R = 10 m \\ \sin α = \frac{1}{2}, a = 30^{\circ}, θ = 90 - α = 60^{\circ} \\ \frac{mv}{qB} = R \Rightarrow v = \frac{RqB}{m} = \frac{10 \times 10^{- 6} \times 10}{5 \times 10^{- 5}} = 2 m / s \end{array}$