There is a uniform magnetic field $\vec{B} = - B_{0} \hat{k}$ from x = 0 to $x = l .$ In the region from $x = l$ to $x = 2 l$ there is a uniform electric field $\vec{E} = - E_{0} \hat{j} .$ A positively charged particle of charge q and mass m moving along x–axis enters the region of magnetic field with velocity $\vec{v} = v_{0} \hat{i} .$ After passing through the region of magnetic and electric field it emerges parallel to its initial direction of motion with speed v as shown in the figure. Then (Given $l = \frac{m v_{0}}{2 q B_{0}}$ )

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$\frac{E_{0}}{B_{0}} = \frac{3 v_{0}}{4}$

$\frac{E_{0}}{B_{0}} = \frac{\sqrt{3} v_{0}}{2}$

Total time of its motion in the regions of electric and magnetic fields is $(π + 2 \sqrt{3}) \frac{m}{6 q B_{0}} .$

$v = \frac{v_{0}}{2}$

answer is B, C.

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Detailed Solution

$\sin θ = \frac{l}{m v_{0} / q B_{0}} = \frac{1}{2}$
$θ = 30^{0}$
Time interval of its motion in magnetic field
Question Image
$t_{1} = \frac{θ}{ω} = \frac{π / 6}{q B_{0} / m} = \frac{π m}{6 q B_{0}}$
Further, $l = \frac{v_{0}^{2} \sin 60^{0}}{2 q E_{0} / m}$
$\Rightarrow \frac{m v_{0}}{2 q B_{0}} \frac{m v_{0}^{2} \sin 60^{0}}{2 q E_{0}} \Rightarrow E_{0} = \frac{\sqrt{3} v_{0} B_{0}}{2}$
$\Rightarrow t_{2} = \frac{m v_{0}}{2 q} \frac{2}{\sqrt{3} v_{0} B_{0}} = \frac{m}{\sqrt{3} q B_{0}}$
Total time $t = t_{1} + t_{2} = \frac{m}{6 q B_{0}} (π + 2 \sqrt{3})$