A particle of specific charge (charge/mass) α starts moving from the origin under the action of an electric field E→=E0 i^ and magnetic field B→=B0 k^. Its velocity at (x0,y0,0) is (4 i^−3 j^). The value of x0 is :

The work done by magnetic force is WB = 0, because, magnetic force is always perpendicular to instantaneous displacement. WE=qE→ . S→=qE0i^ . (x0 i^+y0 j^)=qE0x0 The speed of particle at (x0, y0) is v=42+(−3)2=5m/s According to work – energy theorem, W=ΔT=12mv2−12mu2 WE+WB=12m 52−0 qE0x0=252m x0=25m2qE0=252E0α

A particle of specific charge (charge/mass) α starts moving from the origin under the action of an electric field E→=E0 i^ and magnetic field B→=B0 k^. Its velocity at (x0,y0,0) is (4 i^−3 j^). The value of x0 is :

The work done by magnetic force is WB = 0, because, magnetic force is always perpendicular to instantaneous displacement. WE=qE→ . S→=qE0i^ . (x0 i^+y0 j^)=qE0x0 The speed of particle at (x0, y0) is v=42+(−3)2=5m/s According to work – energy theorem, W=ΔT=12mv2−12mu2 WE+WB=12m 52−0 qE0x0=252m x0=25m2qE0=252E0α

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A particle of specific charge (charge/mass) $α$ starts moving from the origin under the action of an electric field $\vec{E} = E_{0} \hat{i}$ and magnetic field $\vec{B} = B_{0} \hat{k}$ . Its velocity at $(x_{0}, y_{0}, 0)$ is $(4 \hat{i} - 3 \hat{j})$ . The value of $x_{0}$ is :

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$\frac{16 α B_{0}}{E_{0}}$

$\frac{25}{2 α E_{0}}$

$\frac{13}{2} \frac{α E_{0}}{B_{0}}$

$\frac{5 α}{2 B_{0}}$

answer is C.

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Detailed Solution

The work done by magnetic force is W_B = 0, because, magnetic force is always perpendicular to instantaneous displacement.
$W_{E} = q \vec{E} . \vec{S} = q E_{0} \hat{i} . (x_{0} \hat{i} + y_{0} \hat{j}) = q E_{0} x_{0}$
The speed of particle at (x0, y0) is
$v = \sqrt{4^{2} + {(- 3)}^{2}} = 5 m / s$
According to work – energy theorem,
$W = Δ T = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} W_{E} + W_{B} = \frac{1}{2} m 5^{2} - 0 q E_{0} x_{0} = \frac{25}{2} m x_{0} = \frac{25 m}{2 q E_{0}} = \frac{25}{2 E_{0} α}$