A uniform electric field E is present horizontally along the paper throughout region but uniform magnetic field $$B0$$ is present horizontally (perpendicular$$ to plane of paper in inward $$direction) right to the line AB as shown. A charge particle having charge q and mass m is projected vertically upward and crosses the line AB after time $$t0$$. The speed of projection if particle moves after $$t0$$ with constant velocity. (given$$ qE $$=$$ $$mg) is nto. Find the value of ‘n’. (g$$ $$=$$ $$9.8$$ $$m/s2)

Question

A uniform electric field E is present horizontally along the paper throughout region but uniform magnetic field $$B0$$ is present horizontally (perpendicular$$ to plane of paper in inward $$direction) right to the line AB as shown. A charge particle having charge q and mass m is projected vertically upward and crosses the line AB after time $$t0$$. The speed of projection if particle moves after $$t0$$ with constant velocity. (given$$ qE $$=$$ $$mg) is nto. Find the value of ‘n’. (g$$ $$=$$ $$9.8$$ $$m/s2)

Infinity Learn · Accepted Answer

When crosses $$ABqvB0cos$$⁡θ$$=mgqvB0sin$$⁡θ$$=qEtan$$⁡θ$$=qEmg=1$$θ$$=$$π$$4$$ along horizontal vcos⁡θ$$=qEmt0u$$−$$gt0=vsin$$⁡θ$$u=g+qEmt0=2gt0$$

Motion of a charged particle

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When crosses AB
$\begin{array}{l} {qvB}_{0} \cos θ = mg \\ {qvB}_{0} \sin θ = qE \\ \tan θ = \frac{qE}{mg} = 1 \\ θ = \frac{π}{4} \\ along horizontal vcos θ = \frac{qE}{m} t_{0} \end{array}$
$\begin{array}{l} u - {gt}_{0} = vsin θ \\ u = (g + \frac{qE}{m}) t_{0} = 2 {gt}_{0} \end{array}$

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Motion of a charged particle

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When crosses ABqvB0cos⁡θ=mgqvB0sin⁡θ=qEtan⁡θ=qEmg=1θ=π4 along horizontal vcos⁡θ=qEmt0u−gt0=vsin⁡θu=g+qEmt0=2gt0

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When crosses AB
$\begin{array}{l} {qvB}_{0} \cos θ = mg \\ {qvB}_{0} \sin θ = qE \\ \tan θ = \frac{qE}{mg} = 1 \\ θ = \frac{π}{4} \\ along horizontal vcos θ = \frac{qE}{m} t_{0} \end{array}$
$\begin{array}{l} u - {gt}_{0} = vsin θ \\ u = (g + \frac{qE}{m}) t_{0} = 2 {gt}_{0} \end{array}$