A particle having a charge q=1C and mass m=0.5 kg is projected on a rough horizontal plane (x,y) from a point (15m,0,0) with a initial velocity v→=20j^ m/s. In space a uniform electric field E→=25(−k^) V/m and magnetic field B→=10(−k^) T . The acceleration due to gravity g=10 m /sec2 and co–efficient of friction μ=0.5. The particle moves on a spiral path and stops. Total time of journey is y3 sec. Find y.

mdvdt=−μ(mg+qE)mv2r=qvB⇒v=qBmrDifferentiating wrt time,dvdt=qBm×drdt∫0tdt=−1μ∫a0qBmg+qEdr⇒t=23 sechere initial radius is a=mvqB=0.5×201×10=1mAlternate:mdvdt=−μ(mg+qE)⇒∫u0dv=−μ(mg+qE)m∫0tdt⇒t=23s

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A particle having a charge $q = 1 C$ and mass $m = 0.5 k g$ is projected on a rough horizontal plane $(x, y)$ from a point $(15 m, 0, 0)$ with a initial velocity $\vec{v} = 20 \hat{j} m / s$ . In space a uniform electric field $\vec{E} = 25 (- \hat{k}) V / m$ and magnetic field $\vec{B} = 10 (- \hat{k}) T$ . The acceleration due to gravity $g = 10 m / \sec^{2}$ and co–efficient of friction $μ = 0.5$ . The particle moves on a spiral path and stops. Total time of journey is $\frac{y}{3} \sec$ . Find $y$ .

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Detailed Solution

$m \frac{d v}{d t} = - μ (m g + q E)$

$\frac{m v^{2}}{r} = q v B \Rightarrow v = \frac{q B}{m} r$

Differentiating wrt time,

$\frac{d v}{d t} = \frac{q B}{m} \times \frac{d r}{d t}$

$\int_{0}^{t} d t = - \frac{1}{μ} \int_{a}^{0} \frac{q B}{m g + q E} d r \Rightarrow t = \frac{2}{3} \sec$

here initial radius is $a = \frac{m v}{q B} = \frac{0.5 \times 20}{1 \times 10} = 1 m$

Alternate:

$m \frac{d v}{d t} = - μ (m g + q E)$

$\Rightarrow \int_{u}^{0} d v = - \frac{μ (m g + q E)}{m} \int_{0}^{t} d t$

$\Rightarrow t = \frac{2}{3} s$

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