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Question Number 53342 by aseerimad last updated on 20/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19
$pls clarify direction of E^→ =electric field (is it +ve x axis ) V^→ =velocity proton(is it +ve x axis B^→ =magnectic induction (is it −z axis) Lorentz force F^→ =qE^→ +V^→ ×B^→ after clarification... F_E ^→ =q_(proton) ×(Ei^→ ) force due to electric field F_E ^→ =qEi^→ F_m ^→ =q{Vi^→ ×(−Bk^→ )}=(−1)qVB(i^→ ×k^→ ) =(−1)qVB(−j) =qVB(j) electric[force along +ve x axis magnetic force +ve y axis so proton move in x−y plane net force=(qE)i+(qVB)j S^→ =ix+jy ←path way =i(u_x t+(1/2)×(((qE)/m))t^2 )+j(u_y t+(1/2)(((qVB)/m))t^2 =i{Vt+(1/2)(((qE)/m))t^2 }+j{(1/2)(((qVB)/m))t^2 } ^→$
$p l s c l a r i f y d i r e c t i o n o f$ $\vec{E} = e l e c t r i c f i e l d (i s i t + v e x a x i s)$ $\vec{V} = v e l o c i t y p r o t o n (i s i t + v e x a x i s$ $\vec{B} = m a g n e c t i c i n d u c t i o n (i s i t - z a x i s)$ $L o r e n t z f o r c e$ $\vec{F} = q \vec{E} + \vec{V} \times \vec{B}$ $a f t e r c l a r i f i c a t i o n . . .$ ${\vec{F}}_{E} = q_{p r o t o n} \times (E \vec{i}) f o r c e d u e t o e l e c t r i c f i e l d$ ${\vec{F}}_{E} = q E \vec{i}$ ${\vec{F}}_{m} = q {V \vec{i} \times (- B \vec{k})} = (- 1) q V B (\vec{i} \times \vec{k})$ $= (- 1) q V B (- j)$ $= q V B (j)$ $e l e c t r i c [f o r c e a l o n g + v e x a x i s$ $m a g n e t i c f o r c e + v e y a x i s$ $s o p r o t o n m o v e i n x - y p l a n e$ $n e t f o r c e = (q E) i + (q V B) j$ $\vec{S} = i x + j y \leftarrow p a t h w a y$ $= i (u_{x} t + \frac{1}{2} \times (\frac{q E}{m}) t^{2}) + j (u_{y} t + \frac{1}{2} (\frac{q V B}{m}) t^{2}$ $= i {V t + \frac{1}{2} (\frac{q E}{m}) t^{2}} + j {\frac{1}{2} (\frac{q V B}{m}) t^{2}}$ $\vec{}$
Commented by aseerimad last updated on 21/Jan/19

$s r r y . v e l o c i t y o f p r o t o n a l o n g x a x i s .$ $e l e c t r i c f i e l d a l o n g x a x i s a n d$ $m a g n e t i c f i e l d a l o n g - z a x i s$
Commented by aseerimad last updated on 21/Jan/19
May the Almighty reward you with goodness!
Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

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