Question-53342

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

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

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 \dots

{\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

b l e s s i n g s h o w e r t o a l l \dots