First Electric then Magnetic!

Electricity and Magnetism Level 4

A particle having charge 1 C 1 \text{ C} and mass of 1 kg 1 \text{ kg} is released from rest at origin. There are electric and magnetic fields given by

E ⃗ = ( 10 i ^ ) N / C for x ≤ 1.8 m \displaystyle \vec{E} = (10 \hat{i})N/C \quad \text{for} \quad x \leq 1.8 \text{ m}

B ⃗ = ( − 5 k ^ ) T for 1.8 m ≤ x ≤ 2.4 m \displaystyle \vec{B} = (-5 \hat{k})T \quad \text{for} \quad 1.8 \text{ m} \leq x \leq 2.4 \text{ m}

A screen is placed parallel to y \text y - z \text z plane at x = 3 m x = 3 \text{ m} . The y-coordinate of particle where it collides with the screen can be written as

a ( b − c ) d \dfrac{a(b - \sqrt{c})}{d}

Where gcd ( a , d ) = 1 \text{gcd}(a,d) = 1 and c is square free. Find a + b + c + d a+b+c+d .

Details and Assumptions :

  • Neglect gravity.


The answer is 13.

Krishna Sharma by Krishna Sharma , 24, India

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2 solutions

M o t i o n _ I n _ E l e c t r i c _ f i e l d a = 10 m s 2 d v d t = 10 v = 10 t . . . ( 1 ) 1.8 = 0 + 5 t 2 t = 3 5 s e c v ( a t x = 1.8 ) = 6 m s M o t i o n _ I n _ M a g n e t i c _ f i e l d R = m . v q . B = 1.2 m R e g i o n o f m a g n e t i c f i e l d = 0.6 m T h e r e f o r e , a n g l e o f e m e r g e n c e = 30 . B y a p p l y i n g s i m p l e g e o m e t r y , w e c a n d e t e r m i n e c o o r d i n a t e o f p o i n t o f e m e r g e n c e . ( h , k ) ≡ ( 2.4 , 6 − 3 3 5 ) e q u a t i o n o f l i n e p a s s i n g t h r o u g h ( h , k ) t a n 30 = y − 6 − 3 3 5 3 − 2.4 y = 2 ( 3 − 3 ) 5 Motion\_ In\_ Electric\_ field\\ a\quad =\quad 10\frac { m }{ { s }^{ 2 } } \\ \frac { dv }{ dt } =10\\ v=10t\quad ...(1)\\ 1.8\quad =\quad 0\quad +\quad 5{ t }^{ 2 }\\ t\quad =\quad \frac { 3 }{ 5 } sec\\ v(at\quad x=1.8)\quad =\quad 6\frac { m }{ s } \\ Motion\_ In\_ Magnetic\_ field\\ R\quad =\quad \frac { m.v }{ q.B } \quad =\quad 1.2m\\ Region\quad of\quad magnetic\quad field\quad =\quad 0.6m\\ Therefore,\quad angle\quad of\quad emergence\quad =\quad { 30 }^{ . }\\ By\quad applying\quad simple\quad geometry,\quad we\quad can\quad determine\quad \\ coordinate\quad of\quad point\quad of\quad emergence.\\ (h,k)\quad \equiv \quad (2.4,\frac { 6-3\sqrt { 3 } }{ 5 } )\\ equation\quad of\quad line\quad passing\quad through\quad (h,k)\\ tan30\quad =\quad \frac { y\quad -\quad \frac { 6-3\sqrt { 3 } }{ 5 } }{ 3-2.4 } \\ y\quad =\quad \frac { 2(3-\sqrt { 3 } ) }{ 5 } \\ \\

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Aishwary Omkar
May 18, 2015

hey this came in my 17 may test Sorry not a solution cuz therewas no space for comments

0 Helpful 0 Interesting 0 Brilliant 0 Confused

where did you get this

Aishwary Omkar - 6 years ago

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