Long path
Diagram shows combination of 3 cuboidal spaces (1,2,3)
1 and 2 contain uniform electric field E, where as 2 contains uniform magnetic field B
Find out value of E such that the particle enters the magnetic field parallel to x-axis and just passes through point P along the electric field E at that point.
DETAILS AND ASSUMPTIONS
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Neglect the effect of gravity and consider x-axis parallel to L
M = √ 3 k g
V = 2 s e c m
q = 1 C
L = 2 m
θ = 3 0
The answer is 1.5.
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1 solution
Also why you complicated the solution. You could directly use
Half Range = L = 2 g e f f v 2 s i n ( 2 θ )
where g e f f = m q E and get the answer
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I've used this only to obtain relation between r and L
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But in this question since the values of R and B are not given Talking about R is waste of time.
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@Ronak Agarwal – You're right but i wanted to describe complete motion in all 3 spaces that's why
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@Kïñshük Sïñgh – So you should ask other things in the questions too. Infact this question can be made much harder by mixing all the things a bit.
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@Ronak Agarwal – Yeah, I'm thinking of this thing only.. This question could be made much harder than this..I'll try to improve these flaws in my future questions.. Do you want to give some suggestions??
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@Kïñshük Sïñgh – You try my question real life applications in geometry.
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@Ronak Agarwal – I don't have any knowledge about latitude and all.. So first of all I've to prefer Internet then I'll try to attempt
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@Kïñshük Sïñgh – Infact I have given the reference of wikipedia in the solution.
@Kïñshük Sïñgh – Also you don't told me you have a B2 subscription.
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@Ronak Agarwal – Yea... But it's of no use...:D
Easy question.
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Yeah i kno...this is the first question which I've made
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How did you attempted the question "solve for trajectory again" what was your approach in solving it.
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@Ronak Agarwal – Earlier i was not aware of such type of questions but when you have posted that 'solve for trajectory' i tried that but i was not able to solve it.. Then, i went through your solution. So my approach was similar for 'solve for trajectory again'
In space 1, motion of particle will be projectile motion just the gravitational force will be replaced by force due to electric field. Therefore, can say that
L = 2 q E m v 2 s i n 2 Θ where acceleration is equal to m q E
In space 2 motion will be circular due to magnetic field as shown in the figure. Therefore, Max height reach by the particle will be equal to R/2
2 R = 2 q E m V 2 ( s i n Θ ) 2 E = q R m V 2 ( s i n Θ ) 2
As we know, t a n Θ = L R (by using formula of max height and and half range) Therefore using these and eliminating R
E = 2 q L m V 2 s i n 2 Θ = 1 . 5