anonymous
  • anonymous
Pls Explain motion in a magnetic field ,when velocity is at an angle with the field .
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
??
anonymous
  • anonymous
Okay, Can you tell me the Magnetic Field Force ? I forgot sorry
anonymous
  • anonymous
F = q (V x B )

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anonymous
  • anonymous
Hmm Right so do you know cross product ?
anonymous
  • anonymous
i'm a bit confused about it
anonymous
  • anonymous
hmm I won't be able to explain through words but I will try finding some videos fro you. btw do you know about mit ocw ?
anonymous
  • anonymous
nope
anonymous
  • anonymous
what is that?
anonymous
  • anonymous
http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/
anonymous
  • anonymous
oh thanks i'll check it out :)
anonymous
  • anonymous
if you have any doubts please do post them I have to study magnetic field too so your doubts can be helpful to me. sorry If I sound mean
anonymous
  • anonymous
no not at all :) .. i'm going through those videos .. it covers most of my portions ...thank you so much
anonymous
  • anonymous
: ) no problem
anonymous
  • anonymous
|dw:1316022392752:dw| can you tell me how we get v || =\[v \cos \theta\] and v perpendicular = \[v \sin \theta\]
anonymous
  • anonymous
are u here ishaan ?
anonymous
  • anonymous
yea I am ! :D
anonymous
  • anonymous
can you tell me how we get v || =vcosθ and v perpendicular = vsinθ
anonymous
  • anonymous
Okay...I will try |dw:1316022839246:dw|
anonymous
  • anonymous
I think Parallel, Vertical all are taken with respect to x-axis so \(v_{parallel}\) is \(v\cos\theta\)
anonymous
  • anonymous
the angle θ is b/w your vsinθ and v
anonymous
  • anonymous
hmm like this |dw:1316023174038:dw|
anonymous
  • anonymous
yes
anonymous
  • anonymous
Okay ..|dw:1316023222420:dw|
anonymous
  • anonymous
how did u get it ? :(
anonymous
  • anonymous
hmm do you know about sin ratios hmm okay Let me try again \[sin\theta = \frac{Perpendicular }{Hyp}\] so now try to apply this in the triangle the upper triangle
anonymous
  • anonymous
ok let me see..
anonymous
  • anonymous
|dw:1316023483789:dw|
anonymous
  • anonymous
Got It Got It :)
anonymous
  • anonymous
Yay!!
anonymous
  • anonymous
\[\cos\theta =\frac{base}{hyp}\]
anonymous
  • anonymous
thanks a ton :) yea i know it :)
anonymous
  • anonymous
now i'll never forget it :)
anonymous
  • anonymous
!! yay (Again ) :D
anonymous
  • anonymous
ok so vcosθ is Magnetic field and it is given in my notes that thus charge q will hav two motions linear and circular and hence resultant path is helix its axis along direction B |dw:1316023925945:dw|
anonymous
  • anonymous
|dw:1316024026575:dw|
anonymous
  • anonymous
how is it helix ? and how does it have linear and circular motions ?
anonymous
  • anonymous
hmm I think that is where electric field comes to effect
anonymous
  • anonymous
magnetic field makes it to move in circular motion but electric field makes it to move froward
anonymous
  • anonymous
so hence helix is formed
anonymous
  • anonymous
ok thankyou sooooooooooooooooooooooooooooo much for your help yaaayyyy :)
anonymous
  • anonymous
Resolve the velocity into two components: one component parallel to the magnetic field; the other component perpendicular to it. The (assumed charged) particle will maintain the velocity that is parallel to the magnetic field, but will simultaneously move in circles in the plane perpendicular to the magnetic field. The combined effect is that it will "spiral" in a helix "following" the magnetic field.

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