Pls Explain motion in a magnetic field ,when velocity is at an angle with the field .

- anonymous

Pls Explain motion in a magnetic field ,when velocity is at an angle with the field .

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- anonymous

??

- anonymous

Okay, Can you tell me the Magnetic Field Force ? I forgot sorry

- anonymous

F = q (V x B )

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- anonymous

Hmm Right so do you know cross product ?

- anonymous

i'm a bit confused about it

- 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

nope

- anonymous

what is that?

- anonymous

http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/

- anonymous

oh thanks i'll check it out :)

- 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

no not at all :) .. i'm going through those videos .. it covers most of my portions ...thank you so much

- anonymous

: ) no problem

- anonymous

|dw:1316022392752:dw|
can you tell me how we get
v || =\[v \cos \theta\] and v perpendicular = \[v \sin \theta\]

- anonymous

are u here ishaan ?

- anonymous

yea I am ! :D

- anonymous

can you tell me how we get
v || =vcosθ
and v perpendicular = vsinθ

- anonymous

Okay...I will try
|dw:1316022839246:dw|

- anonymous

I think Parallel, Vertical all are taken with respect to x-axis so \(v_{parallel}\) is \(v\cos\theta\)

- anonymous

the angle θ is b/w your vsinθ and v

- anonymous

hmm like this |dw:1316023174038:dw|

- anonymous

yes

- anonymous

Okay ..|dw:1316023222420:dw|

- anonymous

how did u get it ? :(

- 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

ok let me see..

- anonymous

|dw:1316023483789:dw|

- anonymous

Got It Got It :)

- anonymous

Yay!!

- anonymous

\[\cos\theta =\frac{base}{hyp}\]

- anonymous

thanks a ton :) yea i know it :)

- anonymous

now i'll never forget it :)

- anonymous

!! yay (Again ) :D

- 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

|dw:1316024026575:dw|

- anonymous

how is it helix ? and how does it have linear and circular motions ?

- anonymous

hmm I think that is where electric field comes to effect

- anonymous

magnetic field makes it to move in circular motion but electric field makes it to move froward

- anonymous

so hence helix is formed

- anonymous

ok thankyou sooooooooooooooooooooooooooooo much for your help yaaayyyy :)

- 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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