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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, Can you tell me the Magnetic Field Force ? I forgot sorry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm Right so do you know cross product ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm a bit confused about it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm 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
 4 years ago
Best ResponseYou've already chosen the best response.0http://ocw.mit.edu/courses/physics/802electricityandmagnetismspring2002/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh thanks i'll check it out :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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
 4 years ago
Best ResponseYou've already chosen the best response.0no not at all :) .. i'm going through those videos .. it covers most of my portions ...thank you so much

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1316022392752:dw can you tell me how we get v  =\[v \cos \theta\] and v perpendicular = \[v \sin \theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you tell me how we get v  =vcosθ and v perpendicular = vsinθ

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay...I will try dw:1316022839246:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think Parallel, Vertical all are taken with respect to xaxis so \(v_{parallel}\) is \(v\cos\theta\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the angle θ is b/w your vsinθ and v

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm like this dw:1316023174038:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay ..dw:1316023222420:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did u get it ? :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1316023483789:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\cos\theta =\frac{base}{hyp}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks a ton :) yea i know it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now i'll never forget it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1316024026575:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how is it helix ? and how does it have linear and circular motions ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm I think that is where electric field comes to effect

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0magnetic field makes it to move in circular motion but electric field makes it to move froward

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so hence helix is formed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok thankyou sooooooooooooooooooooooooooooo much for your help yaaayyyy :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Resolve 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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