## hermionegranger Group Title Pls Explain motion in a magnetic field ,when velocity is at an angle with the field . 2 years ago 2 years ago

1. hermionegranger Group Title

??

2. Ishaan94 Group Title

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

3. hermionegranger Group Title

F = q (V x B )

4. Ishaan94 Group Title

Hmm Right so do you know cross product ?

5. hermionegranger Group Title

i'm a bit confused about it

6. Ishaan94 Group Title

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 ?

7. hermionegranger Group Title

nope

8. hermionegranger Group Title

what is that?

9. Ishaan94 Group Title
10. hermionegranger Group Title

oh thanks i'll check it out :)

11. Ishaan94 Group Title

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

12. hermionegranger Group Title

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

13. Ishaan94 Group Title

: ) no problem

14. hermionegranger Group Title

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

15. hermionegranger Group Title

are u here ishaan ?

16. Ishaan94 Group Title

yea I am ! :D

17. hermionegranger Group Title

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

18. Ishaan94 Group Title

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

19. Ishaan94 Group Title

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

20. hermionegranger Group Title

the angle θ is b/w your vsinθ and v

21. Ishaan94 Group Title

hmm like this |dw:1316023174038:dw|

22. hermionegranger Group Title

yes

23. Ishaan94 Group Title

Okay ..|dw:1316023222420:dw|

24. hermionegranger Group Title

how did u get it ? :(

25. Ishaan94 Group Title

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

26. hermionegranger Group Title

ok let me see..

27. Ishaan94 Group Title

|dw:1316023483789:dw|

28. hermionegranger Group Title

Got It Got It :)

29. Ishaan94 Group Title

Yay!!

30. Ishaan94 Group Title

$\cos\theta =\frac{base}{hyp}$

31. hermionegranger Group Title

thanks a ton :) yea i know it :)

32. hermionegranger Group Title

now i'll never forget it :)

33. Ishaan94 Group Title

!! yay (Again ) :D

34. hermionegranger Group Title

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|

35. hermionegranger Group Title

|dw:1316024026575:dw|

36. hermionegranger Group Title

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

37. Ishaan94 Group Title

hmm I think that is where electric field comes to effect

38. Ishaan94 Group Title

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

39. Ishaan94 Group Title

so hence helix is formed

40. hermionegranger Group Title

ok thankyou sooooooooooooooooooooooooooooo much for your help yaaayyyy :)

41. physics.commando Group Title

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.