Magnetic field and inductor

  • Thread starter Vibhor
  • Start date
  • #26
971
40
It's much easier to get v from your solution for i by using vBl = Ldi/dt.

That is how I got displacement in post#23 .

v = (Limcosωt)/(Blω)

Now putting v = dx/dt and integrating I get the expression for displacement .

Now in the expression v = (Limcosωt)/(Blω) , if I put t=0 , I get v0 = (Lim)/(Blω) which I think is correct value of v0 .

But from the expression v = v0 + (Biml)(cosωt-1)/(mω) , how do we get value of v0 ?

I am trying to find value of v0 from the expression of v of post#24 .
 
  • #27
TSny
Homework Helper
Gold Member
13,324
3,599
Now in the expression v = (Limcosωt)/(Blω) , if I put t=0 , I get v0 = (Lim)/(Blω) which I think is correct value of v0
Yes. So, v = v0cosωt where v0 = Lim/(Blω).

v0 is not an unknown, it's considered as given. So, v = v0cosωt is the final expression for v as a function of time.

The expression v0 = Lim/(Blω) can be used to obtain im.

But from the expression v = v0 + (Biml)(cosωt-1)/(mω) , how do we get value of v0 ?

I am trying to find value of v0 from the expression of v of post#24 .
I'm a little confused here. Does v0 here represent the initial speed or does it represent an arbitrary constant of integration? The initial speed is not an unknown.
 
  • #28
971
40
You are right Sir :smile:

This was a good problem . Thank you very much .
 

Related Threads on Magnetic field and inductor

Replies
8
Views
3K
  • Last Post
Replies
1
Views
909
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
11
Views
10K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
4K
Replies
6
Views
375
Top