# Magnetic field and inductor

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 .

TSny
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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.

Vibhor
You are right Sir

This was a good problem . Thank you very much .