QUESTION OF THE DAY(electricity and magnetism)

the vector solution is very difficult ... because when force acts on the proton...it will come closer to the wire ..

∫B.dl=u₀i

B.2Î r=u₀ i

where 'r' is a variable=distance of proton from the wire( ..since a force is acting on the proton and getting it closer to the wire)

so B is inversely proportional to 'r'
then how to solve this qn with vector method? it is very difficult

What do you exactly want to solve for?

there is no charge density on the wire ...from where does the electric field arises ..?

The electric field does not arise from anywhere. Rather, it had always been there in the new reference frame. Let me explain.
First of all, you should note that the magnetic field in the two reference frames are equal. The reason is that the current in the wire is same in both the reference frame. Though I stated the invariance of magnetic field with reference to the field of current carrying wire, it turns out that the magnetic field due to any source remains the same in all reference frames (well at least classically).

Lets analyze a general situation. Take a reference frame S in which there is a magnetic field \vec{B} and an electric field \vec{E}. Now consider another reference frame S' which is moving with a velocity \vec{v}_0 relative to S. Our task is to relate the electric and magnetic field \vec{E}' and \vec{B}', respectively, measured in S' to the same quantities in S. As already stated the magnetic field remains invariant. Therefore, \vec{B}' = \vec{B}.

Suppose a charged particle having a charge q is moving with a velocity \vec{v} relative to S. The Lorentz force acting on it is \vec{F} = q(\vec{E}+\vec{v}\times \vec{B}).
On the other hand, the Lorentz force as measured is S' is
\vec{F}' = q(\vec{E}'+\vec{v}\,'\times \vec{B}')
Here \vec{v}\,' is the velocity of q as measured in S'. Since the net force must be same in the two reference frame, we have \vec{F}' = \vec{F}. That is
q(\vec{E}+\vec{v}\times \vec{B}) = q(\vec{E}'+\vec{v}\,'\times \vec{B}') --------- (1)
But we have \vec{B}' = \vec{B} and \vec{v}\,'=\vec{v}-\vec{v}_0
So from (1) we get
\vec{E}+\vec{v}\times \vec{B} = \vec{E}' + \vec{v}\times \vec{B} -\vec{v}_0\times \vec{B}
And hence,
\boxed{\vec{E}'=\vec{E}+\vec{v}_0\times \vec{B}}
That implies the electric field as measured in S' is different than the electric field as measured in S.

thank u so much sir ...now i undestand that electric fireld will always be present because of the charged particle not because of the wire :)

and a very important result
E'=E+v0 X B :D

By the way, don't you think that the current should change since the drift speed of the electrons are different in the new reference frame?

sir i think THE increased ELectric field in the new frame of reference will compensate the POSSIBLE loss of current ...

since E' > E

because of this ...the possible loss of current in the new frame ...has to be compensated by the factor E' > E

...am i correct sir ?