A particle of mass `m` having a charge q enters into a circular region of radius `R` with velocity `v` directed towards the centre. The strength of magnetic field is `B`. Find the deviation in the path of the particle.

Answer

`(a) B_(0)=(mu_(0)i_(0)a^(2))/(2piC(b^(2)-a^(2)))" "(d)B=0`

Explanation

`(a)` On the axis of rod due to `i_(1) vec(B_` is zero on axis
`i_(2)=j pi a^(2)` and `j=(i)/(pi(b^(2)-a^(2)))`
Due to infinite wire we know that `B=(mu_(0)i)/(2pid)`
Hence due to `i_(2)` at the axis of rod
`B_(0)=(mu_(0)i_(2))/(2piC)=(mu_(0)(jpia^(2)))/(2piC)=(mu_(0)(i)/(pi(b^(2)-a^(2))pia^(1)))/(2piC)`
`B_(0)=(mu_(0)i_(2)a^(2))/(2piC(b^(2)-a^(2)))B_(0)=(mu_(0)ia^(2))/(2piC(b^(2)-a^(2)))`
Similarly we can find on the axis of hole

`(b)j=(i)/(pi(b^(2)-a^(2))),intBdl=mu_(0)i_(i n),`
`B2piC=mu_(0)(1)/(pi(b^(2)_a^(2)))piC^(2),B=(mu_(0)iC)/(2pi(b^(2)-a^(2)))`