A long solenoid with radius `2cm` carries a current of `2A`. The solenoid is `70cm` long and is composed of `300` turns of wire. Assuming ideal solenoid model, calculate the flux linked with a circular surface if
(a) it has a radius of `1cm` and is perpendicular to the axis of the solenoid:
(i) inside,
(ii) outside.
(b)it has a radius of `3cm` and is perpendicular to the axis of the solenoid with its center lying on the axis of the solenoid.
( c ) it has a radius greater than `2cm` and axis of the solenoid subtend an angle of `60^(@)` with the normal to the area (the center of the circular surface being on the axis of the solenoid).
(d) the plane of the circular area is parallel to the axis of the solenoid.

Explanation

For an ideal solenoid: `B_(out) = 0` and `B_("in") = mu_(0)nI`.
So, here `B_(out) = 0` and `B_("in") = 10^(-7)[4pi(300//0.7) xx 2] = 1.076 xx 10^(-3)T` and for constant field `phi_(B) = int vec(B). dvec(s) = BS cos theta`.
(a) As in this situation, `theta = 0` and `s = pi xx (1 xx 10^(-2))^(2)m`
(i) `phi_("in") = (1.076 xx 10^(-3)) xx (pi xx 10^(-4))cos0^(@) = 3.38 xx 10^(-7)Wb` ltbr. (ii). `phi_(out) = 0 xx (pi xx 10^(-4))cos 0 = 0` (as B_(out) = 0)`
(b) In this situation, `theta = 0` [Fig. 3.3(a)].
So, `phi_(b) = phi_(in) + phi_(out) = b_(in)[pi xx(2 xx 10^(-2))^(2)] + b_(out)[pi(3^(2) - 2^(2)) xx 10^(-4)]`
`=(1.076 xx 10^(-3)) xx (pi xx 4 xx 10^(-4)) + 0[pi xx 5 xx 10^(-4)]`
`=13.52 xx 10^(-7)Wb`
( c ) In this situation, `theta = 60^(@)[Fig. 3.3(b)]`, so
`phi_(c) = phi_(out) = B_("in") xx (pi xx 4 xx 10^(-4))cos 60 + 0`
`[as B_(out) = 0]`
or `phi_(c) = phi_(b) cos 60 = (13.52 xx 10^(-7)) xx cos 60`
`= 6.76 xx 10^(-7))Wb`
(d) In this situation, as `theta = 90^(@) [Fig. 3.3( c ) ]`
`phi_(d) = B_(in) xx S xx cos 90 = 0`
(a)
(b)
(c )