$\vec{B}$ can only have Z-components.

And it can depend only on r.

$\oint \vec{B} . \vec{dl} = \mu I_{enc} = 0$

$\int_a^b \vec{B} . \vec{dl} + \int_b^c \vec{B} . \vec{dl} + \int_c^d \vec{B} . \vec{dl}$

$+ \int_d^a \vec{B} . \vec{dl}=0$

$\int_a^b \vec{B} . \vec{dl} + \int_c^d \vec{B} . \vec{dl} = 0$

$B(r_1) \int_a^b dl + B(r_2) \int_c^d dl = 0$

$B(r_1) \int_a^b dl = B(r_2) \int_d^c dl = 0$

$\rightarrow B(r_1) = B(r_2)$

For $r_2 \rightarrow \infin$

$B(r_2) \rightarrow 0$

B = 0 to point outside the solenoid.

$\oint \vec{B} . \vec{dl} = \mu_a I_{enc}$

Number of turns per unit length : N

Number of turns : Nl

Total current enclosed by the loop

• = N l I

$\oint \vec{B} . \vec{dl} = \mu_0 N I l$

$B \int_a^b dl = \mu_0 NIl$

$Bl = \mu_0 NIl$

$\rightarrow B = \mu _0 NI$

$\vec{B} = \mu_0 NI \hat{k}$

N = Number of turns per unit length

$\vec{B}$ along axis

$\vec{B} = \frac{\mu_0 I N}{2} \frac{R^2}{(R^2+2^2)^{3 / 2}} \hat{k}$

Number of turns = Ndz

$d \vec{B} = \frac{\mu_0 I N d z}{2} \frac{a^2}{(a^2 + z^2)^{3 / 2}} \hat{k}$

$\vec{B} = \frac{\mu_0 N I}{2} a^2 \int_0^L \frac{dz}{(a^2+z^2)^{3 / 2}} \hat{k}$

$Z = a \tan \theta$

$dz = a \sec ^2 \theta d \theta$

$(a^2 + z^2) = a^2 \sec ^2 \theta$

$\int \frac{d z}{(a^2 + z^2)^{3 / 2}} = \int \frac{a \sec ^2 \theta d \theta}{a^3 \sec ^3 \theta}$

$= \frac{1}{a^2} \int \cos \theta d \theta$

$= \frac{1}{a^2} \sin \theta |_0 ^{\tan ^{-1} {L / a}}$

$= \frac{1}{a^2} \sin [\tan ^{-1}(\frac{L}{a})]$

$\vec{B} = \frac{\mu_0 N I}{2} a^2 . \frac{1}{a^2} \sin [\tan ^{-1}(\frac{L}{a})] \hat{a}$

$= \frac{\mu_0 N I}{2} \sin [\tan ^{-1}(\frac{L}{a})] \hat{k}$

L = 20cm

a = 3cm

Total number of turns = 500

Current = 5A

Number of turns per unit length = $\frac{500}{20} = 25cm^{-1}$

$B = \mu_0 NI = 4 \pi \times 10^{-7} \times 2500 \times 5 \simeq 0.016 T (\simeq Centre)$

Path 1:

$\oint \vec{B} . \vec{dl} = 0$

$B.2 \pi r_1 = 0$

$\rightarrow B = 0$

Path 2:

$\oint \vec{B} . \vec{dl} = 0$

$B.2 \pi r_2 = 0$

B = 0

Path 3:

$\oint \vec{B} . \vec{dl} = \mu_0 I_{evc} = \mu_0 N_t I$

$N_t$ : Total number of turns in the torroid

$B.2 \pi r = \mu_0 N_t I$

$B = \frac{\mu_0 N_t I}{2 \pi r}$

$\vec{F}_B = q(\vec{v} \times \vec{B})$

$\vec{F} = q\vec{E} + q(\vec{v}\times \vec{B})$

Particles will executed a circular motion :-

$F = |q| v B$

Centripetal Force $= \frac{mv^2}{R}$

$\frac{mv^2}{R} = |q| vB$

$R = \frac{mv}{|q|B}$

$w = \frac{v}{R} = \frac{|q|B}{m}$

Number of revolution per unit time

$f = \frac{w}{2 \pi} = \frac{|q|B}{2 \pi m}$

CYCLOTRON FREQUENCY

$qE = qvB$

$v = \frac{E}{B}$

Calculate $\vec{B}$ along the axis of the solenoid using biot - savart law

Plot a systematic diagram of $\vec{B}$ unit position