Class 12th

According to modified Ampere's law

$?B.dI={\mu }_0\left(I_C+I_D\right)$

For Loop  $L_1{I}_C\ne 0$ and $I_D=0$

For Loop  $L_2{I}_C=0$ and $I_D\ne 0$

Due to  $KCL{I}_C=I_D$

(Material)                      (Susceptibility ( $\chi$  )

Diamagnetic                 (II) $0>\chi \ge -1$

Ferromagnetic  (III) $\chi ?1$

Paramagnetic                (IV) $0<\chi <\epsilon$

Non-magnetic               (I) $\chi =0$

The magnitude of magnetic field due to circular coil of $N$  turns is given by

$B_C=\frac{{\mu }_{0}iN}{2R}$

$=\frac{4\pi *10^{-7}*7*100}{2*0.1}$

$=4.4*10^{-3} T$

$=4.4mT$

Divided into 10 parts

$R=\frac{\rho l}{A}$

$R^{\prime }=\frac{\rho l}{10A}=\frac{R}{10}$

$R_S=5*\frac{R}{10} [series]$

$R_S=50$

$R_P=\frac{R}{50}\left[ parallel \right]$

$R_{eq }=R_S+R_P$

$=52\Omega$

For uniformly charged spherical shell,

$V=\frac{kq}{R}\left( For r\le R\right)$

$\therefore V_C=V_P$

$V_C-V_P= Zero$

According to transformer ratio,

$\frac{V_S}{V_P}=\frac{N_S}{N_P}=2:1$

Energy difference $\Delta E=\frac{hc}{\lambda }$

$\therefore \lambda \propto \frac{1}{\Delta E}$

${(\Delta E)}_{6-2}>{(\Delta E)}_{5-2}>{(\Delta E)}_{4-2}>{(\Delta E)}_{3-2}$

${\lambda }_{6-2}<{\lambda }_{5-2}<{\lambda }_{4-2}<{\lambda }_{3-2}$
 

$Y_1=\overline{A·A}$

$=\overline{A}$

$Y_2=\overline{B+B}$

$=\overline{B}$

$Y=\overline{Y_1+Y_2}$

$=\overline{\overline{A}+\overline{B}}$

$=\overline{\overline{A}}·\overline{\overline{B}}$

$=A·B is similar to output of AND Gate$

(A) If $c$  is the velocity of light

so,   $E=h\nu$   (Energy of photon)

(B) Velocity of photon is equal to velocity of light i.e. c.

(C)  $\lambda =\frac{h}{p}$

$p=\frac{h}{\lambda }$

$p=\frac{hv}{c}$

(D) In photon-electron collision both total energy and total momentum are conserved.

At central point on screen, path difference is zero for all wavelength. So, central bright fringe is white and other fringes depend on wavelength as $\beta =\frac{\lambda D}{d}$ .

Therefore, other fringes will be coloured.