Class 12th

From ampere's circuital law

$?\stackrel{?}{B}.\stackrel{?}{dl}={\mu }_0{i}_{enc}$

$={\mu }_0\left(3+1\right)=4{\mu }_0$

The relation between $\alpha$ and $\beta$  is given as

$\beta =\frac{\alpha }{1-\alpha }$

${\beta }_1=\frac{\frac{20}{21}}{1-\frac{20}{21}}=20$

${\beta }_2=\frac{\frac{100}{101}}{1-\frac{100}{101}}=100$

$\stackrel{?}{F}=e{E}_0\stackrel{?}{j}$

${\stackrel{?}{a}}_y=\frac{e{E}_0}{m}\stackrel{?}{j}$

$=v_0\stackrel{?}{i}+\left(\frac{e{E}_{0}t}{m}\right)\stackrel{?}{j}$

$\left|\stackrel{?}{v}\right|=\sqrt[]{v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}=v_0\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}$

$\lambda =\frac{{\lambda }_0}{\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}}$

From Lenz law, current in the square loop will be in anticlockwise sense.

Direction of  is perpendicular to equipotential surface.

For diamagnetic ; $\chi <0andl=\chi H$

At resonance V_L = V_C

_{$\Rightarrow cos?=\frac{V_R}{\sqrt[]{V_R^2+{\left(V_L-V_C\right)}^2}}=1$}

Density of nuclei is order of 10¹⁷ kg m^-3.

Photocell works on photoelectric effect.

$C_1=1\mu F{V}_1=100V?U_1=\frac{1}{2}*1*{10}^{-6}*{\left(100\right)}^2=5*{10}^{-3}J$  

$C_2=1\mu F$ , common potential $V=\frac{C_1{V}_1}{C_1{C}_2}=\frac{100}{2}=50V$

Final electrostatic energy stored in both the capacitors.

$=\frac{1}{2}\left(C_1+C_2\right)V^2=\frac{1}{2}*2*{\left(50\right)}^2*{10}^{-6}$

$=2.5*{10}^{-3}J$

$Energylost=2.5*{10}^{-3}$

$\%lossofenergy=\frac{2.5*{10}^{-3}}{5*{10}^{-3}}*100$

$=50\%$