Physics

for $\pi R^{2}Area\to \text{'}l\text{'}$ current\

1 unit Area ® $\frac{l}{\pi R^2}$  

              $\pi r^2\to \frac{l}{\pi R^2}*\pi r^2$  

$i=\frac{l{r}^2}{R^2}$              

 

Now, consider Amperian loop of radius small 'r' ln Amperian loop magnetic field will be tangential to the amperian loop.

${\displaystyle ?\overrightarrow{B}.d\overrightarrow{i}={\mu }_0{l}_{enclosed}}$           (Ampere circuital law)

$B=\frac{{\mu }_0}{2\pi }\frac{l}{R^2}r$

$B\propto r$  

Based on theory

$V_{rms}=\sqrt[]{\frac{3RT}{M}}$

& $v_P=\sqrt[]{\frac{2RT}{M}}$

$v_{rms}=\sqrt[]{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}_P$

Let, Rain drop moving with terminal velocity v_t in the air F_b (Bnoyancy force) = $\frac{4}{3}\pi r^3{\rho }_{air}g$  

$mg=\frac{4}{3}\pi r^3{\rho }_{air}g$

$F_v=\left(visconsforce\right)=6\pi \eta r{v}_T$

$F_b+F_v=mg$

$\Rightarrow \frac{4}{3}\pi r^3{\rho }_{air}g+6\pi \eta r{v}_T=\frac{4}{3}\pi r^3\rho g$

$v_T=\frac{2}{9}\frac{r^2}{\eta }\left(\rho -{\rho }_{air}\right)$

$v_T\propto r^2$

$g^1=\frac{GM}{{\left(R+h\right)}^2}$

$g^1=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

Given h = D = 2R

$g^1=\frac{g}{{\left(1+\frac{2R}{R}\right)}^2}$

$g^1=\frac{g}{9}$

$\overrightarrow{r}=2\widehat{i}+\widehat{j}+2\widehat{k}$

$\overrightarrow{\tau }=\overrightarrow{r}*\overrightarrow{F}$

$=\left(2\widehat{i}+\widehat{j}+2\widehat{k}\right)*\left(3\widehat{i}+4\widehat{j}-2\widehat{k}\right)=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill -\widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill -2\hfill \end{array}\right|$

$\overrightarrow{\tau }=-10\widehat{i}+10\widehat{j}+5\widehat{k}$

$\left|\widehat{A}+\widehat{B}\right|=\sqrt[]{A^2+B^2+2ABcos\theta }$

$=\sqrt[]{1+1+2cos\theta }$

$=\sqrt[]{2\left(1+cos\theta \right)}$

= $\sqrt[]{2*2co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\left|\widehat{A}+\widehat{B}\right|=2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(1)

$\left|\widehat{A}-\widehat{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }$

$=\sqrt[]{1+1-2cos\theta }$

$=2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(2)

(2) $÷$ $$ (1)

$\frac{\left|\widehat{A}-\widehat{B}\right|}{\left|\widehat{A}+\widehat{B}\right|}=\frac{2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow \left|\widehat{A}-\widehat{B}\right|=\left|\widehat{A}+\widehat{B}\right|tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\overrightarrow{A}*\overrightarrow{A}=AAsin\theta \widehat{n}$

$=AAsin0°\widehat{n}$

= 0 [since Angle between the vectors are zero degree]

$\overrightarrow{A}*\overrightarrow{A}=0$

The direction of the net magnetic force is always perpendicular to the plane of current-carrying conductors. The direction of the magnetic force is determined through the right-hand rule. 

The right-hand thumb rule states that if you wrap a wire in a way that the thumb points towards the flow of current. Then your curled fingers represent the direction of the field lines of the magnetic field