Physics

A U-shaped magnet is another type of bar magnet. This is known as a horseshoe magnet since its shapes resembles the object called horseshoe. The two ends of this U-shaped magnet are its magnetic poles. Configration of these magnetic poles creates a concentrated magnetic field. This concentrated magnetic field makes it much more stronger than a simple and straight bar magnet.

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Based on theory.

$\frac{l_1}{l_2}=\frac{4}{1}$                

$\Rightarrow l_{max}={\left(\sqrt[]{l_1}+\sqrt[]{l_2}\right)}^2=l_1+l_2+2\sqrt[]{l_1{l}_2}$

$\Rightarrow \frac{l_{max}+l_{min}}{l_{max}-l_{min}}=\frac{5}{x}$

$\Rightarrow \frac{1}{2}*\frac{l_2\left(1+\frac{l_1}{l_2}\right)}{\sqrt[]{l_1{l}_2}}=\frac{5}{x}$

$\Rightarrow \frac{1}{2}\frac{1}{\sqrt[]{4}}=\frac{1}{x}\Rightarrow x=4$               

When electric field vector is completely removed and incident on Brewster's angle then only refraction takes place.

The two points in a bar magnet are its North pole and South pole. A bar magnet has magnetic field lines around it. These points give these field lines a travelling direction. For example, the magnetic field lines emerge form the North pole of the bar magnet and then they curve toward South pole. This makes a complete loop of magnetic field lines around the bar magnet.

we know,   $C=\frac{1}{\sqrt[]{{\mu }_0\in o}}=3*10^8,Given,{\in }_r=1$

$v=\frac{1}{\sqrt[]{{\mu }_0{\in }_r{\in }_o{\mu }_r}}=2*10^8$            

$\Rightarrow \frac{3*10^8}{2*10^8}=\sqrt[]{{\in }_r{\mu }_r}$               

$={\left(1.5\right)}^2={\in }_r{\mu }_r$              

${\mu }_r=2.25$               

V = 210 sin 3000 t

$\omega$ = 3000

$X_L=\omega L=30$                

$X_C=\frac{1}{\omega c}=\frac{40}{3}$                

$tan?=\frac{V_L-V_C}{V_R}=\frac{X_L-X_C}{R}$                

$\begin{array}{l}tan?=0.17\\ ?=ta{n}^{-1}\left(0.17\right)\end{array}$                

  $B=\eta {\mu }_{0}l$

 if l ->2l

$\eta \to \raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$                

$B^1=\frac{n}{2}{\mu }_{0}2l$

$B^1=B$                

Kindly go through the solution

 $\begin{array}{l}T\text{?}\to Absolutetemperature\\ x\to susceptibilities\end{array}\}x\propto \frac{1}{T}$

Given

E₁ = E₂ = E

$I=\frac{2E}{R=r_1+r_2}$   - (i)

Potential drop across second cell is

  $V_A-E_2+I{r}_2=v_B$

According to question V_A – V_B = 0

E₂ -lr₂ = 0

$\Rightarrow R=r_2-r_1$        

$\Delta v={\displaystyle \int \overrightarrow{E}.d\overrightarrow{x}}$  

$\Delta v=\frac{\sigma }{\in o}\frac{d}{2}$

$C_{New}=\frac{Q}{\Delta v}=\frac{\sigma A2\in o}{\sigma d}=\frac{2\in oA}{d}$

$C_{New}=2{\begin{array}{l}C\\ original\\ \end{array}}_{}$

$\frac{C_{New}}{C_{Original}}=2:1$