Physics Moving Charges and Magnetism

Answer (1)

$eE=e\frac{dV}{dr}=m{\omega }^{2}r$

$?V=\int dV=\underset{R}{\overset{2R}{\int }}\frac{m}{e}{\omega }^{2}rdr$

$=?V=\frac{3m{\omega }^2{R}^2}{2e}$

$B=N\frac{{\mu }_{0}l}{2\mu }$

$B\alpha \frac{N}{\mu }$

$\frac{{\beta }_x}{{\beta }_y}=\frac{N_x}{r_x}*\frac{r_y}{N_y}$

$\frac{B_x}{B_y}=\frac{200}{20}*\frac{20}{400}=\frac{1}{2}$

 $\overrightarrow{a}\perp \overrightarrow{B}$

$\Rightarrow \overrightarrow{a}.\overrightarrow{B}=0$

$\Rightarrow \left(2\alpha -12\right)=0$

= 6

As we know that radius of circular path in magnetic field is given as $r=\frac{mv}{qB}=\frac{\sqrt[]{2mK}}{qB},so$  

  $r=\frac{mv}{qB}=\frac{\sqrt[]{2mK}}{qB},so$

$\frac{r_d}{r_{\alpha }}=\sqrt[]{\frac{m_d}{m_{\alpha }}}*\frac{q_{\alpha }}{q_d}=\frac{1}{\sqrt[]{2}}*2=\sqrt[]{2}$               

$B_1=\frac{{\mu }_{0}l}{4\pi y}\left[sin{\theta }_1+sin90°\right]-------\left(i\right)$

B due to OX wire

$B_2=\frac{{\mu }_{0}l}{4\pi x}\left[sin{\theta }_2+sin90°\right]$       

As in the diagram direction of B₁ and B₂ are in downward direction

So B = B₁ + B₂

$B=\frac{{\mu }_{0}l}{4\pi y}\left(sin{\theta }_1+1\right)+\frac{{\mu }_{0}l}{4\pi x}\left(sin{\theta }_2+1\right)$

$B=\frac{{\mu }_{0}l}{4\pi xy}\left[\left(x+y\right)+\sqrt[]{x^2+y^2}\right]$  

Using magnetic field due to straight wire :

$B=\frac{{\mu }_{0}i}{4\pi r}\left(sin\alpha +sin\beta \right)$

$=\frac{10^{-7}*1.5}{\left(\frac{0.09}{2\sqrt[]{3}}\right)}*\left(sin60°+sin60°\right)=10^{-5}T$

So, magnetic field due to three wires

= 3 * 10^-5 T

inside the plane

$B_1=\frac{{\mu }_{0}l}{4\pi y}\left[sin{\theta }_1+sin90°\right]-------\left(i\right)$

B due to OX wire

$B_2=\frac{{\mu }_{0}l}{4\pi x}\left[sin{\theta }_2+sin90°\right]$       

As in the diagram direction of B₁ and B₂ are in downward direction

So B = B₁ + B₂

$B=\frac{{\mu }_{0}l}{4\pi y}\left(sin{\theta }_1+1\right)+\frac{{\mu }_{0}l}{4\pi x}\left(sin{\theta }_2+1\right)$

$B=\frac{{\mu }_{0}l}{4\pi xy}\left[\left(x+y\right)+\sqrt[]{x^2+y^2}\right]$  

Let us consider an elementary ring of radius r and thickness dr in which current is flowing.

So, No. of turns in this elementary ring

$dN=\left(\frac{N}{b-a}\right)dr$

$\therefore {\left(dB\right)}_{atcentre}=\frac{{\mu }_{0}ldN}{2r}$

$B=\frac{{\mu }_{0}lN}{2\left(b-a\right)}\mathcal{l}n\frac{b}{a}$

In triangle $N_t=\frac{240}{3a}=8$ ${}_{}\frac{}{}$

In square Ns = $\frac{24a}{4a}=6$ $\frac{}{}$

$\frac{M_t}{Ms}=\frac{N_{t}l{A}_t}{NslAs}=\frac{8*\raisebox{1ex}{$\sqrt[]{3}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.*a^2}{6*a^2}\Rightarrow Mt/Ms=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt[]{3}$}\right.\Rightarrow y=3$

$R=\frac{\sqrt[]{2Em}}{B.q}$

$\Rightarrow R*\frac{\sqrt[]{m}}{q}$

$\Rightarrow \frac{R_1}{R_2}=\frac{\sqrt[]{4}}{2}*\frac{3}{\sqrt[]{16}}=\frac{3}{4}$

$\Rightarrow R_2=\frac{4R_1}{3}$

$sin\theta =\frac{d}{R}\Rightarrow \theta \alpha \frac{1}{2}\Rightarrow {\theta }_2<{\theta }_1$