Physics Moving Charges and Magnetism

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

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

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

= 6

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

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

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

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

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 :

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

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

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

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

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

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

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

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

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

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

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

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

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

2nqV = $\frac{1}{2}m{v}^2$

$n=\frac{m{v}^2}{4qV}$

$=\frac{1.67\times 10^{-27}\times {\left(0.5\times 10^8\right)}^2}{4\times 1.6\times 10^{-19}\times 12\times 10^3}=543$  

$B_{axis}=\frac{{\mu }_{0}Nl{a}^2}{2{\left(a^2+r^2\right)}^{3/2}}$

$B_{cent}=\frac{{\mu }_{0}NI}{2a}$

Fractional change = $\frac{B_{centre}-B_{axis}}{B_{centre}}$

$=1-\frac{B_{axis}}{B_{centre}}$

$=1-\left(1-\frac{3r^2}{2a^2}\right)\left[r<<a\right]$

$=\frac{3r^2}{2a^2}$

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

1 unit Area $\frac{l}{\pi R^2}$ $\frac{}{{}^{}}$

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

$i=\frac{l{r}^2}{R^2}$

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)

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

$B\propto r$