Moving Charges and Magnetism Test - 38

Magnetic force provides necessary centripetal force,
$$qVB=\dfrac{mv^{2}}{r}$$

$$qVB=\dfrac{2KE}{r}$$

$$V=\dfrac{2KE}{qBr}$$

=$$\dfrac{2\times 10}{10^{-4}\times 0.1}$$

=$$2.0 \times10^{6}ms^{-1}$$

Force on charge particle in magnetic field is given by,
$$\bar{F}=q\left ( \vec{V}\times \vec{B} \right )$$

Since wire hangs in the air, So
 $$mg=Bil$$
$$\dfrac{m}{l}=\dfrac{Bi}{g}$$
$$ 20\times 10^{-3}=\bigg (\dfrac{\mu _{0}i}{g2\pi r}\bigg )i$$
$$i=98A$$

$$|B|_{net}=0$$

$$\bar{F}=q\left ( \underset{V}{\rightarrow}\times \underset{B}{\rightarrow} \right )$$
$$m\bar{a}=q(\bar{V}\times \bar{B})$$
$$\therefore \bar{a}.\bar{B}=0$$
$$(\hat{i}+7\hat{j}).(7\hat{i}-3\hat{j})=0$$
$$7x -21=0$$
$$x=3$$

Force due to two current carrying conductor is given by,
$$F=\dfrac{\mu_0I_1I_2}{2\pi d}l$$
$$F^1=\dfrac{\mu_0I_1\times2\times I_2}{2\pi \times3d}l$$ (direction is reversed)
$$F^1=\dfrac{-2F}{3}$$

$$F = I LB$$

Magnetic field due to a moving charge is $$\vec{B} = \dfrac{\mu_0}{4\pi}q \left( \dfrac{\vec{v} \times\vec{r}}{r^3} \right)$$
where $$\vec{v}$$ is velocity of the charge q and $$\vec{r}$$ is the position vector joining the particle position to the point on the circumference
$$ \therefore \vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{qvrsin\theta}{r^3}= \dfrac{\mu_0 q}{4\pi}\dfrac{vR}{\left(\sqrt{x^2 + R^2} \right)^3}= \dfrac{\mu_0 q}{4\pi}\dfrac{vR}{\left(x^2 + R^2 \right)^\dfrac{3}{2}}$$

Magnetic field due to $$10A$$ wire at a distance of $$1cm
\\ =\dfrac{\mu _{0}I}{2\pi r}=\dfrac{2\times 10^{-7}\times 10}{1\times 10^{-2}}=20\times 10^{-5}T$$
                                                                                     Now , $$  mg=Bil$$
                                                                         $$10^{-3}\times 9.8=20\times 10^{-5}\times i\times 10^{-1}$$
                                                                          $$i=490A$$

Let magnetic field strength be $$B$$

Charge $$q$$

Mass $$m$$ and 

Radius of the circular path be $$r$$

$$\vec { B=B\hat { k }  } $$

$$\vec { v } =v\cos \theta \hat { i } +v\sin \theta \hat { j } \\ \vec { F } =\left( v\cos \theta \hat { i } +v\sin \theta \hat { j }  \right) \times Bq\hat { k } \\ \Rightarrow \vec { F } =Bqv\left( \cos \theta \hat { i } +\sin \theta \hat { j }  \right) \\ \theta =\omega t\\ { a }_{ x }=\dfrac { Bqv }{ m } \sin \omega t\\ \dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } =\dfrac { Bqv }{ m } \sin \omega t$$

Integrating

$$\Rightarrow x=\dfrac { Bqv }{ m{ \omega  }^{ 2 } } \sin \left( \omega t+\theta  \right) $$

So magnitude of maximum horizontal displacement suffered by the particle in the field will be 

$${ x }_{ max }=\dfrac { Bqv }{ m{ \omega  }^{ 2 } } $$

Again,

$${ x }_{ max }=\dfrac { Bqv }{ m{ \omega  }^{ 2 } } \\ Bqv=m{ \omega  }^{ 2 }r\\ \omega =\dfrac { Bq }{ m } \\ { x }_{ max }=\dfrac { v }{ \omega  } $$

$$v=r\omega \\ \Rightarrow { x }_{ max }=r$$

This means for a particle to reach maximum displacement the field must have a minimum length of $$r$$

since length of field is $$1.5r$$ the particle cannot come out of field horizontally so to come out of it must reverse its direction and thus displacement.

Thus the particle will deflect by $$180°$$