Consider a long thin conducting wire carrying a uniform current I… - Vidyadip

MCQ

Consider a long thin conducting wire carrying a uniform current I. A particle having mass " M " and charge " q " is released at a distance " a " from the wire with a speed $v_{0}$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance $x$ from the wire. The value of x is [ $\mu_{0}$ is vacuum permeability]

A
$a\left[1-\frac{m v_{o}}{2 q \mu_{o} I}\right]$
B
$\frac{a}{2}$
C
$\mathrm{a}\left[1-\frac{m v_{o}}{q \mu_{o} I}\right]$
D
$\mathrm{ae}^{\frac{-4 \pi \mathrm{mv}_{0}}{\mathrm{qu}_{0} \mathrm{I}}}$

✓

Answer

$\mathrm{A} \rightarrow \mathrm{B}$
$\overrightarrow{\mathrm{V}}=-\mathrm{v}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{v}_{\mathrm{y}} \hat{\mathrm{j}}$
$\overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}(-\hat{\mathrm{k}})$
$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\frac{\mu_{0} \operatorname{Iq}}{2 \pi \mathrm{r}}\left[-\mathrm{v}_{\mathrm{x}} \hat{\mathrm{j}}-\mathrm{v}_{\mathrm{y}} \hat{\mathrm{i}}\right]$
$\mathrm{a}_{\mathrm{x}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \cdot \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}$
$\mathrm{a}_{\mathrm{y}}=-\frac{\mu_{0} \text { Iq }}{2 \pi \mathrm{~m}} \cdot \frac{\mathrm{v}_{\mathrm{x}}}{\mathrm{r}}$
$\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{dr}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}$
$\frac{v_{x} \mathrm{dv}_{\mathrm{x}}}{\mathrm{v}_{\mathrm{y}}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{dr}}{\mathrm{r}}$
$\int_{0}^{v_{0}} \frac{v_{x} d v_{x}}{\sqrt{v_{0}^{2}-v_{x}^{2}}}=-\frac{\mu_{0} I q^{2}}{2 \pi m_{a}} \int_{\mathrm{x}} \frac{d r}{r}$
Let, $z^{2}=v_{0}{ }^{2}-v_{x}^{2}$
$2 \mathrm{zd} \mathrm{z}=-2 \mathrm{v}_{\mathrm{x}} \mathrm{dv} \mathrm{v}_{\mathrm{x}}$
$\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\sqrt{\mathrm{v}_{0}^{2}-\mathrm{v}_{\mathrm{x}}^{2}}}=\frac{-\mathrm{zdz}}{\mathrm{z}}=-\mathrm{dz}$
then integral becomes
$-\int_{\mathrm{v}_{0}}^{0} \mathrm{dz}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \ln \frac{\mathrm{x}_{1}}{\mathrm{a}}$
$v_{0}=-\frac{\mu_{0} I q}{2 \pi m} \ln \frac{x_{1}}{a}$
$x_{1}=a e^{-\frac{2 \pi \mathrm{mv}_{0}}{\mu_{0} \mathrm{I}_{1}}}$
For B $\rightarrow$ C
$\overrightarrow{\mathrm{v}}=-\mathrm{v}_{\mathrm{x}} \hat{\mathrm{i}}-\mathrm{v}_{\mathrm{y}} \hat{\mathrm{j}}$
$\overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}(-\hat{\mathrm{k}})$
$\vec{F}=q(\vec{v} \times \vec{B})=\frac{\mu_{0} I q}{2 \pi r}\left(-v_{x} \hat{j}+v_{y} \hat{i}\right)$
$\mathrm{a}_{\mathrm{x}}=+\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}$
$a_{y}=-\frac{\mu_{0} I q}{2 \pi m} \cdot \frac{v_{x}}{r}$
$\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{dr}}=\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}$
$\int_{v_{0}}^{0} \frac{v_{x} \mathrm{dv}_{\mathrm{x}}}{\sqrt{\mathrm{v}_{0}^{2}-\mathrm{v}_{\mathrm{x}}^{2}}}=\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \int_{\mathrm{x}_{1}}^{\mathrm{x}} \frac{\mathrm{dr}}{\mathrm{r}}$
$\frac{\mu_{0} \operatorname{Iq}}{2 \pi m} \ln \frac{x}{x_{1}}=-\int_{0}^{v_{0}} d z=-v_{0}$
$x=x_{1} e^{-\frac{2 \pi m v_{0}}{\mu_{0} \mathrm{Iq}^{2}}}$
From equation 1 and 2
$X=a e^{-\frac{4 \pi \mathrm{mv}_{0}}{\mu_{0} I_{q}}}$

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