Question

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 \(\mathrm{v}_{\mathrm{0}}\) 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]

(1) \(a\left[1-\frac{m v_{0}}{2 q \mu_{0} I}\right]\)

(2) \(\frac{a}{2}\)

(3) \(a\left[1-\frac{m v_{o}}{q \mu_{0} I}\right]\)

(4) \(\mathrm{ae}^{\frac{-4 \pi m v_{0}}{\mathrm{q} \mathrm{\mu}_0\mathrm{I}}}\)

Answer 1

Correct option is : (4) \(a\mathrm{e}^{\frac{-4 \pi m v_{0}}{\mathrm{q} \mathrm{\mu}_0\mathrm{I}}}\)

  

\(\mathrm{A} \rightarrow \mathrm{B}\) 

\(\vec{V}=-v_{x} \hat{i}+v_{y} \hat{j}\) 

\(\vec{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}(-\hat{\mathrm{k}})\) 

\(\vec{\mathrm{F}}=\mathrm{q}(\vec{\mathrm{v}} \times \vec{\mathrm{B}})=\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{r}}\left[-\mathrm{v}_{\mathrm{x}} \hat{\mathrm{j}}-\mathrm{v}_{\mathrm{y}} \hat{\mathrm{i}}\right]\) 

\(\mathrm{a}_{\mathrm{x}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{m}} \cdot \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}\) 

 \(a_{y}=-\frac{\mu_{0} I q}{2 \pi m} \cdot \frac{v_{x}}{r}\) 

\(\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{dr}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}\) 

\(\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{v}_{\mathrm{y}}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{m}} \frac{\mathrm{dr}}{\mathrm{r}}\) 

 \(\int\limits_0^{v_0} \frac{v_xdv_x}{\sqrt{v_0^2 - v_x^2}} = - \frac{\mu_0Iq}{2\pi m} \int\limits_a^{x_1} \frac{dr}{r}\) 

\( Let, \ z^{2}=v_{0}{ }^{2}-v_{x}{ }^{2}\) 

\(2 \mathrm{zdz}=-2 \mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}\) 

\(z d z=-v_{x} d_{x}\) 

\(\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\sqrt{\mathrm{v}_{0}^{2}-\mathrm{v}_{\mathrm{x}}^{2}}}=\frac{-\mathrm{zdz}}{\mathrm{z}}=-\mathrm{dz}\) 

then integral becomes

\(-\int\limits_{\mathrm{v}_{0}}^{0} \mathrm{dz}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{m}} \ln \frac{\mathrm{x}_{1}}{\mathrm{a}}\) 

\(\mathrm{v}_{0}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{m}} \ln \frac{\mathrm{x}_{1}}{\mathrm{a}}\) 

\(\mathrm{X}_{1}=\mathrm{a} \mathrm{e}^{-\frac{2 \pi \mathrm{mv}_{0}}{\mu_{0} \mathrm{Iq}}}\ ...(1)\) 

For  \(\mathrm{B} \rightarrow \mathrm{C}\) 

\(\vec{v}=-v_{x} \hat{i}-v_{y} \hat{j}\) 

\(\vec{\mathrm{B}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}(-\hat{\mathrm{k}})\) 

\(\vec{F}=q(\vec{v} \times \vec{B})=\frac{\mu_{0} I q}{2 \pi r}\left(-v_{x} \hat{j}+v_{y} \hat{i}\right)\) 

\( \mathrm{a}_{\mathrm{x}}=+\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}} \quad \mathrm{a}_{\mathrm{y}}=-\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{~m}} \cdot \frac{\mathrm{v}_{\mathrm{x}}}{\mathrm{r}}\) 

\(\frac{\mathrm{v}_{\mathrm{x}} \mathrm{dv}_{\mathrm{x}}}{\mathrm{dr}}=\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{m}} \frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{r}}\) 

\( \int\limits_{v_{0}}^{0} \frac{v_{x} \mathrm{dv}_{\mathrm{x}}}{\sqrt{\mathrm{v}_{0}^{2}-\mathrm{v}_{\mathrm{x}}^{2}}}=\frac{\mu_{0} \mathrm{Iq}}{2 \pi m} \int\limits_{\mathrm{x}_{1}}^{\mathrm{x}} \frac{\mathrm{dr}}{\mathrm{r}}\) 

\(\frac{\mu_{0} \mathrm{Iq}}{2 \pi \mathrm{m}} \ln \frac{\mathrm{x}}{\mathrm{x}_{1}}=-\int\limits_{0}^{\mathrm{v}_{0}} \mathrm{dz}=-\mathrm{v}_{0}\) 

\(\mathrm{x}=\mathrm{x}_{1} \mathrm{e}^{-\frac{2 \pi \mathrm{mv}_{0}}{\mu_{0} \mathrm{I} \mathrm{q}}} \ldots (2)\) 

 From equation 1 and 2

\(X=a e^{-\frac{4 \pi \mathrm{mv}_{0}}{\mu_{0} \mathrm{I}_{\mathrm{q}}}}\)