Question 34: Magnetism and Matter

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 $v_0$ 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]

Accepted Answer

For $$A 	o B$$
$$\overrightarrow V  =  - {v_x}\widehat i + {v_y}\widehat j$$
$$\overrightarrow B  = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} ight)$$
$\because$ $$\overrightarrow F  = q\left( {\overrightarrow v  	imes \overrightarrow B } ight)$$
So, $$\overrightarrow F  = {{q{\mu _0}I} \over {2\pi r}}\left[ {\left[ {\left( { - {v_x}} ight)\widehat i + {v_y}\widehat j} ight] 	imes \left( { - \widehat k} ight)} ight]$$
$$ = {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i 	imes \widehat k} ight) - {v_y}\left( {\widehat j 	imes \widehat k} ight)} ight]$$
$$\overrightarrow F  = {{q{\mu _0}I} \over {2\pi r}}\left[ { - {v_x}\widehat j - {v_y}\widehat i} ight]$$
$$ \Rightarrow {F_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}$$
$$ \Rightarrow m{a_x} = {{ - {\mu _0}Iq} \over {2\pi r}}{v_y}$$
$$ \Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}$$
$$ \Rightarrow {{{v_x}d{v_x}} \over {{v_y}}} = {{ - {\mu _0}Iq} \over {2\pi m}}{{dr} \over r}$$
Since, $$v_x^2 + v_y^2 = v_0^2$$
$$ \Rightarrow {v_y} = \sqrt {v_0^2 - v_x^2} $$
$$ \Rightarrow \int_0^{{v_0}} {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{ - {\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} } $$
Let $$v_0^2 - v_x^2 = {z^2}$$
$$ \Rightarrow  - 2{v_x}d{v_x} = 2zdz$$
when $${v_x} = 0,\,z = {v_0}$$
$${v_x} = {v_0},\,z = 0$$
$$ \Rightarrow \int_{{v_0}}^0 {{{zdz} \over z} =  - {{{\mu _0}Iq} \over {2\pi m}}\int_a^{x'} {{{dr} \over r}} } $$
$$ \Rightarrow  - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{{x'} \over a}} ight)$$
$$ \Rightarrow \ln \left( {{{x'} \over a}} ight) =  - {{2\pi m{v_0}} \over {{\mu _0}Iq}}$$
$$ \Rightarrow x' = ac{{ - 2\pi m{v_0}} \over {{\mu _0}Iq}}$$ .... (1)
Now, for $$B 	o C$$
$$\overrightarrow v  =  - {v_x}\widehat i - {v_y}\widehat j$$
$$\overrightarrow B  = {{{\mu _0}I} \over {2\pi r}}\left( { - \widehat k} ight)$$
$$\overrightarrow F  = q\left( {\overrightarrow v  	imes \overrightarrow B } ight) = {{q{\mu _0}I} \over {2\pi r}}\left[ {{v_x}\left( {\widehat i 	imes \widehat k} ight) + {v_y}\left( {\widehat j 	imes \widehat k} ight)} ight]$$
$$ \Rightarrow \overrightarrow F  = {{{\mu _0}Iq} \over {2\pi r}}\left[ { - {v_x}\widehat j + {v_y}\widehat i} ight]$$
$$ \Rightarrow {F_x} = {{{\mu _0}Iq} \over {2\pi r}}{v_y}$$
$$ \Rightarrow {v_x}{{d{v_x}} \over {dr}} = {{{\mu _0}Iq} \over {2\pi m}}{{{v_y}} \over r}$$
$$\int_{{v_0}}^0 {{{{v_x}d{v_x}} \over {\sqrt {v_0^2 - v_x^2} }} = {{{\mu _0}Iq} \over {2\pi m}}\int_{x'}^x {{{dr} \over r}} } $$
Let $$v_0^2 - v_x^2 = {z^2}$$
$$ - 2{v_x}d{v_x} = 2zdz$$
$${v_x} = {v_0} \Rightarrow z = 0$$
$${v_x} = 0 \Rightarrow z = {v_0}$$
$$ \Rightarrow  - \int_0^{{v_0}} {{{zdz} \over z} = {{{\mu _0}Iq} \over {2\pi m}}\left[ {\ln r} ight]_{x'}^x} $$
$$ \Rightarrow  - {v_0} = {{{\mu _0}Iq} \over {2\pi m}}\ln \left( {{x \over {x'}}} ight)$$
$$ \Rightarrow x = x'{e^{ - {{2\pi m{v_0}} \over {{\mu _0}Iq}}}}$$ .... (2)
From (1) & (2), $$x = a{e^{ - {{4\pi m{v_0}} \over {{\mu _0}Iq}}}}$$

Answer

$a\left[1-\frac{m v_o}{2 q \mu_0 \mathrm{I}}\right]$

Answer

$a e^{-\frac{4 \pi \mathrm{mv}_o}{q \mu_o \mathrm{I}}}$

Answer

$a\left[1-\frac{\mathrm{mv}}{\mathrm{q}} \mu_{\mathrm{o}} \mathrm{I}\right]$

Answer

$\frac{a}{2}$

Magnetism and Matter

Choose the correct answer: