Two particles, each with mass m are placed at a separation d in a uniform magnetic field B, as shown in the figure. They have opposite charges of equal magnitude q. At time t = 0, the particles are projected towards each other, each with a speed v. Suppose the Coulomb force between the charges is switched off. 1. Find the maximum value v_m of the projection speed, so that the two particles do not collide. 2. What would be the minimum and maximum separation between the particles if v=v_ m 2? 3. At what instant will a collision occur between the particles if v = 2v_m? 4. Suppose v = 2v_m and the collision between the particles is completely inelastic. Describe the motion after the collision.

Two particles, each with mass m are placed at a separation d in a…

Question

Two particles, each with mass m are placed at a separation d in a uniform magnetic field B, as shown in the figure. They have opposite charges of equal magnitude q. At time t = 0, the particles are projected towards each other, each with a speed v. Suppose the Coulomb force between the charges is switched off.

Find the maximum value $v_m$ of the projection speed, so that the two particles do not collide.
What would be the minimum and maximum separation between the particles if $\text{v}=\text{v}_{\text{m}}\sqrt{2}?$
At what instant will a collision occur between the particles if $v = 2v_m?$
Suppose $v = 2v_m$ and the collision between the particles is completely inelastic. Describe the motion after the collision.

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Answer

The particulars will not collide if,

$\text{d}=\text{r}_1+\text{r}_2$

$\Rightarrow\text{d}=\frac{\text{mV}_\text{m}}{\text{qB}}+\frac{\text{mV}_\text{m}}{\text{qB}}$
$\Rightarrow\text{d}=\frac{2\text{mV}_\text{m}}{\text{qB}}$
$\text{V}_\text{m}=\frac{\text{qBd}}{2\text{m}}$

$\text{V}=\frac{\text{V}_\text{m}}{2}$

$\text{d}_1'=\text{r}_1+\text{r}_2=\Big(\frac{\text{m}\times\text{qBd}}{2\times2\text{m}\times\text{qB}}\Big)=\frac{\text{d}}{2}$ (min. dist.)
Max. distance $\text{d}_2'=\text{d}+2\text{r}=\text{d}+\frac{\text{d}}{2}=\frac{3\text{d}}{2}$

$\text{V}=2\text{V}_\text{m}$

$\text{r}_1'=\frac{\text{m}_2\text{V} _\text{m}}{\text{qB}}=\frac{\text{m}\times2\times\text{qBd}}{2\text{n}\times\text{qB}}$
$\text{r}_2=\text{d}$
$\therefore$ The arc is $\frac{1}{6}$

$\text{V}_\text{m}=\frac{\text{qBd}}{2\text{m}}$

The particles will collide at point P. At point p, both the particles will have motion m in upward direction. Since the particles collide inelastically the stick together.
Distance l between centres $=\text{d},\sin\theta=\frac{1}{2\text{r}}$
Velocity upward $=\text{v}\cos90-\theta=\text{V}\sin\theta=\frac{\text{Vl}}{2\text{r}}$
$\frac{\text{mv}^2}{\text{r}}=\text{qvB}$
$\Rightarrow\text{r}=\frac{\text{mv}}{\text{qB}}$
$\text{V}\sin\theta=\frac{\text{vl}}{2\text{r}}=\frac{\text{vl}}{2\frac{\text{mv}}{\text{qb}}}=\frac{\text{qBd}}{2\text{m}}=\text{V}_\text{m}$
Hence the combined mass will move with velocity $V_m$.

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