A non-uniform magnetic field exists in the free space and given as $\vec{B} = \frac{B_{0}}{ℓ^{2}} x^{2} \hat{k}$ , where $B_{0}$ and $ℓ$ are positive constants. A particle having positive charge ‘q’ and mass ‘m’ is projected with speed ‘ $v_{0}$ ’ along positive x-axis from the origin. The maximum distance of the charged particle from the y- axis before it turns back due to the magnetic field is D. (Ignore any interaction other than magnetic field). If $D = {(\frac{α m ℓ^{2} v_{0}}{q B_{0}})}^{1 / β}$ , where $α, β$ are positive integers then find the value of $α t i m e s β$ ?

Difficult

Solution

Since, magnetic filed and velocity are perpendicular, speed of particle will never change. At the moment of turning back, the direction of velocity becomes parallel to y axis.
$\begin{array}{l} \vec{F} = q \vec{v} \times \vec{B} = q (v_{x} \hat{i} + v_{y} \hat{j}) (\frac{B_{0}}{ℓ^{2}} x^{2} \hat{k}) \\ \Rightarrow \vec{F} = (q v_{x} \frac{B_{0}}{ℓ^{2}} x^{2} \hat{j}) + (q v_{y} \frac{B_{0}}{ℓ^{2}} x^{2} \hat{i}) = F_{y} \hat{j} + F_{x} \hat{i} \\ \Rightarrow F_{y} = q \frac{B_{0}}{ℓ^{2}} x^{2} v_{x} \\ \Rightarrow m \frac{d v_{y}}{d t} = q \frac{B_{0}}{ℓ^{2}} x^{2} \frac{d x}{d t} \\ \Rightarrow \int_{0}^{v_{0}} d v_{y} = \frac{q B_{0}}{m ℓ^{2}} \int_{0}^{x_{\max}} x^{2} d x \\ \Rightarrow v_{0} = \frac{q B_{0}}{m ℓ^{2}} \frac{x_{\max}^{3}}{3} \\ \Rightarrow x_{\max} = {(\frac{3 m ℓ^{2} v_{0}}{q B_{0}})}^{1 / 3} \end{array}$
$∴ α = 3, β = 3$
So, $α t i m e s β = 9$

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